a) ĐK: \(x\ge-2\)
\(\sqrt{x^2+2x+1}=x+2\Leftrightarrow\sqrt{\left(x+1\right)^2}=x+2\Leftrightarrow\left|x+1\right|=x+2\)(*)
_ Nếu \(x\ge-1\) thì (*)\(\Leftrightarrow x+1=x+2\Leftrightarrow1=2\left(ktm\right)\)
_ Nếu \(-2\le x\le-1\) thì (*)\(\Leftrightarrow-x-1=x+2\Leftrightarrow-2x=3\Leftrightarrow x=-\dfrac{3}{2}\left(tm\right)\)
Vậy S={\(-\dfrac{3}{2}\)}
b) ĐK: \(x\ge-\dfrac{3}{2}\)
\(\sqrt{x^2+4}=\sqrt{2x+3}\Leftrightarrow\left(\sqrt{x^2+4}\right)^2=\left(\sqrt{2x+3}\right)^2\Leftrightarrow x^2+4=2x+3\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\left(tm\right)\)
Vậy S={1}
a,\(\sqrt{x^2+2x+1}=x+2\Leftrightarrow\sqrt{\left(x+1\right)^2}=x+2\)
\(\Leftrightarrow\left|x+1\right|=x+2\)
Nếu x \(\ge-1\)=> \(x+1=x+2\)
\(\Leftrightarrow1=2\)(vô lý)
Nếu x<-1=>x+1=-2-x
\(\Leftrightarrow x=\dfrac{-3}{2}\)
b,\(ĐK:\sqrt{2x+3}\ge0
\Leftrightarrow x\ge\dfrac{-3}{2}\)
Bình phương 2 vế
\(\Leftrightarrow x^2+4=2x+3\)
\(\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\)
\(x-1=0\Leftrightarrow x=1\)
