Көрсеткіштік теңдеулер (Рустюмова 3.1.6(13-24))

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Теңдеулерді әртүрлі әдіспен шешіңіз.

№ 13 Теңдеуді шешіңіз: $${3^{x - 5}} + {3^{x - 7}} + {3^{x - 9}} = 45,5 + 22,75 + 11,375 + \ldots $$

Шешуі: $${3^{x - 9}} \cdot \left( {{3^4} + {3^2} + 1} \right) = 45,5 + 22,75 + 11,375 + \ldots $$ $${3^{x - 9}} \cdot 91 = \frac{{91}}{2} + \frac{{91}}{4} + \frac{{91}}{8} + \cdots $$ $${3^{x - 9}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots $$ $${b_1} = \frac{1}{2},\quad q = \frac{1}{2}$$ $$\boxed{{{S_n} = \frac{{{b_1}}}{{1 - q}}}}$$ $${S_n} = \frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}} = 1$$ $${3^{x - 9}} = 1,\quad x - 9 = 0,\quad x = 9$$

№ 14 Теңдеуді шешіңіз: $${2^{|x + 1|}} - \left| {{2^x} - 1} \right| = 1 + {2^x}$$

Шешуі: $$\boxed{1.} \quad {2^{|x + 1|}} - {2^x} + 1 = 1 + {2^x}$$ $${2^{|x + 1|}} + {2^x} - 1 = 1 + {2^x}$$ $$|x + 1| = x + 1$$ $$1)\,x + 1 = x + 1$$ $$x + 1 = x + 1$$ $$0 = 0$$ $$2)\,x + 1 = - x - 1$$ $$2x = - 2$$ $$x \ne - 1$$ $$\boxed{2.} \quad {2^{|x + 1|}} + {2^x} - 1 = 1 + {2^x}$$ $${2^{|x + 1|}} = 2$$ $$|x + 1| = 1$$ $$1)\,x + 1 = 1,\quad x = 0$$ $$2)\,x + 1 = - 1,\quad x = - 2$$ $$\text{Жауабы: }x \in \left\{ { - 2} \right\} \cup \left[ {0; + \infty } \right)$$

№ 15 Теңдеуді шешіңіз: $$f'(x) = f(0) \cdot {e^{ - 2x}},\quad f(x) = \frac{{{e^{ - 2x}}}}{{1 - 8x}}$$

Шешуі: $$f(0) = \frac{{{e^{ - 2 \cdot 0}}}}{{1 - 8 \cdot 0}} = 1$$ $$f'(x) = \frac{{ - 2 \cdot {e^{ - 2x}} \cdot (1 - 8x) + 8 \cdot {e^{ - 2x}}}}{{{{(1 - 8x)}^2}}} = $$ $$ = \frac{{ - 2 \cdot {e^{ - 2x}} + 16x \cdot {e^{ - 2x}} + 8 \cdot {e^{ - 2x}}}}{{{{(1 - 8x)}^2}}} = \frac{{6 \cdot {e^{ - 2x}} + 16x \cdot {e^{ - 2x}}}}{{{{(1 - 8x)}^2}}}$$ $$\frac{{6 \cdot {e^{ - 2x}} + 16x \cdot {e^{ - 2x}}}}{{{{(1 - 8x)}^2}}} = {e^{ - 2x}}$$ $$6 + 16x = {(1 - 8x)^2}$$ $$6 + 16x = 1 - 16x + 64{x^2}$$ $$64{x^2} - 32x - 5 = 0$$ $$D = 1024 + 1280 = 2304 = {48^2}$$ $${x_{1,2}} = \frac{{32 \pm 48}}{{64}} = \left\langle \begin{array}{l}\frac{{80}}{{64}} = \frac{5}{4}\\ - \frac{{16}}{{64}} = - \frac{1}{4}\end{array} \right.$$ $${x_1} = \frac{5}{4};\quad {x_2} = - \frac{1}{4}$$

№ 16 Теңдеуді шешіңіз: $${x^2} \cdot {2^{\sqrt {x + 2} }} - {2^{\sqrt {x + 2} + 4}} = {x^2} \cdot {2^x} - {2^{x + 4}}$$

Шешуі: $${2^{\sqrt {x + 2} }}\left( {{x^2} - {2^4}} \right) = {2^x}\left( {{x^2} - {2^4}} \right)$$ $${2^{\sqrt {x + 2} }}\left( {{x^2} - 16} \right) - {2^x}\left( {{x^2} - 16} \right) = 0$$ $$\left( {{x^2} - 16} \right)\left( {{2^{\sqrt {x + 2} }} - {2^x}} \right) = 0$$ $$1)\quad {x^2} - 16 = 0,\quad x = \pm 4$$ $$2)\quad {2^{\sqrt {x + 2} }} - {2^x} = 0$$ $$\sqrt {x + 2} = x,\quad x + 2 = {x^2}$$ $${x^2} - x - 2 = 0$$ $$D = 1 + 8 = 9$$ $${x_{1,2}} = \frac{{1 \pm 3}}{2} = \left\langle \begin{array}{l}2\\ - 1\end{array} \right.$$ $$x = 2$$

№ 17 Теңдеуді шешіңіз: $${x^2} \cdot {3^{\sqrt x }} + {3^{2x + 1}} = {3^{2 + \sqrt x }} + {x^2} \cdot {3^{2x - 1}}$$

Шешуі: $${x^2} \cdot {3^{\sqrt x }} + {3^{2x}} \cdot 3 - 9 \cdot {3^{\sqrt x }} - \frac{{{x^2}}}{3} \cdot {3^{2x}} = 0$$ $${3^{\sqrt x }}\left( {{x^2} - 9} \right) + \frac{{{3^{2x}}}}{3}\left( {9 - {x^2}} \right) = 0$$ $$\left( {{x^2} - 9} \right)\left( {{3^{\sqrt x }} - {3^{2x - 1}}} \right) = 0$$ $$1)\quad {x^2} - 9 = 0,\quad x = \pm 3$$ $$2)\quad {\mkern 1mu} {3^{\sqrt x }} - {3^{2x - 1}} = 0$$ $$\sqrt x = 2x - 1$$ $$x = 4{x^2} - 4x + 1$$ $$4{x^2} - 5x + 1 = 0$$ $$D = 25 - 16 = 9$$ $${x_{1,2}} = \frac{{5 \pm 3}}{{2 \cdot 4}} = \left\langle \begin{array}{l}1\\ - \frac{1}{4}\end{array} \right.$$ $$\text{Жауабы:} x = 3,\quad x = 1$$

№ 18 Теңдеуді шешіңіз: $${4^{2 - x + \sqrt {4 - 5x + 2{x^2}} }} - 5 \cdot {2^{1 - x + \sqrt {4 - 5x + 2{x^2}} }} = 6$$

Шешуі:$${4^2} \cdot {4^{ - x + \sqrt {4 - 5x + 2{x^2}} }} - 5 \cdot 2 \cdot {2^{ - x + \sqrt {4 - 5x + 2{x^2}} }} = 6$$ $${2^{ - x + \sqrt {4 - 5x + 2{x^2}} }} = m,\quad 16{m^2} - 10m - 6 = 0$$ $$8{m^2} - 5m - 3 = 0$$ $$D = 25 + 4 \cdot 8 \cdot 3 = 121$$ $${m_{1,2}} = \frac{{5 \pm 11}}{{8 \cdot 2}} = \left\langle \begin{array}{l}1\\ - \frac{6}{{16}} = - \frac{3}{8}\end{array} \right.$$ $$1)\quad {2^{ - x + \sqrt {4 - 5x + 2{x^2}} }} \ne - \frac{3}{8}$$ $$2)\quad {2^{ - x + \sqrt {4 - 5x + 2{x^2}} }} = 1$$ $$ - x + \sqrt {4 - 5x + 2{x^2}} = 0$$ $$\sqrt {4 - 5x + 2{x^2}} = x,\quad 4 - 5x + 2{x^2} = {x^2}$$ $${x^2} - 5x + 4 = 0,$$ $$D = 25 - 16 = 9$$ $${x_{1,2}} = \frac{{5 \pm 3}}{2} = \left\langle \begin{array}{l}4\\1\end{array} \right.$$ $${x_1} = 4,\quad {x_2} = 1$$

№ 19 Теңдеуді шешіңіз: $${3^{\sqrt {2x + 5} }} = \frac{{{3^{\sqrt {12x + 25} }}}}{{{3^{\sqrt {5x + 6} }}}}$$

Шешуі: $${3^{\sqrt {2x + 5} }} \cdot {3^{\sqrt {5x + 6} }} = {3^{\sqrt {12x + 25} }}$$ $${3^{\sqrt {2x + 5} + \sqrt {5x + 6} }} = {3^{\sqrt {12x + 25} }}$$ $${\left( {\sqrt {2x + 5} + \sqrt {5x + 6} } \right)^2} = {\left( {\sqrt {12x + 25} } \right)^2}$$ $$2x + 5 + 5x + 6 + 2\sqrt {2x + 5} \cdot \sqrt {5x + 6} = 12x + 25$$ $${\left( {2\sqrt {(2x + 5)(5x + 6)} } \right)^2} = {\left( {5x + 14} \right)^2}$$ $$4\left( {10{x^2} + 12x + 25x + 30} \right) = 25{x^2} + 140x + 196$$ $$40{x^2} + 148x + 120 - 25{x^2} - 140x - 196 = 0$$ $$15{x^2} + 8x - 76 = 0$$ $$D = 64 + 4 \cdot 15 \cdot 76 = 4624 = {68^2}$$ $${x_{1,2}} = \frac{{ - 8 \pm 68}}{{2 \cdot 15}} = \left\langle \begin{array}{l} - \frac{{38}}{{15}} = - 2\frac{8}{{15}}\\2\end{array} \right.$$ $$x = 2,\quad x \ne - 2\frac{8}{{15}}$$

№ 20 Теңдеуді шешіңіз: $$\sqrt[x]{{81}} - \sqrt[x]{{{9^{x + 1}}}} + 18 = 0$$

Шешуі: $$\sqrt[x]{{{9^2}}} - 9\sqrt[x]{9} + 18 = 0$$ $$\sqrt[x]{9} = m,\quad {m^2} - 9m + 18 = 0$$ $$D = 81 - 4 \cdot 18 = 9$$ $${m_{1,2}} = \frac{{9 \pm 3}}{2} = \left\langle \begin{array}{l}6\\3\end{array} \right.$$ $$1)\,\sqrt[x]{9} = 6,\quad {3^{\frac{2}{x}}} = 6,\quad \frac{2}{x} = {\log _3}6,$$ $$x \in N,\quad x \ne {\log _3}6$$ $$2)\,\sqrt[x]{9} = 3$$ $${x = 2}$$

№ 21 Теңдеуді шешіңіз: $${2^{\cos x}} \cdot {2^{{{\cos }^2}x}} \cdot {2^{{{\cos }^3}x}} \cdot \ldots = 2$$

Шешуі: $${2^{\cos x + {{\cos }^2}x + {{\cos }^3}x + \ldots }} = 2$$ $$\cos x + {\cos ^2}x + {\cos ^3}x + \ldots = 1$$ $${b_1} = \cos x,\;\;\;{\mkern 1mu} q = \cos x,\quad (\cos x \lt 1)$$ $$1 = \frac{{\cos x}}{{1 - \cos x}};\quad 1 - \cos x = \cos x$$ $$2\cos x = 1;\quad \cos x = \frac{1}{2}$$ $$x = \pm \frac{\pi }{3} + 2\pi k$$

№ 22 Теңдеуді шешіңіз: $${2^{3 + |x - 1|}} = 16 \cdot {4^{ - 0,5x}}$$

Шешуі: $${2^{3 + |x - 1|}} = {2^{4 - x}}$$ $$3 + |x - 1| = 4 - x$$ $$|x - 1| = 1 - x$$ $$\left[ \begin{array}{l}\left\{ \begin{array}{l} - \left( {x - 1} \right) = 1 - x\\x - 1 \lt 0\end{array} \right.\\\left\{ \begin{array}{l}x - 1 = 1 - x\\x - 1 \ge 0\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l} - x + 1 = 1 - x\\x \lt 1\end{array} \right.\\\left\{ \begin{array}{l}2x = 2\\x \ge 1\end{array} \right.\end{array} \right. \Rightarrow \\ \Rightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}0 = 0\\x \lt 1\end{array} \right.\\\left\{ \begin{array}{l}x = 1\\x \ge 1\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}x \lt 1\\x = 1\end{array} \right.$$ Жауабы $x \in ( - \infty ;1]$

№ 23 Теңдеуді шешіңіз: $${3^{3x - 1}} + {27^x} = {2^{2x + 1}} + 7 \cdot {4^x}$$

Шешуі: $${27^x} \cdot \frac{1}{3} + {27^x} = {4^x} \cdot 2 + 7 \cdot {4^x}$$ $$\frac{4}{3} \cdot {27^x} = 9 \cdot {4^x},\quad \left| { \div \left( {\frac{4}{3} \cdot {4^x}} \right)} \right.$$ $$\frac{{{{27}^x}}}{{{4^x}}} = \frac{9}{{\frac{4}{3}}},\quad \left( {\frac{{27}}{4}} \right) = \frac{{27}}{4}$$ $$x = 1$$

№ 24 Теңдеуді шешіңіз: $$\frac{{{2^x}}}{{{5^{x - 1}}}} + 3 = \frac{{{5^x}}}{{{2^{x - 1}}}}$$

Шешуі: $$5 \cdot \frac{{{2^x}}}{{{5^x}}} + 3 = 2 \cdot \frac{{{5^x}}}{{{2^x}}}$$ $$5 \cdot {\left( {\frac{2}{5}} \right)^x} + 3 - 2 \cdot {\left( {\frac{5}{2}} \right)^x} = 0$$ $${\left( {\frac{5}{2}} \right)^x} = m,\quad 5 \cdot \frac{1}{m} + 3 - 2m = 0$$ $$5 + 3m - 2{m^2} = 0$$ $$2{m^2} - 3m - 5 = 0$$ $$D = 9 + 4 \cdot 2 \cdot 5 = 49$$ $${m_{1,2}} = \frac{{3 \pm 7}}{{2 \cdot 2}} = \left\langle \begin{array}{l}\frac{5}{2}\\ - 1\end{array} \right.$$ $${\left( {\frac{5}{2}} \right)^x} = \frac{5}{2};\quad x = 1$$

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