Câu hỏi của buithehagiang

Question

Bài 1: Tìm x

a) $\left(\frac{1}{8}\right)^x.27^{2.x}=\left(-9\right)^4$

b) $5^x\left(5^3\right)^2=625$

c)$\left(4x-1\right)^{30}=\left(4x-1\right)^{20}$

Trần Thanh Phương · Accepted Answer

a) $\left(\frac{1}{81}\right)^x\cdot27^{2x}=\left(-9\right)^4$

$\Leftrightarrow\frac{1}{3^{4x}}\cdot3^{6x}=9^4$

$\Leftrightarrow\frac{3^{6x}}{3^{4x}}=3^8$

$\Leftrightarrow3^{2x}=3^8$

$\Leftrightarrow2x=8$

$\Leftrightarrow x=4$

b) $5^x\cdot\left(5^3\right)^2=625$

$\Leftrightarrow5^{x+6}=5^4$

$\Leftrightarrow x+6=4$

$\Leftrightarrow x=-2$

c) $\left(4x-1\right)^{30}=\left(4x-1\right)^{20}$

$\Leftrightarrow\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0$

$\Leftrightarrow\left(4x-1\right)^{20}\cdot\left[\left(4x-1\right)^{10}-1\right]=0$

$\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\left(4x-1\right)^{10}=1=\left(\pm1\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\x=\frac{1}{2}\x=0\end{matrix}\right.$

Vậy....Câu trả lời: a) $\left(\frac{1}{81}\right)^x\cdot27^{2x}=\left(-9\right)^4$