\(\sqrt{x+3}+\sqrt{x-1}=2\) (ĐK: \(x>0\))
\(\Leftrightarrow\left(\sqrt{x+3}+\sqrt{x-1}\right)^2=4\)
\(\Leftrightarrow x+3+x-1+2\sqrt{\left(x+3\right)\left(x-1\right)}=4\)
\(\Leftrightarrow2x+2\sqrt{\left(x+3\right)\left(x-1\right)}=2\)
\(\Leftrightarrow2\left(x+\sqrt{\left(x+3\right)\left(x-1\right)}\right)=2\)
\(\Leftrightarrow x+\sqrt{\left(x+3\right)\left(x-1\right)}=1\)
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x-1\right)}=1-x\) (ĐK: \(x\le1\))
\(\Leftrightarrow\left(\sqrt{\left(x+3\right)\left(x-1\right)}\right)^2=\left(1-x\right)^2\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=1-2x+x^2\)
\(\Leftrightarrow x^2+2x-3=1-2x+x^2\)
\(\Leftrightarrow x^2+2x-3-1+2x-x^2=0\)
\(\Leftrightarrow4x-4=0\)
\(\Leftrightarrow4\left(x-1\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\left(TM\right)\)
Vậy PT có nghiệm là \(x=1\)
