Ta có:

\(\left(x^2+8x+12\right)\left(x^2-7x+12\right)=16x^2\)

\(\Leftrightarrow\left(x^2-7x+12+15x\right)\left(x^2-7x+12\right)=16x^2\)

Đặt \(x^2-7x+12=a\)

Khi đó phương trình trở thành:

\(\left(a+15x\right)a=16x^2\)

\(\Leftrightarrow a^2+15ax=16x^2\)

\(\Leftrightarrow a^2+15ax-16x^2=0\)

\(\Leftrightarrow a^2-ax+16ax-16x^2=0\)

\(\Leftrightarrow a\left(a-x\right)+16x\left(a-x\right)=0\)

\(\Leftrightarrow\left(a-x\right)\left(a+16x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a-x=0\\a+16x=0\end{cases}}\)

+) Với \(a-x=0\Leftrightarrow x^2-7x+12-x=0\Leftrightarrow x^2-8x+12=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

+) Với \(a+16x=0\Leftrightarrow x^2-7x+12+16x=0\Leftrightarrow x^2+9x+12=0\)(vô nghiệm)

Vậy tập hợp nghiệm của phương trình là \(S=\left\{2;6\right\}\)

\(T=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\left(\frac{\sqrt{x}+1}{\sqrt{x-1}}+\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)

\(\Rightarrow T=\frac{x-1}{\sqrt{x}}\left(\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x-1}\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}\right)\)

\(\Rightarrow T=\frac{x-1}{\sqrt{x}}.\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{x-1}\)

\(\Rightarrow T=\frac{x-1}{\sqrt{x}}.\frac{2x+2}{x-1}\)

\(\Rightarrow T=\frac{2x+2}{\sqrt{x}}\)

\(T=8\Leftrightarrow\frac{2x+2}{\sqrt{x}}=8\)

\(\Leftrightarrow x+1=4\sqrt{x}\)

\(\Leftrightarrow x^2+2x+1=8x\)

\(\Leftrightarrow x^2-6x+1=0\)

\(\Delta=\left(-6\right)^2-4.1.1=36-4=32,\sqrt{\Delta}=\sqrt{32}\)

Vậy pt có 2 nghiệm phân biệt x₁; x₂

\(x_1=\frac{6+\sqrt{32}}{2}=3+\sqrt{8}\);\(x_2=\frac{6-\sqrt{32}}{2}=3-\sqrt{8}\)

\(M=\frac{x+\sqrt{x^2-2x}}{x-\sqrt{x^2-2x}}-\frac{x-\sqrt{x^2-2x}}{x+\sqrt{x^2-2x}}\left(x< 0;x\ge2\right)\)

\(=\frac{\left(x+\sqrt{x^2-2x}\right)\left(x+\sqrt{x^2-2x}\right)}{x^2-\sqrt{x^2-2x}^2}-\frac{\left(x-\sqrt{x^2-2x}\right)\left(x-\sqrt{x^2-2x}\right)}{x^2-\sqrt{x^2-2x}^2}\)

\(=\frac{x^2+x\sqrt{x^2-2x}+x\sqrt{x^2-2x}+x^2-2x}{x^2-x^2-2x}-\frac{x^2-x\sqrt{x^2-2x}-x\sqrt{x^2-2x}+x^2-2x}{x^2-x^2-2x}\)

\(=\frac{2x^2+2x\sqrt{x^2-2x}-2x}{-2x}-\frac{2x^2-2\sqrt{x^2-2x}-2x}{-2x}\)

\(=\frac{2x^2+2x\sqrt{x^2-2x}-2x-2x^2+2x\sqrt{x^2-2x}+2x}{-2x}\)

\(=\frac{4x\sqrt{x^2-2x}}{-2x}=-2x\sqrt{x^2-2x}\)

Chỉ làm thử thôi nhé-.-

\(B=\left(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+2+4\sqrt{x-2}}\right):\sqrt{\frac{4}{x^2}-\frac{4}{x}+1}\left(đk:x\ge2\right)\)

\(=\left(\sqrt{x-2-2\sqrt{x-2}.2+2^2}+\sqrt{x-2+2\sqrt{x-2}.2+2^2}\right):\sqrt{\frac{4}{x^2}-\frac{4x}{x^2}+\frac{x^2}{x^2}}\)

\(=[\left(\sqrt{\left(\sqrt{x-2}-2\right)^2}+\sqrt{\left(\sqrt{x-2}+2\right)^2}\right):\sqrt{\frac{4-4x+x^2}{x^2}}\)

\(=\left(|\sqrt{x-2}-2|+|\sqrt{x-2}+2|\right):\sqrt{\frac{\left(2-x\right)^2}{x^2}}\)

\(=\left(\sqrt{x-2}-2+\sqrt{x-2}+2\right).\frac{x}{2-x}\)

\(=2\sqrt{x-2}.\frac{x}{2-x}=\frac{2x\sqrt{x-2}}{2-x}\)