Question

[Nguyễn Diệu Hoa]

Tính:

a) \(\frac{2}{3+2\sqrt{2}}-\frac{7}{1-2\sqrt{2}}+\frac{4}{\sqrt{5}-1}+\sqrt{8}-2\)

b) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{5}}\)

c)  \(\sqrt{4-2\sqrt{3}}+2\sqrt{3}\)

d) \(A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)

e) \(B=\sqrt{\frac{2}{2+\sqrt{3}}}\)

Các bạn giải đc câu nào thì giải dùm mk nha!!! Thanks :))))

Answer 1

[nguyễn văn thành]

\(a,\frac{2}{3+2\sqrt{2}}-\frac{7}{1-2\sqrt{2}}+\frac{4}{\sqrt{5}-1}+\sqrt{8}-2\)

\(=\frac{2.\left(3-2\sqrt{2}\right)}{9-8}-\frac{7.\left(1+2\sqrt{2}\right)}{1-8}+\frac{4.\left(\sqrt{5}+1\right)}{5-1}+2\sqrt{2}-2\)

\(=6-4\sqrt{2}-\frac{7.\left(1+2\sqrt{2}\right)}{-7}+\frac{4.\left(\sqrt{5}+1\right)}{4}+2\sqrt{2}-2\)

\(=6-4\sqrt{2}+1+2\sqrt{2}+\sqrt{5}+1+2\sqrt{2}-2\)

\(=6+\sqrt{5}\)

\(b,\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{5}}\)

\(=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{5}}{4-5}\)

\(=\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{5}}{-1}\)

\(=-1+\sqrt{2}+\sqrt{3}-\sqrt{2}-2+\sqrt{5}\)

\(=-3+\sqrt{3}+\sqrt{5}\)

\(c,\sqrt{4-2\sqrt{3}}+2\sqrt{3}\)

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

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

\(=-1+3\sqrt{3}\)

\(d,A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)

\(=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}+\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)

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

\(=\frac{\sqrt{3}-1}{\sqrt{2}}+\frac{\sqrt{3}+1}{\sqrt{2}}\)

\(=\frac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}\)

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

\(=\sqrt{6}\)

\(e,B=\sqrt{\frac{2}{2+\sqrt{3}}}\)

Ta có \(\frac{2}{2+\sqrt{3}}=\frac{2.\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}\)

Thay lại ta được \(\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

.... Đúng thì ủng hộ nha ....
 Kết bạn với mình ... ;) ;)