Көрсеткіштік теңдеулер (Рустюмова 3.1.4)

 +/-  - Есептің жауабын көрсету/көрсетпеу.

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Біртекті теңдеулерді шешіңіз.

№ 1 Теңдеуді шешіңіз: $${9^x} + {6^x} = {2^{2x + 1}}$$

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

№ 2 Теңдеуді шешіңіз: $$125 \cdot {25^x} - 70 \cdot {10^x} + 8 \cdot {4^x} = 0$$

Шешуі: $$125 \cdot {5^{2x}} - 70 \cdot {(5 \cdot 2)^x} + 8 \cdot {2^{2x}} = 0$$ $$125 \cdot {5^{2x}} - 70 \cdot {5^x} \cdot {2^x} + 8 \cdot {2^{2x}} = 0,\quad \left| { \div {2^{2x}}} \right.$$ $$\frac{{125 \cdot {5^{2x}}}}{{{2^{2x}}}} - \frac{{70 \cdot {5^x} \cdot {2^x}}}{{{2^{2x}}}} + \frac{{8 \cdot {2^{2x}}}}{{{2^{2x}}}} = 0$$ $$125 \cdot {\left( {\frac{5}{2}} \right)^{2x}} - 70 \cdot {\left( {\frac{5}{2}} \right)^x} + 8 = 0$$ $${\left( {\frac{5}{2}} \right)^x} = m$$ $$125{m^2} - 70m + 8 = 0$$ $$D = 70 - 4 \cdot 125 \cdot 8 = 900 = {30^2}$$ $${m_{1,2}} = \frac{{70 \pm 30}}{{125 \cdot 2}} = \left\langle \begin{array}{l}\frac{{100}}{{250}} = \frac{2}{5}\\\frac{{40}}{{250}} = \frac{4}{{25}}\end{array} \right.$$ $${\left( {\frac{5}{2}} \right)^x} = \frac{2}{5};\quad {\left( {\frac{5}{2}} \right)^x} = \frac{4}{{25}}$$ $$x = - 1;\quad x = - 2$$

№ 3 Теңдеуді шешіңіз: $$3 \cdot {4^x} - 5 \cdot {6^x} + 2 \cdot {9^x} = 0$$

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

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

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

№ 5 Теңдеуді шешіңіз: $${3^{2{x^2} - 6x + 3}} + {6^{{x^2} - 3x + 1}} = {2^{2{x^2} - 6x + 3}}$$

Шешуі: $${3^{2{x^2} - 6x}} \cdot {3^3} + {(3 \cdot 2)^{{x^2} - 3x}} \cdot 6 = {2^{2{x^2} - 6x}} \cdot {2^3},\quad \div \left( {{2^{2{x^2} - 6x}}} \right)$$ $$27 \cdot \frac{{{3^{2{x^2} - 6x}}}}{{{2^{2{x^2} - 6x}}}} + \frac{{{3^{{x^2} - 3x}} \cdot {2^{{x^2} - 3x}} \cdot 6}}{{{2^{2\left( {{x^2} - 3x} \right)}}}} = \frac{{8 \cdot {2^{2{x^2} - 6x}}}}{{{2^{2{x^2} - 6x}}}}$$ $$27 \cdot {\left( {\frac{3}{2}} \right)^{2{x^2} - 6x}} + {\left( {\frac{3}{2}} \right)^{{x^2} - 3x}} \cdot 6 - 8 = 0$$ $${\left( {\frac{3}{2}} \right)^{{x^2} - 3x}} = m$$ $$27{m^2} + 6m - 8 = 0$$ $$D = 36 + 4 \cdot 27 \cdot 8 = 900 = {30^2}$$ $${m_{1,2}} = \frac{{ - 6 \pm 30}}{{27 \cdot 2}} = \left\langle \begin{array}{l}\frac{{ - 36}}{{54}} = - \frac{2}{3}\\\frac{{24}}{{54}} = \frac{4}{9}\end{array} \right.$$ $${\left( {\frac{3}{2}} \right)^{{x^2} - 3x}} \ne - \frac{2}{3};\quad {\left( {\frac{3}{2}} \right)^{{x^2} - 3x}} = \frac{4}{9}$$ $${x^2} - 3x = - 2$$ $${x^2} - 3x + 2 = 0$$ $$D=9-8=1$$ $${x_{1,2}} = \frac{{3 \pm 1}}{2} = \left\langle \begin{array}{l}2\\1\end{array} \right.$$ $${x_1} = 1;\quad {x_2} = 2$$

№ 6 Теңдеуді шешіңіз: $$8 \cdot {9^x} + {6^{x + 1}} = 27 \cdot {4^x}$$

Шешуі: $$8 \cdot {3^{2x}} + {6^x} \cdot 6 = 27 \cdot {2^{2x}}$$ $$8 \cdot {3^{2x}} + 6 \cdot {(3 \cdot 2)^x} - 27 \cdot {2^{2x}} = 0$$ $$8 \cdot {3^{2x}} + 6 \cdot {3^x} \cdot {2^x} - 27 \cdot {2^{2x}} = 0,\quad \left| { \div \left( {{2^{2x}}} \right)} \right.$$ $$8 \cdot \frac{{{3^{2x}}}}{{{2^{2x}}}} + \frac{{6 \cdot {3^x} \cdot {2^x}}}{{{2^{2x}}}} - \frac{{27 \cdot {2^{2x}}}}{{{2^{2x}}}} = 0$$ $$8 \cdot {\left( {\frac{3}{2}} \right)^{2x}} + 6 \cdot {\left( {\frac{3}{2}} \right)^x} - 27 = 0$$ $${\left( {\frac{3}{2}} \right)^x} = m$$ $$8{m^2} + 6m - 27 = 0$$ $$D = 36 + 4 \cdot 8 \cdot 27 = 900 = {30^2}$$ $${m_{1,2}} = \frac{{ - 6 \pm 30}}{{2 \cdot 8}} = \left\langle \begin{array}{l}\frac{{ - 36}}{{16}} = - \frac{9}{4}\\\frac{{24}}{{16}} = \frac{3}{2}\end{array} \right.$$ $${\left( {\frac{3}{2}} \right)^x} = \frac{3}{2}$$ $$x = 1$$

№ 7 Теңдеуді шешіңіз: $$4 \cdot {2^{2x}} - {6^x} = 18 \cdot {3^{2x}}$$

Шешуі: $$4 \cdot {2^{2x}} - {3^x} \cdot {2^x} - 18 \cdot {3^{2x}} = 0,\quad \left| { \div \left( {{3^{2x}}} \right)} \right.$$ $$4 \cdot \frac{{{2^{2x}}}}{{{3^{2x}}}} - \frac{{{3^x} \cdot {2^x}}}{{{3^{2x}}}} - \frac{{18 \cdot {3^{2x}}}}{{{3^{2x}}}} = 0$$ $$4 \cdot {\left( {\frac{2}{3}} \right)^{2x}} - {\left( {\frac{2}{3}} \right)^x} - 18 = 0$$ $${\left( {\frac{2}{3}} \right)^x} = m$$ $$4{m^2} - m - 18 = 0$$ $$D = 1 + 4 \cdot 4 \cdot 18 = 289 = {17^2}$$ $${m_{1,2}} = \frac{{1 \pm 17}}{{4 \cdot 2}} = \left\langle \begin{array}{l}\frac{{18}}{8} = \frac{9}{4}\\ - 2\end{array} \right.$$ $${\left( {\frac{2}{3}} \right)^x} = \frac{9}{4};$$ $$x = - 2$$

№ 8 Теңдеуді шешіңіз: $$3 \cdot {16^x} + {36^x} = 2 \cdot {81^x}$$

Шешуі: $$3 \cdot {4^{2x}} + {4^x} \cdot {9^x} - 2 \cdot {9^{2x}} = 0,\quad \left| { \div \left( {{9^{2x}}} \right)} \right.$$ $$3 \cdot \frac{{{4^{2x}}}}{{{9^{2x}}}} + \frac{{{4^x} \cdot {9^x}}}{{{9^{2x}}}} - \frac{{2 \cdot {9^{2x}}}}{{{9^{2x}}}} = 0$$ $$3 \cdot {\left( {\frac{4}{9}} \right)^{2x}} + {\left( {\frac{4}{9}} \right)^x} - 2 = 0$$ $${\left( {\frac{4}{9}} \right)^x} = m$$ $$3{m^2} + m - 2 = 0$$ $$D = 1 + 4 \cdot 3 \cdot 2 = 25 = {5^2}$$ $${m_{1,2}} = \frac{{ - 1 \pm 5}}{{3 \cdot 2}} = \left\langle \begin{array}{l} - 1\\\frac{2}{3}\end{array} \right.$$ $${\left( {\frac{4}{9}} \right)^x} = \frac{2}{3};\quad 2x = 1;\quad x = \frac{1}{2}$$

№ 9 Теңдеуді шешіңіз: $${2^{2x + 1}} + {3^{2x + 1}} = 5 \cdot {6^x}$$

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

№ 10 Теңдеуді шешіңіз: $${4^x} + {6^x} = 2 \cdot {9^x}$$

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

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