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= \(2^2-2.2.\left(\sqrt{5}\right)^2+\left(\sqrt{5}\right)^2\)
\(=\left(2-\sqrt{5}\right)^2\)
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14) \(\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{2^2-2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{2-\sqrt{3}}\right)\cdot\left(\sqrt{2-\sqrt{3}}\right)\)
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i) \(=x\left(x-5\right)+3\left(x-5\right)=\left(x-5\right)\left(x+3\right)\)
j) \(=x\left(x-1\right)+8\left(x-1\right)=\left(x-1\right)\left(x+8\right)\)
k) \(=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
l) \(=x\left(x-3\right)-2\left(x-3\right)=\left(x-3\right)\left(x-2\right)\)
m) \(=x\left(x^2-4x+4\right)=x\left(x-2\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-1-y\right)\left(x-1+y\right)\)
\(xy-y\sqrt{x}+\sqrt{x}-1\\ =\left(xy-y\sqrt{x}\right)+\left(\sqrt{x}-1\right)\\ =\sqrt{x}y\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)\\ =\sqrt{x}y\left(\sqrt{x}-1\right)\)
sai rồi