Question

giải phương trình: \(\sqrt{x}+\sqrt{x+7}+2\sqrt{x^2+7x}=35-2x\)

Answer 1

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow2x+7+2\sqrt{x^2+7x}+\sqrt{x}+\sqrt{x+7}-42=0\)

Đặt \(\sqrt{x}+\sqrt{x+7}=t>0\)

\(\Rightarrow2x+7+2\sqrt{x^2+7x}=t^2\)

Pt trở thành:

\(t^2+t-42=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-7\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\sqrt{x+7}=6\)

\(\Leftrightarrow2x+7+2\sqrt{x^2+7x}=36\)

\(\Leftrightarrow2\sqrt{x^2+7x}=29-2x\) (\(x\le\frac{29}{2}\))

\(\Leftrightarrow4\left(x^2+7x\right)=\left(29-2x\right)^2\)

\(\Leftrightarrow144x-841=0\Rightarrow x=\frac{841}{144}\)