\(\left(\sqrt{x}+\sqrt{y}\right)\left(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\right)\)

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

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

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

\(=\frac{xy-y^2}{y}\)

\(=\frac{y\left(x-y\right)}{y}\)

= x - y (đpcm)

\(A=\frac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(A=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(A=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(A=\frac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(A=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(A=\frac{4}{x-1}\)

b) \(\frac{4}{x-1}=7\)

\(\Leftrightarrow4=7.\left(x-1\right)\)

\(\Leftrightarrow\frac{4}{7}=x-1\)

\(\Leftrightarrow\frac{4}{7}+1=x\)

\(\Leftrightarrow\frac{11}{7}=x\)

\(\Rightarrow x=\frac{11}{7}\)

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

\(\left(\sqrt{x}-1\right)^2\ge0\Leftrightarrow-\left(\sqrt{x}-1\right)^2\le0\)

\(\Leftrightarrow-\left(\sqrt{x}-1\right)^2-2\le-2\)

\(\Leftrightarrow\frac{1}{-\left(\sqrt{x}-1\right)^2-2}\ge\frac{1}{-2}=\frac{-1}{2}\)

\("="\Leftrightarrow x=1\)

Vậy biểu thức B đạt giá trị nhỏ nhất là -1/2 khi x=1

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 \(S=\left\{2;6\right\}\)