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\(x+2\sqrt{x-1}=\left(x-1\right)+2\sqrt{x-1}+1=\left(\sqrt{x-1}+1\right)^2\)
\(x-4\sqrt{x-2}+2=\left(x-2\right)-4\sqrt{x-2}+4=\left(\sqrt{x-2}-2\right)^2\)
\(x+2\sqrt{x-1}=\left(\sqrt{x-1}+1\right)^2\)
\(x-4\sqrt{x-2}+2=\left(\sqrt{x-2}+4\right)^2\)
Ta có : \(M=7\sqrt{x-1}-\sqrt{x^3-x^2}+x-1\)
\(=7\sqrt{x-1}-\sqrt{x^2\left(x-1\right)}+x-1\)
\(=7\sqrt{x-1}-x\sqrt{x-1}+\left(\sqrt{x-1}\right)^2\)
\(=\sqrt{x-1}\left(7-x+\sqrt{x-1}\right)\)
\(=\sqrt{x-1}\left(\sqrt{x-1}+2\right)\left(\sqrt{x-1}-3\right)\)
\(x+7\sqrt{x}+10=\left(\sqrt{x}+2\right)\left(\sqrt{x}+5\right)\)
d: \(=-\left(x+\sqrt{x}-12\right)=-\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)\)
\(x-\sqrt{x}-6=\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\)
\(2x+5\sqrt{x}-3=\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\)
\(2+\sqrt{3}+\sqrt{6}+\sqrt{8}=2+\sqrt{3}+\sqrt{6}+2\sqrt{2}\)
\(=2+\sqrt{3}+\sqrt{2}\left(2+\sqrt{3}\right)=\left(2+\sqrt{3}\right)\left(\sqrt{2}+1\right)\)
\(2+\sqrt{3}+\sqrt{6}+\sqrt{8}=\left(\sqrt{2}+1\right)\left(2+\sqrt{3}\right)\)
\(M=7\sqrt{x-1}-\sqrt{x^2\left(x-1\right)}+\left(\sqrt{x-1}\right)^2=\sqrt{x-1}\left(7-x+\sqrt{x-1}\right)\)
\(=\sqrt{x-1}\left(6-\left(x-1\right)+\sqrt{x-1}\right)\)( đến đây bạn có thể đặt \(\sqrt{x-1}=t\),t>=0 rồi giải)
\(=-\sqrt{x-1}\left(\sqrt{x-1}-3\right)\left(\sqrt{x-1}+2\right)\)