Ta có:
\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\Rightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\Rightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2\Rightarrow x-\frac{2}{9}=\frac{4}{9}\)
\(\Rightarrow x=\frac{4}{9}+\frac{2}{9}\Rightarrow x=\frac{6}{9}\Rightarrow x=\frac{2}{3}\)
Đề : \(\left(\frac{x-2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)
Giải \(\frac{\left(x-2\right)^3}{9^3}=\frac{2^6}{3^6}\Rightarrow\frac{x^3-8-6x\left(x-2\right)}{\left(3^2\right)^3}=\frac{64}{3^6}\Rightarrow\frac{x^3-8-6x\left(x-2\right)}{3^6}=\frac{64}{3^6}\)
\(\Rightarrow x^3-8-6x^2+12x=\frac{64.3^6}{3^6}=64\)
\(\Rightarrow x^3-6x^2+12x-8-64=0\Rightarrow x^3-6x^2+12x-72=0\)
\(\Rightarrow x^2\left(x-6\right)+12\left(x-6\right)=0\Rightarrow\left(x-6\right)\left(x^2+12\right)=0\Rightarrow\orbr{\begin{cases}x-6=0\Rightarrow x=6\\x^2+12=0\Rightarrow x^2=-12\end{cases}}\)
loại trường hợp x^2 =-12 vì bình phương luôn cho ta số dương
Chúc bạn hoc tốt . nha . T I C K cho mình nha cảm ơn
