Көрсеткіштік теңдеулер (Рустюмова 3.1.6(1-12))

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Теңдеулерді әртүрлі әдіспен шешіңіз.

№ 1 Теңдеуді шешіңіз: $${27^x} - 13 \cdot {9^x} + 13 \cdot {3^{x + 1}} - 27 = 0$$

Шешуі: $${3^{3x}} - 13 \cdot {3^{2x}} + 13 \cdot {3^x} \cdot 3 - {3^3} = 0$$ $${3^{3x}} - {3^3} - 13 \cdot {3^x}\left( {{3^x} - 3} \right) = 0$$ $$\left( {{3^x} - 3} \right)\left( {{3^{2x}} + 3 \cdot {3^x} + 9} \right) - 13 \cdot {3^x}\left( {{3^x} - 3} \right) = 0$$ $$\left( {{3^x} - 3} \right)\left( {{3^{2x}} + 3 \cdot {3^x} + 9 - 13 \cdot {3^x}} \right) = 0$$ $$1)\,{3^x} - 3 = 0;$$ $${3^x} = 3;$$ $$x = 1$$ $$2)\,{3^{2x}} - 10 \cdot {3^x} + 9 = 0$$ $${3^x} = m$$ $${m^2} - 10m + 9 = 0$$ $$D = 100 - 4 \cdot 9 = 64 = {8^2}$$ $${m_{1,2}} = \frac{{10 \pm 8}}{2} = \left\langle \begin{array}{l} 9\\1\end{array} \right.$$ $${3^x} = 9;\quad {3^x} = 1$$ $$x = 2;\quad x = 0$$

№ 2 Теңдеуді шешіңіз: $$3 \cdot {4^x} + \frac{1}{3} \cdot {9^{x + 2}} = 6 \cdot {4^{x + 1}} - \frac{1}{2} \cdot {9^{x + 1}}$$

Шешуі: $$3 \cdot {4^x} - 6 \cdot 4 \cdot {4^x} = - \frac{1}{2} \cdot {9^x} \cdot 9 - \frac{1}{3} \cdot {9^x} \cdot {9^2}$$ $${4^x}(3 - 24) = {9^x} \cdot \left( { - \frac{9}{2} - 27} \right)$$ $${4^x}( - 21) = {9^x} \cdot \left( { - \frac{{63}}{2}} \right),\quad \left| { \div \left( { - 21 \cdot {9^x}} \right)} \right.$$ $$\frac{{{4^x}}}{{{9^x}}} = - \frac{{63}}{2}:( - 21)$$ $${\left( {\frac{4}{9}} \right)^x} = \frac{3}{2}$$ $${\left( {\frac{2}{3}} \right)^{2x}} = {\left( {\frac{2}{3}} \right)^{ - 1}}$$ $$2x = - 1$$ $$x = - \frac{1}{2}$$

№ 3 Теңдеуді шешіңіз: $$2 \cdot {12^x} - {3^{x + 1}} + {4^{x + 1}} - 6 = 0$$

Шешуі: $$2 \cdot {(3 \cdot 4)^x} - {3^x} \cdot 3 + {2^{2x}} \cdot 4 - 6 = 0$$ $$2 \cdot {3^x} \cdot {2^{2x}} - {3^x} \cdot 3 + {2^{2x}} \cdot 4 - 6 = 0$$ $$2 \cdot {2^{2x}}\left( {{3^x} + 2} \right) - 3\left( {{3^x} + 2} \right) = 0$$ $$\left( {{3^x} + 2} \right)\left( {2 \cdot {2^{2x}} - 3} \right) = 0$$ $$1)\,{3^x} + 2 = 0$$ $$x \in \emptyset $$ $$2)\,2 \cdot {2^{2x}} - 3 = 0$$ $${2^{2x + 1}} = 3$$ $$2x + 1 = {\log _2}3$$ $$x = \frac{{{{\log }_2}3 - 1}}{2}$$

№ 4 Теңдеуді шешіңіз: $$2 \cdot {3^{x + 1}} - 5 \cdot {9^{x - 2}} = 81$$

Шешуі: $$2 \cdot {3^x} \cdot 3 - 5 \cdot {3^{2x}} \cdot \frac{1}{{{3^4}}} = 81$$ $$6 \cdot {3^x} - \frac{5}{{81}} \cdot {3^{2x}} - 81 = 0$$ $$\frac{5}{{81}} \cdot {3^{2x}} - 6 \cdot {3^x} + 81 = 0$$ $$D = 36 - 4 \cdot \frac{5}{{81}} \cdot 81 = 16 = {4^2}$$ $${x_{1,2}} = \frac{{6 \pm 4}}{{2 \cdot \frac{5}{{81}}}} = \left\langle \begin{array}{l}81\\\frac{{81}}{5}\end{array} \right.$$ $$1)\,{3^x} = 81;\quad x = 4;$$ $$2)\,{3^x} = \frac{{81}}{5}$$ $$x = {\log _3}\frac{{81}}{5} = {\log _3}81 - {\log _3}5$$ $$x = 4 - {\log _3}5$$

№ 5 Теңдеуді шешіңіз: $$\frac{{{{10}^x} + {{10}^{ - x}}}}{{{{10}^x} - {{10}^{ - x}}}} = 5$$

Шешуі: $$\frac{{{{10}^x} + \frac{1}{{{{10}^x}}}}}{{{{10}^x} - \frac{1}{{{{10}^x}}}}} = 5$$ $${10^x} + \frac{1}{{{{10}^x}}} = 5 \cdot {10^x} - \frac{5}{{{{10}^x}}}$$ $$\frac{1}{{{{10}^x}}} + \frac{5}{{{{10}^x}}} = 5 \cdot {10^x} - {10^x}$$ $$\frac{6}{{{{10}^x}}} = 4 \cdot {10^x}$$ $$\frac{6}{4} = {10^{2x}}$$ $${10^{2x}} = \frac{3}{2}$$ $$2x = {\log _{10}}\frac{3}{2}$$ $$x = \frac{1}{2} \cdot {\log _{10}}\frac{3}{2}$$ $$x = \frac{1}{2}\lg 1,5$$

№ 6 Теңдеуді шешіңіз: $${2^{|3x - 5|}} = 4 \cdot {8^{|x - 1|}}$$

Шешуі: $${2^{|3x - 5|}} = {2^{2 + 3|x - 1|}}$$ $$|3x - 5| = 2 + 3|x - 1|$$ $$|3x - 5| - 2 - 3|x - 1| = 0$$ $$3x - 5 = 0$$ $$x = \frac{5}{3};\quad x - 1 = 0,\quad x = 1.$$ Көрсеткіштік теңдеулер (Рустюмова 3.1.6(1-12)) $$1)\quad ( - \infty ;1)$$ $$ - (3x - 5) - 2 + 3(x - 1) = 0$$ $$ - 3x + 5 - 2 + 3x - 3 = 0$$ $$0 = 0\quad ( - \infty ;1)$$ $$2)\quad \left[ {1;\frac{5}{3}} \right)$$ $$ - (3x - 5) - 2 - 3(x - 1) = 0$$ $$ - 3x + 5 - 2 - 3x + 3 = 0$$ $$ - 6x + 6 = 0$$ $$x = 1$$ $$3)\quad \left[ {\frac{5}{3}; + \infty } \right)$$ $${3x - 5 - 2 - (3x - 3) = 0}$$ $${3x - 7 - 3x + 3 = 0}$$ $${ - 4 \ne 0}$$ Жауабы: $x \in ( - \infty ;1]$

№ 7 Теңдеуді шешіңіз: $${4^{\sin x}} + {2^{5 - 2\sin x}} = 18$$

Шешуі: $${2^{2\sin x}} + {2^5} \cdot \frac{1}{{{2^{2\sin x}}}} = 18$$ $${2^{4\sin x}} + {2^5} = 18 \cdot {2^{2\sin x}}$$ $${2^{2\sin x}} = m;\quad {m^2} + 32 - 18m = 0$$ $${m^2} - 18m + 32 = 0$$ $$D = 324 - 4 \cdot 32 = 196$$ $${m_{1,2}} = \frac{{18 \pm 14}}{2} = \left\langle \begin{array}{l}16\\2\end{array} \right.$$ $$1)\,{2^{2\sin x}} = 2;\quad 2\sin x = 1;\quad \sin x = \frac{1}{2}$$ $$x = {\left( { - 1} \right)^k}\arcsin \frac{1}{2} + \pi k$$ $$x = {\left( { - 1} \right)^k}\frac{\pi }{6} + \pi k$$ $$2)\,{2^{2\sin x}} = 16$$ $$2\sin x = 4$$ $$\sin x \ne 2$$

№ 8 Теңдеуді шешіңіз: $${\left( {\sqrt {2 + \sqrt 3 } } \right)^x} + {\left( {\sqrt {2 - \sqrt 3 } } \right)^x} = 4$$

Шешуі: $${\left( {\sqrt {2 + \sqrt 3 } } \right)^x} + {\left( {\sqrt {2 - \sqrt 3 } } \right)^x} = 4,\quad \left| { \times {{\left( {\sqrt {2 + \sqrt 3 } } \right)}^x}} \right.\quad $$ $${\left( {\sqrt {2 + \sqrt 3 } } \right)^x}{\left( {\sqrt {2 + \sqrt 3 } } \right)^x} + {\left( {\sqrt {2 - \sqrt 3 } } \right)^x}{\left( {\sqrt {2 + \sqrt 3 } } \right)^x} = 4{\left( {\sqrt {2 + \sqrt 3 } } \right)^x}$$ $${(2 + \sqrt 3 )^x} + 1 = 4{\left( {\sqrt {2 + \sqrt 3 } } \right)^x}$$ $${\left( {\sqrt {2 + \sqrt 3 } } \right)^x} = m,\quad {\left( {2 + \sqrt 3 } \right)^x} = {m^2}$$ $${m^2} - 4m + 1 = 0$$ $$D = 16 - 4 = 12$$ $${m_{1,2}} = \frac{{4 \pm 2\sqrt 3 }}{2} = \left\langle \begin{array}{l}2 + \sqrt 3 \\2 - \sqrt 3 \end{array} \right.$$ $$1)\,{\left( {\sqrt {2 + \sqrt 3 } } \right)^x} = 2 + \sqrt 3 $$ $${\left( {2 + \sqrt 3 } \right)^{\frac{x}{2}}} = {\left( {2 + \sqrt 3 } \right)^1}$$ $$\frac{x}{2} = 1,\quad x = 2$$ $$2)\,{\left( {\sqrt {2 + \sqrt 3 } } \right)^x} = 2 - \sqrt 3 $$ $${\left( {\sqrt {2 + \sqrt 3 } } \right)^x} = \left( {2 - \sqrt 3 } \right) \cdot \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}}$$ $${\left( {2 + \sqrt 3 } \right)^{\frac{x}{2}}} = \frac{{4 - 3}}{{2 + \sqrt 3 }}$$ $${\left( {2 + \sqrt 3 } \right)^{\frac{x}{2}}} = {\left( {2 + \sqrt 3 } \right)^{ - 1}}$$ $$\frac{x}{2} = - 1,\quad x = - 2$$

№ 9 Теңдеуді шешіңіз: $${\left( {\sqrt {3 - 2\sqrt 2 } } \right)^x} + {\left( {\sqrt {3 + 2\sqrt 2 } } \right)^x} = 6$$

Шешуі: $$3 - 2\sqrt 2 = {(a + b)^2}$$ $$3 - 2\sqrt 2 = {a^2} + 2ab + {b^2}$$ $${a^2} + {b^2} = 3$$ $$ - 2ab = - 2\sqrt 2 $$ $$ab = \sqrt 2 $$ $$a = \sqrt 2 ,\quad b = 1$$ $${\left( {\sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} } \right)^x} + {\left( {\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} } \right)^x} = 6$$ $${\left( {\sqrt 2 - 1} \right)^x} + {\left( {\sqrt 2 + 1} \right)^x} = 6,\quad \left| { \times {{\left( {\sqrt 2 + 1} \right)}^x}} \right.$$ $$1 + {\left( {\sqrt 2 + 1} \right)^{2x}} = 6{\left( {\sqrt 2 + 1} \right)^x}$$ $${\left( {\sqrt 2 + 1} \right)^x} = m,\quad 1 + {m^2} = 6m$$ $${m^2} - 6m + 1 = 0$$ $$D = 36 - 4 = 32$$ $${m_{1,2}} = \frac{{6 \pm 4\sqrt 2 }}{2} = \left\langle \begin{array}{l}3 + \sqrt 2 \\3 - \sqrt 2 \end{array} \right.$$ $$1)\,{\left( {\sqrt 2 + 1} \right)^x} = 3 + 2\sqrt 2 $$ $${\left( {\sqrt 2 + 1} \right)^x} = {\left( {\sqrt 2 + 1} \right)^2}$$ $$x = 2$$ $$2)\,{\left( {\sqrt 2 + 1} \right)^x} = 3 - 2\sqrt 2 $$ $$\frac{{{{\left( {\sqrt 2 + 1} \right)}^x}{{\left( {\sqrt 2 - 1} \right)}^x}}}{{{{\left( {\sqrt 2 - 1} \right)}^x}}} = {\left( {\sqrt 2 - 1} \right)^2}$$ $$\frac{1}{{{{\left( {\sqrt 2 - 1} \right)}^x}}} = \frac{1}{{{{\left( {\sqrt 2 - 1} \right)}^{ - 2}}}}$$ $$x = - 2$$

№ 10 Теңдеуді шешіңіз: $$3 \cdot {8^{x - 1}} - {2^{2x - 1}} - {2^{x + 2}} = 0$$

Шешуі: $$3 \cdot {8^x} \cdot \frac{1}{8} - {2^{2x}} \cdot \frac{1}{2} - {2^x} \cdot 4 = 0$$ $$\frac{3}{8} \cdot {2^{3x}} - \frac{1}{2} \cdot {2^{2x}} - 4 \cdot {2^x} = 0$$ $$\frac{1}{8} \cdot {2^x}\left( {3 \cdot {2^{2x}} - 4 \cdot {2^x} - 32} \right) = 0$$ $$1)\frac{1}{8} \cdot {2^x} = 0;\quad x \in \emptyset $$ $$2)\,3 \cdot {2^{2x}} - 4 \cdot {2^x} - 32 = 0$$ $${2^x} = m$$ $$3{m^2} - 4m - 32 = 0$$ $$D = 16 + 4 \cdot 3 \cdot 32 = 400 = {20^2}$$ $${m_{1,2}} = \frac{{4 \pm 20}}{{3 \cdot 2}} = \left\langle \begin{array}{l}\frac{{24}}{6} = 4\\\frac{{ - 16}}{{3 \cdot 2}} = - \frac{8}{3}\end{array} \right.$$ $${2^x} = 4$$ $$x = 2$$

№ 11 Теңдеуді шешіңіз: $${0,5^{2 + 4 + 6 + … + 2x}} = {0,5^{72}}$$

Шешуі: $$2 + 4 + 6 + \ldots + 2x = 72$$ $${a_1} = 2,\quad d = 2$$ $$ \boxed{{{S_n} = \frac{{2{a_1} + (n - 1) \cdot d}}{2} \cdot n}}$$ $$72 = \frac{{2 \cdot 2 + (n - 1) \cdot 2}}{2} \cdot n$$ $$72 = \frac{{2(2 + n - 1) \cdot n}}{2}$$ $$72 = n + {n^2},\quad n^2 + n - 72 = 0$$ $$D = 1 + 4 \cdot 72 = 289 = {17^2}$$ $${n_{1,2}} = \frac{{ - 1 \pm 17}}{2} = \left\langle \begin{array}{l}8\\ - 9\end{array} \right.$$ $$\boxed{{{a_n} = {a_1} + (n - 1) \cdot d}}$$ $${a_8} = 2 + 7 \cdot 2 = 16$$ $$16 = 2x,\quad x = 8$$

№ 12 Теңдеуді шешіңіз: $$6 \cdot \sqrt[x]{9} - 13\sqrt[x]{6} + 6 \cdot \sqrt[x]{4} = 0$$

Шешуі: $$6 \cdot {3^{\frac{2}{x}}} - {13 \cdot 3^{\frac{1}{x}}} \cdot {2^{\frac{1}{x}}} + 6 \cdot {2^{\frac{2}{x}}} = 0,\quad \left| \div \right.\left( {{3^{\frac{2}{x}}}} \right)$$ $$\frac{{6 \cdot {3^{\frac{2}{x}}}}}{{{3^{\frac{2}{x}}}}} - \frac{{13 \cdot {3^{\frac{1}{x}}}{2^{\frac{1}{x}}}}}{{{3^{\frac{2}{x}}}}} + \frac{{6 \cdot {2^{\frac{2}{x}}}}}{{{3^{\frac{2}{x}}}}} = 0$$ $$6 - 13 \cdot \frac{{{2^{\frac{1}{x}}}}}{{{3^{\frac{1}{x}}}}} + 6 \cdot {\left( {\frac{2}{3}} \right)^{\frac{2}{x}}} = 0$$ $${\left( {\frac{2}{3}} \right)^{\frac{1}{x}}} = m,\quad 6 - 13m + 6{m^2} = 0$$ $$6{m^2} - 13m + 6 = 0$$ $$D = {( - 13)^2} - 4 \cdot 6 \cdot 6 = 169 - 144 = 25$$ $${m_{1,2}} = \frac{{13 \pm 5}}{{6 \cdot 2}} = \left\langle \begin{array}{l}\frac{3}{2}\\\frac{2}{3}\end{array} \right.$$ $$1)\,{\left( {\frac{2}{3}} \right)^{\frac{1}{x}}} = \frac{3}{2},\quad \frac{1}{x} = - 1;\quad x = - 1.$$ $$2)\,{\left( {\frac{2}{3}} \right)^{\frac{1}{x}}} = \frac{2}{3},\quad \frac{1}{x} = 1,\quad x = 1.$$

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