Bài 13:
ĐKXĐ: \(x\ge-\frac32\)
Ta có: \(x^2+4x+5=2\sqrt{2x+3}\)
=>\(2x+3-2\sqrt{2x+3}+1+x^2+2x+1=0\)
=>\(\left(\sqrt{2x+3}-1\right)^2+\left(x+1\right)^2=0\)
=>\(\begin{cases}2x+3=1\\ x+1=0\end{cases}\Rightarrow x=-1\) (nhận)
Bài 14: ĐKXĐ: x>=-1/4
\(2x^2+2x+1=\sqrt{4x+1}\)
=>\(4x^2+4x+2=2\sqrt{4x+1}\)
=>\(4x+1-2\sqrt{4x+1}+1+x^2=0\)
=>\(\left(\sqrt{4x+1}-1\right)^2+x^2=0\)
=>\(\begin{cases}\sqrt{4x+1}-1=0\\ x=0\end{cases}\Rightarrow x=0\) (nhận)
Bài 15:
a:
ĐKXĐ: x>=-1/2
\(x^2-6x+26=6\sqrt{2x+1}\)
=>\(x^2-8x+16+2x+10-6\sqrt{2x+1}=0\)
=>\(\left(x-4\right)^2+\left(\sqrt{2x+1}-3\right)^2=0\)
=>\(\begin{cases}x-4=0\\ \sqrt{2x+1}-3=0\end{cases}\Rightarrow\begin{cases}x=4\\ \sqrt{2x+1}=3\end{cases}\)
=>x=4(nhận)
b: ĐKXĐ: x>=2; y>=3; z>=5
Ta có: \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)
=>\(x-2-2\cdot\sqrt{x-2}\cdot1+1+y-3-2\cdot\sqrt{y-3}\cdot2+4+z-5-2\cdot\sqrt{z-5}\cdot3+9=0\)
=>\(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
=>\(\begin{cases}x-2=1\\ y-3=4\\ z-5=9\end{cases}\Rightarrow\begin{cases}x=3\\ y=7\\ z=14\end{cases}\) (nhận)
