Иррационал өрнектер (Сканави (А) 2.1-2.15)

Шешуі: ММЖ: $0<x\ne 1$ $${\displaystyle \frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}}:{\displaystyle \frac{1}{x^2-\sqrt{x}}}={\displaystyle \frac{\sqrt{x}+1}{\sqrt{x}(x+\sqrt{x}+1)}}\cdot {\displaystyle \frac{\sqrt{x}(x\sqrt{x}-1)}{1}}=$$ $$={\displaystyle \frac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}(x+\sqrt{x}+1)(\sqrt{x}-1)}}\cdot {\displaystyle \frac{\sqrt{x}(x\sqrt{x}-1)}{1}}={\displaystyle \frac{x-1}{\sqrt{x}(x\sqrt{x}-1)}}\times$$ $$\times {\displaystyle \frac{\sqrt{x}(x\sqrt{x}-1)}{1}}=x-1$$

Шешуі: ММЖ: $p\ne q$ $$=\left({\displaystyle \frac{1}{{(\sqrt[4]{p}-\sqrt[4]{q})}^2}}+{\displaystyle \frac{1}{{(\sqrt[4]{p}+\sqrt[4]{q})}^2}}\right)\times {\displaystyle \frac{p-q}{\sqrt{p}+\sqrt{q}}}=$$ $$={\displaystyle \frac{{(\sqrt[4]{p}+\sqrt[4]{q})}^2+{(\sqrt[4]{p}-\sqrt[4]{q})}^2}{{(\sqrt{p}-\sqrt{q})}^2}}\cdot {\displaystyle \frac{(\sqrt{p}-\sqrt{q})(\sqrt{p}+\sqrt{q})}{\sqrt{p}+\sqrt{q}}}=$$ $$={\displaystyle \frac{\sqrt{p}+2\sqrt[4]{pq}+\sqrt{q}+\sqrt{p}-2\sqrt[4]{pq}+\sqrt{q}}{\sqrt{p}-\sqrt{q}}}={\displaystyle \frac{2(\sqrt{p}+\sqrt{q})}{\sqrt{p}-\sqrt{q}}}=$$ $$={\displaystyle \frac{2(\sqrt{p}+\sqrt{q})(\sqrt{p}+\sqrt{q})}{(\sqrt{p}-\sqrt{q})(\sqrt{p}+\sqrt{q})}}={\displaystyle \frac{2{(\sqrt{p}+\sqrt{q})}^2}{p-q}}$$

Шешуі: $$X={\displaystyle \frac{(\sqrt{a^2+a\sqrt{a^2-b^2})}-{\left(\sqrt{a^2-a\sqrt{a^2-b^2}}\right)}^2}{2\sqrt{a^{3}b}}}=$$ $$={\displaystyle \frac{{\left(\sqrt{a\left(a+\sqrt{a^2-b^2}\right)}-\sqrt{a\left(a-\sqrt{a^2-b^2}\right)}\right)}^2}{2a\sqrt{ab}}}=$$ $$={\displaystyle \frac{{\left(\sqrt{a}\left(\sqrt{a+\sqrt{a^2-b^2}}-\sqrt{a-\sqrt{a^2-b^2}}\right)\right)}^2}{2a\sqrt{ab}}}=$$ $$={\displaystyle \frac{a\left(a+\sqrt{a^2-b^2}-2\sqrt{a^2-a^2+b^2}+a-\sqrt{a^2-b^2}\right)}{2a\sqrt{ab}}}={\displaystyle \frac{2a-2b}{2\sqrt{ab}}}={\displaystyle \frac{a-b}{\sqrt{ab}}}$$ $$Y=\sqrt{{\displaystyle \frac{a}{b}}}+\sqrt{{\displaystyle \frac{b}{a}}}-2={\displaystyle \frac{\sqrt{a}}{\sqrt{b}}}+{\displaystyle \frac{\sqrt{b}}{\sqrt{a}}}-2={\displaystyle \frac{{(\sqrt{a})}^2-2\sqrt{ab}+{(\sqrt{b})}^2}{\sqrt{ab}}}={\displaystyle \frac{{(\sqrt{a}-\sqrt{b})}^2}{\sqrt{ab}}}$$ онда $$X:Y={\displaystyle \frac{a-b}{\sqrt{ab}}}:{\displaystyle \frac{{(\sqrt{a}-\sqrt{b})}^2}{\sqrt{ab}}}={\displaystyle \frac{(a-b)\cdot \sqrt{ab}}{\sqrt{ab}\cdot {(\sqrt{a}-\sqrt{b})}^2}}={\displaystyle \frac{a-b}{{(\sqrt{a}-\sqrt{b})}^2}}=$$ $$={\displaystyle \frac{(a-b){(\sqrt{a}+\sqrt{b})}^2}{{(\sqrt{a}-\sqrt{b})}^2{(\sqrt{a}+\sqrt{b})}^2}}={\displaystyle \frac{(a-b){(\sqrt{a}+\sqrt{b})}^2}{{((\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}))}^2}}={\displaystyle \frac{(a-b){(\sqrt{a}+\sqrt{b})}^2}{{(a-b)}^2}}=$$ $$={\displaystyle \frac{{(\sqrt{a}+\sqrt{b})}^2}{a-b}}$$

Шешуі: ММЖ: $a\ne -b=-0,04$ $$X={\left({\displaystyle \frac{{(a+b)}^{-n/4}\cdot c^{1/2}}{a^{2-n}{b}^{-3/4}}}\right)}^{4/3}={\displaystyle \frac{{(a+b)}^{-n/3}\cdot c^{2/3}}{a^{(8-4n)/3}{b}^{-1}}}={\displaystyle \frac{b\cdot c^{2/3}}{a^{(8-4n)/3}\cdot {(a+b)}^{n/3}}}$$ $$Y={\left({\displaystyle \frac{b^3{c}^4}{{(a+b)}^{2n}{a}^{16-8n}}}\right)}^{1/6}={\displaystyle \frac{b^{1/2}\cdot c^{2/3}}{{(a+b)}^{n/3}\cdot a^{(8-4n/3)}}}$$ Онда $$X:Y={\displaystyle \frac{b\cdot c^{2/3}}{a^{(8-4n)/3}\cdot {(a+b)}^{n/3}}}:{\displaystyle \frac{b^{1/2}\cdot c^{2/3}}{{(a+b)}^{n/3}\cdot a^{(8-4n)/3}}}=$$ $$={\displaystyle \frac{b\cdot c^{2/3}\cdot {(a+b)}^{n/3}\cdot a^{(8-4n)/3}}{a^{(8-4n)/3}\cdot {(a+b)}^{n/3}\cdot b^{1/2}\cdot c^{2/3}}}=b^{1/2}=(0,04{)}^{1/2}=\sqrt{0,04}=0,2.$$

Шешуі: ММЖ: $\{\begin{array}{l}x\ne 0\\ x\ne 1\\ x\ne 3\end{array}$ $${\displaystyle \frac{2x^{-1/3}}{x^{2/3}-3x^{-1/3}}}-{\displaystyle \frac{x^{2/3}}{x^{5/3}-x^{2/3}}}-{\displaystyle \frac{x+1}{x^2-4x+3}}={\displaystyle \frac{2x^{-1/3}}{x^{-1/3}(x-3)}}-$$ $$-{\displaystyle \frac{x^{2/3}}{x^{2/3}(x-1)}}-{\displaystyle \frac{x+1}{(x-1)(x-3)}}={\displaystyle \frac{2}{x-3}}-{\displaystyle \frac{1}{x-1}}-{\displaystyle \frac{x+1}{(x-1)(x-3)}}=$$ $$={\displaystyle \frac{2x-2-x+3-x-1}{(x-1)(x-3)}}={\displaystyle \frac{0}{(x-1)(x-3)}}=0$$

Шешуі: ММЖ: $\{\begin{array}{l}a\ne b\\ a>0\\ b>0\end{array}$ $${\displaystyle \frac{{(\sqrt{a}+\sqrt{b})}^2-4b}{(a-b):\left(\sqrt{{\displaystyle \frac{1}{b}}}+3\sqrt{{\displaystyle \frac{1}{a}}}\right)}}:{\displaystyle \frac{a+9b+6\sqrt{ab}}{{\displaystyle \frac{1}{\sqrt{b}}}+{\displaystyle \frac{1}{\sqrt{a}}}}}={\displaystyle \frac{a+2\sqrt{ab}+b-4b}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}):\left({\displaystyle \frac{1}{\sqrt{b}}}+{\displaystyle \frac{3}{\sqrt{a}}}\right)}}÷$$ $${\displaystyle \frac{{(\sqrt{a}+3\sqrt{b})}^2}{{\displaystyle \frac{\sqrt{a}+\sqrt{b}}{\sqrt{ab}}}}}={\displaystyle \frac{a+2\sqrt{ab}-3b}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}):{\displaystyle \frac{\sqrt{a}+3\sqrt{b}}{\sqrt{ab}}}}}:{\displaystyle \frac{{(\sqrt{a}+3\sqrt{b})}^2\sqrt{ab}}{(\sqrt{a}+\sqrt{b})}}=$$ $$={\displaystyle \frac{(a+2\sqrt{ab}-3b)(\sqrt{a}+3\sqrt{b})(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})\sqrt{ab}{(\sqrt{a}+3\sqrt{b})}^2\sqrt{ab}}}=$$ $$={\displaystyle \frac{a+2\sqrt{ab}-3b}{ab(a-\sqrt{ab}+3\sqrt{ab-3b})}}={\displaystyle \frac{1}{ab}}$$

Шешуі: ММЖ: $\{\begin{array}{l}m\ne n\\ m>0\\ n>0.\end{array}$ $${\displaystyle \frac{{(\sqrt[4]{m}+\sqrt[4]{n})}^2+{(\sqrt[4]{m}-\sqrt[4]{n})}^2}{2(m-n)}}:{\displaystyle \frac{1}{\sqrt{m^3}-\sqrt{n^3}}}-3\sqrt{mn}=$$ $$={\displaystyle \frac{\sqrt{m}+2\sqrt[4]{mn}+\sqrt{n}+\sqrt{m}-2\sqrt[4]{mn}+\sqrt{n}}{2(\sqrt{m}-\sqrt{n})(\sqrt{m}+\sqrt{n})}}\times$$ $$\times {\displaystyle \frac{(\sqrt{m}-\sqrt{n})\left({(\sqrt{m})}^2+\sqrt{mn}+{(\sqrt{n})}^2\right)}{1}}-3\sqrt{mn}=$$ $$={\displaystyle \frac{2(\sqrt{m}+\sqrt{n})(\sqrt{m}-\sqrt{n})\left({(\sqrt{m})}^2+\sqrt{mn}+{(\sqrt{n})}^2\right)}{2(\sqrt{m}-\sqrt{n})(\sqrt{m}+\sqrt{n})}}-3\sqrt{mn}=$$ $$={(\sqrt{m})}^2+\sqrt{mn}+{(\sqrt{n})}^2-3\sqrt{mn}={(\sqrt{m})}^2-$$ $$-2\sqrt{mn}+{(\sqrt{n})}^2={(\sqrt{m}-\sqrt{n})}^2$$

Шешуі: ММЖ: $\sqrt{2}+3\sqrt[5]{y}\ne 0,\iff y\ne -{\left({\displaystyle \frac{\sqrt{2}}{3}}\right)}^5$ $${\left(\left({\displaystyle \frac{2^{3/2}+27y^{3/5}}{\sqrt{2}+3\sqrt[5]{y}}}+3\sqrt[10]{32y^2}-2\right)\cdot 3^{-2}\right)}^5=$$ $$={\left(\left({\displaystyle \frac{{(\sqrt{2})}^3+{(3\sqrt[5]{y})}^3}{\sqrt{2}+3\sqrt[5]{y}}}+3\sqrt{2}\sqrt[5]{y}-2\right)\cdot {\displaystyle \frac{1}{9}}\right)}^5=$$ $$={\left(\left({\displaystyle \frac{(\sqrt{2}+3\sqrt[5]{y}){(\sqrt{2})}^2-3\sqrt{2}\sqrt[5]{y}+{(3\sqrt[5]{y})}^2)}{\sqrt{2}+3\sqrt[5]{y}}}+3\sqrt{2}\sqrt[5]{y}-2\right)\cdot {\displaystyle \frac{1}{9}}\right)}^5=$$ $$={\left(\left(2+9\sqrt[5]{y^2}-2\right)\cdot {\displaystyle \frac{1}{9}}\right)}^5={\left(9\sqrt[5]{y^2}\cdot {\displaystyle \frac{1}{9}}\right)}^5={\left(\sqrt[5]{y^2}\right)}^5=y^2$$

Шешуі: ММЖ: $0<t\ne 1$ $${\displaystyle \frac{2\sqrt{1+{\displaystyle \frac{1}{4}}{\left(\sqrt{{\displaystyle \frac{1}{t}}}-\sqrt{t}\right)}^2}}{\sqrt{1+{\displaystyle \frac{1}{4}}{\left(\sqrt{{\displaystyle \frac{1}{t}}}-\sqrt{t}\right)}^2}-{\displaystyle \frac{1}{2}}\left(\sqrt{{\displaystyle \frac{1}{t}}}-\sqrt{t}\right)}}={\displaystyle \frac{2\sqrt{1+{\displaystyle \frac{1}{4}}\left({\displaystyle \frac{1}{t}}-2+t\right)}}{\sqrt{1+{\displaystyle \frac{1}{4}}\left({\displaystyle \frac{1}{t}}-2+t\right)-{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1}{\sqrt{t}}}-\sqrt{t}\right)}}}=$$ $$={\displaystyle \frac{2\sqrt{1+{\displaystyle \frac{1}{4}}\cdot {\displaystyle \frac{1-2t+t^2}{t}}}}{\sqrt{1+{\displaystyle \frac{1}{4}}\cdot {\displaystyle \frac{1-2t+t^2}{t}}}-{\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1-t}{\sqrt{t}}}}}={\displaystyle \frac{2\sqrt{{\displaystyle \frac{4t+1-2t+t^2}{4t}}}}{\sqrt{{\displaystyle \frac{4t+1-2t+t^2}{4t}}}-{\displaystyle \frac{1-t}{2\sqrt{t}}}}}=$$ $$={\displaystyle \frac{\sqrt{{\displaystyle \frac{1+2t+t^2}{t}}}}{{\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{1+2t+t^2}{t}}}-{\displaystyle \frac{1-t}{2\sqrt{t}}}}}={\displaystyle \frac{{\displaystyle \frac{1+t}{\sqrt{t}}}}{{\displaystyle \frac{1+t}{2\sqrt{t}}}-{\displaystyle \frac{1-t}{2\sqrt{t}}}}}={\displaystyle \frac{{\displaystyle \frac{1+t}{\sqrt{t}}}}{{\displaystyle \frac{1+t-1+t}{2\sqrt{t}}}}}={\displaystyle \frac{{\displaystyle \frac{1+t}{\sqrt{t}}}}{{\displaystyle \frac{2t}{2\sqrt{t}}}}}={\displaystyle \frac{1+t}{t}}$$

Шешуі: ММЖ: $\{\begin{array}{l}t+4>0,\\ 2-\sqrt{t+4}\ne 0\end{array}\iff \{\begin{array}{l}t>-4,\\ t\ne 0.\end{array}$ $$1+{\displaystyle \frac{2}{\sqrt{t+4}}}+\sqrt{t+4}+{\displaystyle \frac{4}{\sqrt{t+4}}}=t\cdot {\displaystyle \frac{{\displaystyle \frac{\sqrt{t+4}+2}{\sqrt{t+4}}}}{2-\sqrt{t+4}}}+{\displaystyle \frac{{(\sqrt{t+4})}^2+4}{\sqrt{t+4}}}=$$ $$=t\cdot {\displaystyle \frac{\sqrt{t+4}+2}{\sqrt{t+4}(2-\sqrt{t+4}}}+{\displaystyle \frac{{(\sqrt{t+4})}^2+4}{\sqrt{t+4}}}=n$$ $$={\displaystyle \frac{t(\sqrt{t+4}+2)(2+\sqrt{t+4})}{\sqrt{t+4}(2-\sqrt{t+4})(2+\sqrt{t+4})}}+{\displaystyle \frac{{(\sqrt{t+4})}^2+4}{\sqrt{t+4}}}=$$ $$={\displaystyle \frac{t{(\sqrt{t+4}+2)}^2}{\sqrt{t+4}(4-t-4)}}+{\displaystyle \frac{{(\sqrt{t+4})}^2+4}{\sqrt{t+4}}}=$$ $$={\displaystyle \frac{-(t+4+4\sqrt{t+4}+4)}{\sqrt{t+4}}}+{\displaystyle \frac{t+4+4}{\sqrt{t+4}}}=$$ $$={\displaystyle \frac{-t-4\sqrt{t+4}-8+t+8}{\sqrt{t+4}}}=-{\displaystyle \frac{4\sqrt{t+4}}{\sqrt{t+4}}}=-4$$

Шешуі: ММЖ: $\{\begin{array}{l}1+x⩾0,\\ x⩾0,\\ 1-\sqrt{x}\ne 0\end{array}\iff \{\begin{array}{l}x⩾0\\ x\ne 1\end{array}$ $${\left({\displaystyle \frac{1+\sqrt{x}}{\sqrt{1+x}}}-{\displaystyle \frac{\sqrt{1+x}}{1+\sqrt{x}}}\right)}^2-{\left({\displaystyle \frac{1-\sqrt{x}}{\sqrt{1+x}}}-{\displaystyle \frac{\sqrt{1+x}}{1-\sqrt{x}}}\right)}^2={\left({\displaystyle \frac{{(1+\sqrt{x})}^2-{(\sqrt{1+x})}^2}{\sqrt{1+x}(1+\sqrt{x})}}\right)}^2-$$ $$-{\left({\displaystyle \frac{{(1-\sqrt{x})}^2-{(\sqrt{1+x})}^2}{\sqrt{1+x(1-\sqrt{x})}}}\right)}^2={\left({\displaystyle \frac{1+2\sqrt{x}+x-1-x}{\sqrt{1+x}(1+\sqrt{x})}}\right)}^2-$$ $${-{\left({\displaystyle \frac{1-2\sqrt{x}+x-1-x}{\sqrt{1+x}{(1-\sqrt{x})}^2}}\right)}^2=\left({\displaystyle \frac{2\sqrt{x}}{\sqrt{1+x}(1+\sqrt{x}}}\right))}^2-$$ $$-{\left({\displaystyle \frac{-2\sqrt{x}}{\sqrt{1+x}(1-\sqrt{x})}}\right)}^2={\displaystyle \frac{4x}{{(\sqrt{1+x}(1+\sqrt{x}))}^2}}-$$ $$-{\displaystyle \frac{4x}{{(\sqrt{1+x}(1-\sqrt{x}))}^2}}={\displaystyle \frac{4x\left({(1-\sqrt{x})}^2-{(1+\sqrt{x})}^2\right)}{{(\sqrt{1+x}(1+\sqrt{x})(1-\sqrt{x}))}^2}}=$$ $$={\displaystyle \frac{4x(1-2\sqrt{x}+x-1-2\sqrt{x}-x)}{(1+x){(1-x)}^2}}={\displaystyle \frac{-16x\sqrt{x}}{(1+x)(1-x)(1-x)}}={\displaystyle \frac{16x\sqrt{x}}{\left(1-x^2\right)(x-1)}}$$

Шешуі: ММЖ: $\{\begin{array}{l}x>0\\ x\ne 1\end{array}$ $${\displaystyle \frac{x-1}{x+x^{1/2}+1}}:{\displaystyle \frac{x^{0,5}+1}{x^{1,5}-1}}+{\displaystyle \frac{2}{x^{-0,5}}}={\displaystyle \frac{\left(x^{1/2}-1\right)\left(x^{1/2}+1\right)}{x+x^{1/2}+1}}\cdot {\displaystyle \frac{{\left(x^{1/2}\right)}^3-1}{x^{1/2}+1}}+{\displaystyle \frac{2}{{\displaystyle \frac{1}{x^{1/2}}}}}=$$ $$={\displaystyle \frac{\left(x^{1/2}-1\right)\left(x^{1/2}-1\right)\left(x+x^{1/2}+1\right)}{x+x^{1/2}+1}}+2x^{1/2}={\left(x^{1/2}-1\right)}^2+2x^{1/2}=x-2x^{1/2}+$$ $$+1+2x^{1/2}=x+1.$$

Шешуі: ММЖ: $\{\begin{array}{l}a⩾0\\ \sqrt{a}-\sqrt{a-1}\ne 0,\\ {\displaystyle \frac{a+1}{a-1}}⩾0\end{array}\iff \{\begin{array}{l}a⩾0,\\ a>1\end{array}\iff a>1.$ X - бірінші жақшадағы ал, Y өрнек болсын. $$X={\displaystyle \frac{1}{\sqrt{a}+\sqrt{a+1}}}+{\displaystyle \frac{1}{\sqrt{a}-\sqrt{a-1}}}={\displaystyle \frac{\sqrt{a}-\sqrt{a-1}+\sqrt{a}+\sqrt{a+1}}{(\sqrt{a}+\sqrt{a+1})(\sqrt{a}-\sqrt{a-1})}}=$$ $$={\displaystyle \frac{2\sqrt{a}+\sqrt{a+1}-\sqrt{a-1}}{a+\sqrt{a(a+1)}-\sqrt{a(a-1)}-\sqrt{(a-1)(a+1)}}};$$ $$Y=1+{\displaystyle \frac{\sqrt{a+1}}{\sqrt{a-1}}}={\displaystyle \frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a-1}}}.$$ Онда $$X:Y={\displaystyle \frac{2\sqrt{a}+\sqrt{a+1}-\sqrt{a-1}}{a+\sqrt{a(a+1)}-\sqrt{a(a-1)}-\sqrt{(a-1)(a+1)}}}:{\displaystyle \frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a-1}}}=$$ $$={\displaystyle \frac{\sqrt{a-1}(2\sqrt{a}+\sqrt{a+1}-\sqrt{a-1})}{a\sqrt{a+1}+a\sqrt{a}+\sqrt{a}-a\sqrt{a-1}-\sqrt{a-1}+a\sqrt{a-1}-a\sqrt{a}+\sqrt{a}-a\sqrt{a+1}+\sqrt{a+1}}}=$$ $$={\displaystyle \frac{\sqrt{a-1}(2\sqrt{a}+\sqrt{a+1}-\sqrt{a-1})}{2\sqrt{a}+\sqrt{a+1}-\sqrt{a-1}}}=\sqrt{a-1}.$$

Шешуі: $\{\begin{array}{l}x⩾0\\ y⩾0\\ x\ne y\end{array}$ $${\displaystyle \frac{x-y}{x^{3/4}+x^{1/2}{y}^{1/4}}}\cdot {\displaystyle \frac{x^{1/2}{y}^{1/4}+x^{1/4}{y}^{1/2}}{x^{1/2}+y^{1/2}}}\cdot {\displaystyle \frac{x^{1/4}{y}^{-1/4}}{x^{1/2}-2x^{1/4}{y}^{1/4}+y^{1/2}}}=$$ $$={\displaystyle \frac{x-y}{\sqrt[4]{x^3}+\sqrt[4]{x^2}\sqrt[4]{y}}}\cdot {\displaystyle \frac{\sqrt[4]{x^2}\sqrt[4]{y}+\sqrt[4]{x}\sqrt[4]{y^2}}{\sqrt{x}+\sqrt{y}}}\cdot {\displaystyle \frac{\sqrt[4]{x}}{\sqrt[4]{y}\left(\sqrt[4]{x^2}-2\sqrt[4]{x}\sqrt[4]{y}+\sqrt[4]{y^2}\right)}}=$$ $$={\displaystyle \frac{x-y}{\sqrt[4]{x^2}(\sqrt[4]{x}+\sqrt[4]{y})}}\cdot {\displaystyle \frac{\sqrt[4]{x}\sqrt[4]{y}(\sqrt[4]{x}+\sqrt[4]{y})}{\sqrt{x}+\sqrt{y}}}\cdot {\displaystyle \frac{\sqrt[4]{x}}{\sqrt[4]{y}{(\sqrt[4]{x}-\sqrt[4]{y})}^2}}=$$ $$={\displaystyle \frac{x-y}{(\sqrt{x}+\sqrt{y}){(\sqrt[4]{x}-\sqrt[4]{y})}^2}}={\displaystyle \frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{(\sqrt{x}+\sqrt{y}){(\sqrt[4]{x}-\sqrt[4]{y})}^2}}=$$

Шешуі: ММЖ: $\{\begin{array}{l}m\ne \pm n,\\ y\ne 0,\\ y>0,m=2k.\end{array}$ $$\sqrt[n]{{\displaystyle \frac{2n}{y^{m-n}}}}:\sqrt[m]{{\displaystyle \frac{{(m-n)}^2+4mn}{m^2-n^2}}}=y^{{\displaystyle \frac{2n}{n(m-n)}}}:y^{{\displaystyle \frac{m^2-2mn+n^2+4mn}{m(m+n)(m-n)}}}=$$ $$=y^{{\displaystyle \frac{2}{m-n}}}:y^{{\displaystyle \frac{m^2+2mn+n^2}{m(m+n)(m-n)}}}=y^{{\displaystyle \frac{2}{m-n}}}:y^{{\displaystyle \frac{{(m+n)}^2}{m(m+n)(m-n)}}}=y^{{\displaystyle \frac{2}{m-n}}}:y^{{\displaystyle \frac{m+n}{m(m-n)}}}=$$ $$=y^{{\displaystyle \frac{2}{m-n}}-{\displaystyle \frac{m+n}{m(m-n)}}}=y^{{\displaystyle \frac{2m-m-n}{m(m-n)}}}=y^{{\displaystyle \frac{m-n}{m(m-n)}}}=y^{{\displaystyle \frac{1}{m}}}=\sqrt[m]{y}.$$