Question

[Nguyễn Châu Mỹ Linh]

Rút gọn biểu thức:

a) \(A=\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

b) \(B=\left(\frac{\sqrt{x}+1}{x-4}-\frac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right).\frac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\) với \(x>0,x\ne4\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a) Ta có: \(A=\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(=\sqrt{4-2\cdot2\cdot\sqrt{3}+3}-\sqrt{4+2\cdot2\cdot\sqrt{3}\cdot3}\)

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

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

\(=2-\sqrt{3}-\left(2+\sqrt{3}\right)\)(Vì \(2>\sqrt{3}>0\))

\(=2-\sqrt{3}-2-\sqrt{3}\)

\(=-2\sqrt{3}\)

b) Ta có: \(B=\left(\frac{\sqrt{x}+1}{x-4}-\frac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right)\cdot\frac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\)

\(=\left(\frac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}-2\right)}\right)\cdot\frac{\left(\sqrt{x}+2\right)\cdot\left(x-4\right)}{\sqrt{x}}\)

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

\(=\frac{x+3\sqrt{x}+2-x+3\sqrt{x}-2}{\sqrt{x}}\)

\(=\frac{6\sqrt{x}}{\sqrt{x}}=6\)

Answer 2

[Nguyễn Thị Ngọc Hân]

a) Bình phương lên ta đc

\(A^2=7-4\sqrt{3}+7+4\sqrt{3}-2\sqrt{7^2-\left(4\sqrt{3}\right)^2}=14-2=12\)

\(\Rightarrow A=\mp\sqrt{12}\)