b)Ta có:
\(A=\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(=\dfrac{2^{10}.3^8\left(-2\right)}{2^{10}.3^8.2.3}=-\dfrac{2^{11}.3^8}{2^{11}.3^9}=-\dfrac{1}{3}\)
Vậy...
Câu 2: a) Ta có:
\(\left|x-\dfrac{1}{9}\right|-\dfrac{9}{5}=0\Leftrightarrow\left|x-\dfrac{1}{9}\right|=\dfrac{9}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{9}=\dfrac{9}{5}\\x-\dfrac{1}{9}=-\dfrac{9}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{86}{45}\\x=\dfrac{-76}{45}\end{matrix}\right.\)
b)Ta có:
\(\left(3x-2\right)^4=81\Leftrightarrow\left[{}\begin{matrix}3x-2=3\\3x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)
