Question

Giải phương trình:

\(a,x^2-7x+\sqrt{x^2-7x+8}=12\)

b, \(x^2+4x+5=2\sqrt{2x+3}\)

Answer 1

a/ĐKXĐ: ...

\(\Leftrightarrow x^2-7x+8+\sqrt{x^2-7x+8}-20=0\)

Đặt \(\sqrt{x^2-7x+8}=a\ge0\)

\(\Rightarrow a^2+a-20=0\) \(\Rightarrow\left[{}\begin{matrix}a=4\\a=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-7x+8}=4\)

\(\Leftrightarrow x^2-7x-8=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=8\end{matrix}\right.\)

b/ ĐKXĐ: ...

\(\Leftrightarrow x^2+2x+1+2x+3-2\sqrt{2x+3}+1=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\\sqrt{2x+3}-1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)