a) \(\sqrt{x+5}=2x-1\)
Đk: \(x\ge\dfrac{1}{2}\)
Bình phương 2 vế :
\(x+5=4x^2-4x+1\)
\(4x^2-5x-4=0\)
\(\Delta=89\)
=> Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x_1=\dfrac{5-\sqrt{89}}{8}\left(KTM\right)\\x_2=\dfrac{5+\sqrt{89}}{8}\left(TM\right)\end{matrix}\right.\)
Vậy ....
b) \(x+\sqrt{x-5}+7=0\) . ĐK : \(x\ge5\)
Vì \(x\ge5\Rightarrow x+\sqrt{x-5}+7\ge12\)
=> Phương trình vô nghiệm.
a. \(\sqrt{x+5}=2x-1\left(x\ge\dfrac{1}{2}\right)\)
\(\Leftrightarrow x+5=4x^2-4x+1\)
\(\Leftrightarrow4x^2-4x+1-x-5=0\)
\(\Leftrightarrow4x^2-5x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{89}}{8}\left(n\right)\\x=\dfrac{5-\sqrt{89}}{8}\left(l\right)\end{matrix}\right.\)
b. \(x+\sqrt{x-5}+7=0\)
\(\Leftrightarrow\sqrt{x-5}=-7-x\left(ĐK:x\ge5\right)\)
\(\Leftrightarrow x-5=-x^2-14x+49\)
\(\Leftrightarrow x^2+15x-54=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(l\right)\\x=-18\left(l\right)\end{matrix}\right.\)
