Giải phương trình
\(\left(x+3\right)\sqrt{\left(4-x\right)\left(12+x\right)}=28-x\)
\(\sqrt{3x^3-5x^2+5x+5x-2}=\frac{x^2}{2}+x-\frac{1}{2}\)
\(\sqrt{x-2}+\sqrt{10-x}=x^2-12x+40\)
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\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\sqrt{3+x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}\)
\(=\frac{\sqrt{3+x}}{\sqrt{3-x}}\)
\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right)\sqrt{x^2-6x+8}}\)
\(=\frac{\left(x-3\right)\left(x-2\right)+3\sqrt{\left(x-4\right)\left(x-2\right)}}{3\left(x-4\right)+\left(x-3\right)\sqrt{\left(x-4\right)\left(x-2\right)}}\)
\(=\frac{\sqrt{x-2}\left(\left(x-3\right)\sqrt{x-2}+3\sqrt{x-4}\right)}{\sqrt{x-4}\left(3\sqrt{x-4}+\left(x-3\right)\sqrt{x-2}\right)}\)
\(=\frac{\sqrt{x-2}}{\sqrt{x-4}}\)
b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)