đk:x\(\ge2\)
\(pt\Leftrightarrow\frac{x+1-x+2}{\sqrt{x+1}+\sqrt{x-2}}.\left(1+\sqrt{\left(x+1\right)\left(x-2\right)}\right)=3\)
\(\Leftrightarrow\frac{3}{\sqrt{x+1}+\sqrt{x-2}}\left(1+\sqrt{\left(x+1\right)\left(x-2\right)}\right)=3\)
\(\Leftrightarrow\frac{1+\sqrt{\left(1+x\right)\left(x-2\right)}}{\sqrt{x+1}+\sqrt{x-2}}=1\Leftrightarrow1+\sqrt{\left(1+x\right)\left(x-2\right)}=\sqrt{x+1}+\sqrt{x-2}\)
\(\Leftrightarrow1+\left(x+1\right)\left(x-2\right)+2\sqrt{\left(x+1\right)\left(x-2\right)}=x+1+x-2+2\sqrt{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow1+x^2-x-2=2x-1\Leftrightarrow x^2-3x=0\Leftrightarrow\left[\begin{array}{nghiempt}x=0\left(loai\right)\\x=3\left(tm\right)\end{array}\right.\)
vậy pt có no x=3
