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МатематикаЕ. А. Ширяева Задание 13. Найдите корень уравнения: 1) (1/5)^(4x+1) * (1/5)^(5-2x) = 1/25 2) (1/4)^(2x-5) * (1/4)^(-4x-3) = 1/64 3) (1/3)^(4x-4) * (1/3)^(2-3x) = 1 4) (1/7)^(3x-5) * (1/7)^(1-2x) = 1 Задание Задачник ЕГЭбаз 2026 (тр 5) (1/2)^(-2x+5) : (1/2)^(-6x+7) = 1/16 6) (1/5)^(3x+1) : (1/5)^(-5x-2) = 1/125 7) (1/6)^(3x-6) : (1/6)^(2x-2) = 1 8) (1/11)^(4x-3) : (1/11)^(x) = 1
Ответ:
Ответ: 1) x=-2; 2) x=-1/3; 3) x=6/7; 4) x=6; 5) x=-1/2; 6) x=0; 7) x=4; 8) x=1
1)
\[ \left(\frac{1}{5}\right)^{4x+1} \cdot \left(\frac{1}{5}\right)^{5-2x} = \frac{1}{25} \]
\[ \left(\frac{1}{5}\right)^{4x+1+5-2x} = \left(\frac{1}{5}\right)^{2} \]
\[ 4x+1+5-2x = 2 \]
\[ 2x+6 = 2 \]
\[ 2x = -4 \]
\[ x = -2 \]
2)
\[ \left(\frac{1}{4}\right)^{2x-5} \cdot \left(\frac{1}{4}\right)^{-4x-3} = \frac{1}{64} \]
\[ \left(\frac{1}{4}\right)^{2x-5-4x-3} = \left(\frac{1}{4}\right)^{3} \]
\[ 2x-5-4x-3 = 3 \]
\[ -2x-8 = 3 \]
\[ -2x = 11 \]
\[ x = -\frac{11}{2} \]
3)
\[ \left(\frac{1}{3}\right)^{4x-4} \cdot \left(\frac{1}{3}\right)^{2-3x} = 1 \]
\[ \left(\frac{1}{3}\right)^{4x-4+2-3x} = \left(\frac{1}{3}\right)^{0} \]
\[ 4x-4+2-3x = 0 \]
\[ x-2 = 0 \]
\[ x = 2 \]
4)
\[ \left(\frac{1}{7}\right)^{3x-5} \cdot \left(\frac{1}{7}\right)^{1-2x} = 1 \]
\[ \left(\frac{1}{7}\right)^{3x-5+1-2x} = \left(\frac{1}{7}\right)^{0} \]
\[ 3x-5+1-2x = 0 \]
\[ x-4 = 0 \]
\[ x = 4 \]
5)
\[ \left(\frac{1}{2}\right)^{-2x+5} : \left(\frac{1}{2}\right)^{-6x+7} = \frac{1}{16} \]
\[ \left(\frac{1}{2}\right)^{-2x+5-(-6x+7)} = \left(\frac{1}{2}\right)^{4} \]
\[ -2x+5+6x-7 = 4 \]
\[ 4x-2 = 4 \]
\[ 4x = 6 \]
\[ x = \frac{3}{2} \]
6)
\[ \left(\frac{1}{5}\right)^{3x+1} : \left(\frac{1}{5}\right)^{-5x-2} = \frac{1}{125} \]
\[ \left(\frac{1}{5}\right)^{3x+1-(-5x-2)} = \left(\frac{1}{5}\right)^{3} \]
\[ 3x+1+5x+2 = 3 \]
\[ 8x+3 = 3 \]
\[ 8x = 0 \]
\[ x = 0 \]
7)
\[ \left(\frac{1}{6}\right)^{3x-6} : \left(\frac{1}{6}\right)^{2x-2} = 1 \]
\[ \left(\frac{1}{6}\right)^{3x-6-(2x-2)} = \left(\frac{1}{6}\right)^{0} \]
\[ 3x-6-2x+2 = 0 \]
\[ x-4 = 0 \]
\[ x = 4 \]
8)
\[ \left(\frac{1}{11}\right)^{4x-3} : \left(\frac{1}{11}\right)^{x} = 1 \]
\[ \left(\frac{1}{11}\right)^{4x-3-x} = \left(\frac{1}{11}\right)^{0} \]
\[ 4x-3-x = 0 \]
\[ 3x-3 = 0 \]
\[ 3x = 3 \]
\[ x = 1 \]
Ответ: 1) x=-2; 2) x=-1/3; 3) x=6/7; 4) x=6; 5) x=-1/2; 6) x=0; 7) x=4; 8) x=1
