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mình học lớp 9 cho tớ hỏi sửa lớp ở đâu

\(E=\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(3-2\sqrt{2}+\frac{4-2\sqrt{3}}{2}\right)\)

\(=\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left[\left(1-\sqrt{2}\right)^2+\frac{\left(\sqrt{3}-1\right)^2}{2}\right]\)

Ta có : \(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(\sqrt{2}-1\right)^2\)

\(=\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(\sqrt{2}-1\right)\)

và \(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\frac{\left(\sqrt{3}-1\right)^2}{2}\)

\(=\frac{1}{2}.\left(\sqrt{2}+1\right)\left(\sqrt{6}+1\right).\left(3-1\right)\left(\sqrt{3}-1\right)\)

\(=\left(\sqrt{2}+1\right)\left(\sqrt{6}+1\right)\left(\sqrt{3}-1\right)\)

\(\Rightarrow E=\left(\sqrt{6}+1\right)\left[\left(\sqrt{2}+1\right)\left(\sqrt{3}-1\right)+\left(\sqrt{2}-1\right)\left(\sqrt{3}+1\right)\right]\)

\(=\left(\sqrt{6}+1\right)\left(2\sqrt{6}-2\right)=2\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)\)

\(=2\left(6-1\right)=10\)

ĐKXĐ: \(x\ne0\)

\(y=\sqrt{\frac{x^4-6x^2+9+12x^2}{x^2}}+\sqrt{x^2+4x+4-8x}\)

\(y=\sqrt{\frac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}\)

\(y=\sqrt{\frac{\left(x^2+3\right)^2}{x^2}}+\sqrt{\left(x-2\right)^2}\)

\(y=\left|\frac{x^2+3}{x}\right|+\left|x-2\right|\)

Ta có bảng xét dấu:

Với \(x< 0,y=\frac{x^2+3}{-x}+2-x=\frac{2x^2-2x+3}{-x}\)

Với \(0< x\le2,y=\frac{x^2+3}{x}+2-x=\frac{2x+3}{x}\)

Với \(x>2,y=\frac{x^2+3}{x}+x-2=\frac{2x^2-2x+3}{x}\)

- Ta thấy ngay, với cả ba trường hợp thì \(y\in Z\Leftrightarrow x\in U\left(3\right)=\left\{-3;-1;1;3\right\}\)