Question

\(\sqrt{49x^2-28x+4}-3=x\)

Answer 1

\(\Leftrightarrow\left|7x-2\right|=x+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-3\\\left(7x-2-x-3\right)\left(7x-2+x+3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-3\\\left(6x-5\right)\left(8x+1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\left(nhận\right)\\x=-\dfrac{1}{8}\left(nhận\right)\end{matrix}\right.\)

Answer 2

⇔7x-2-3=x

⇔7x-5=x

⇔6x=5

⇔x=5/6

Answer 3

\(\sqrt{49x^2-28x+4}-3=x\\ \Leftrightarrow\sqrt{49x^2-28x+4}=x+3\)

ĐKXĐ:\(\left\{{}\begin{matrix}49x^2-28x+4\ge0\left(luôn.dúng\right)\\x+3\ge0\end{matrix}\right.\Leftrightarrow x\ge-3\)

 

\(\sqrt{49x^2-28x+4}=x+3\\ \Leftrightarrow49x^2-28x+4=x^2+6x+9\\ \Leftrightarrow48x^2-34x-5=0\\ \Leftrightarrow\left(48x^2+6x\right)-\left(40x+5\right)=0\\ \Leftrightarrow6x\left(8x+1\right)-5\left(8x+1\right)=0\\ \Leftrightarrow\left(8x+1\right)\left(6x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{8}\left(tm\right)\\x=\dfrac{5}{6}\left(tm\right)\end{matrix}\right.\)