Question

[Mai Kawakami]

Tính:

1)\(\dfrac{1}{\sqrt{3}-2}\) - \(\dfrac{1}{\sqrt{3}+2}\)

2) \(\dfrac{2}{4-3\sqrt{2}}\) - \(\dfrac{2}{4+3\sqrt{2}}\)

3)\(\sqrt{17-12\sqrt{2}}\) + \(\sqrt{17+12\sqrt{2}}\)

4)\(\sqrt{29-4\sqrt{7}}\)-\(\sqrt{29+4\sqrt{7}}\)

5)\(\sqrt{4+\sqrt{7}}\) - \(\sqrt{4-\sqrt{7}}\)

Answer 1

[Phùng Khánh Linh]

\(1.\dfrac{1}{\sqrt{3}-2}-\dfrac{1}{\sqrt{3}+2}=\dfrac{\sqrt{3}+2+2-\sqrt{3}}{3-4}=-4\)\(2.\dfrac{2}{4-3\sqrt{2}}-\dfrac{2}{4+3\sqrt{2}}=\dfrac{8+6\sqrt{2}+6\sqrt{2}-8}{16-18}=\dfrac{-12\sqrt{2}}{2}-6\sqrt{2}\)\(3.\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}=\sqrt{8-2.2\sqrt{2}.3+9}+\sqrt{8+2.2\sqrt{2}.3+9}=\sqrt{\left(2\sqrt{2}-3\right)^2}+\sqrt{\left(2\sqrt{2}+3\right)^2}=\text{|}2\sqrt{2}-3\text{|}+\text{|}2\sqrt{2}+3\text{|}=4\sqrt{2}\)
\(4.\sqrt{29-4\sqrt{7}}-\sqrt{29+4\sqrt{7}}=\sqrt{28-2.2\sqrt{7}.1+1}-\sqrt{28+2.2\sqrt{7}.1+1}=\sqrt{\left(2\sqrt{7}-1\right)^2}-\sqrt{\left(2\sqrt{7}+1\right)^2}=\text{|}2\sqrt{7}-1\text{|}-\text{|}2\sqrt{7}+1\text{|}=-2\)\(5.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{7+2\sqrt{7}.1+1}-\sqrt{7-2\sqrt{7}.1+1}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{\text{|}\sqrt{7}+1\text{|}-\text{|}\sqrt{7}-1\text{|}}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\dfrac{2\sqrt{2}}{2}\)

Answer 2

[Trần Minh Hoàng]

1)

\(\dfrac{1}{\sqrt{3}-2}-\dfrac{1}{\sqrt{3}+2}\)

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

\(=\dfrac{4}{\left(\sqrt{3}\right)^2-2^2}\)

\(=\dfrac{4}{3-4}=-4\)