Question

BT1: Tính

A = \(\left(\frac{3}{\sqrt{2}+1}+\frac{14}{2\sqrt{2}-1}-\frac{4}{2-\sqrt{2}}\right)\)\(\left(\sqrt{8}+2\right)\)

B = \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

C= \(\sqrt{3+2\sqrt{2}}\)+ \(\sqrt{6-4\sqrt{2}}\)

 

Answer 1

b) \(B=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2+2}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

Answer 2

c) \(C=\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)

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

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

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

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

Answer 3

C = \(\sqrt{3+2\sqrt{3}}\)+\(\sqrt{6-4\sqrt{2}}\)thì làm sao ạ

Answer 4

Dùng hằng đẳng thức \(\left(a\pm b\right)^2=a^2\pm2ab+b^2\)

\(C=\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)

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

\(=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}+\sqrt{2^2-2.2.\sqrt{2}+\left(\sqrt{2}\right)^2}\)

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

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

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

Answer 5

2 căn 3 chứ không phải 2 căn 2