Ta có: \(\sqrt{25x^2-30x+9}=x+7\)
\(\Leftrightarrow\sqrt{25x^2-15x-15x+9}=x+7\)
\(\Leftrightarrow\sqrt{5x\left(5x-3\right)-3\left(5x-3\right)}=x+7\)
\(\Leftrightarrow\sqrt{\left(5x-3\right)^2}=x+7\)
\(\Leftrightarrow5x-3=x+7\)
\(\Leftrightarrow5x-x=3+7\)
\(\Leftrightarrow4x=10\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(x=\dfrac{5}{2}.\)
\(\sqrt{25x^2-30x+9}=x+7\) (ĐK: \(x\ge-7\))
\(\Leftrightarrow\sqrt{\left(5x-3\right)^2}=x+7\)
\(\Leftrightarrow\left|5x-3\right|=x+7\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3=x+7\\5x-3=-x-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(n\right)\\x=-\dfrac{2}{3}\left(n\right)\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{5}{2};-\dfrac{2}{3}\right\}\)
\(\sqrt{25x^2-30x+9}=x+7\left(x\ge-7\right)\)
\(\Rightarrow\sqrt{\left(5x+3\right)^2}=x+7\)
\(\Rightarrow\left|5x+3\right|=x+7\)
Xét trường hợp \(x\ge-\dfrac{5}{3}\) và \(x< \dfrac{5}{3}\) nha
