Em ơi anh chưa thấy đề
ĐKXĐ:\(x\ge\dfrac{1}{5}\)
Đặt \(\sqrt{5x-1}=a,\sqrt{x+2}=b\left(a,b\ge0\right)\)
Ta có hệ:
\(\left\{{}\begin{matrix}a-b=\dfrac{4x-3}{5}\\a^2-b^2=4x-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a-b=\dfrac{4x-3}{5}\\\left(a-b\right)\left(a+b\right)=4x-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=\dfrac{4x-3}{5}\\\dfrac{4x-3}{5}.\left(a+b\right)=4x-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=\dfrac{4x-3}{5}\\a+b=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\a-b+a+b=\dfrac{4x-3}{5}+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\2a=\dfrac{4x+22}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4x+22}{10}+b=5\\a=\dfrac{4x+22}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{28-4x}{10}\\a=\dfrac{4x+22}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+2}=\dfrac{28-4x}{10}\\\sqrt{5x-1}=\dfrac{4x+22}{10}\end{matrix}\right.\)
\(\Leftrightarrow...\)
