K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

a) Ta có: \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)

\(=\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(1-\sqrt{5}\right)}+\frac{8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)

\(=\frac{10-10\sqrt{5}+2\sqrt{10}-10\sqrt{2}+8\sqrt{5}+8\sqrt{2}}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)

\(=\frac{10-2\sqrt{5}+2\sqrt{10}-2\sqrt{2}}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)

\(=\frac{2\sqrt{5}\left(\sqrt{5}-1\right)+2\sqrt{2}\left(\sqrt{5}-1\right)}{-\left(\sqrt{5}-1\right)\left(\sqrt{5}+\sqrt{2}\right)}\)

\(=\frac{2\cdot\left(\sqrt{5}-1\right)\left(\sqrt{5}+\sqrt{2}\right)}{-\left(\sqrt{5}-1\right)\left(\sqrt{5}+\sqrt{2}\right)}\)

\(=\frac{2}{-1}=-2\)

b) Ta có: \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(=\frac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

\(=\frac{-2}{\sqrt{6}}-\frac{1}{\sqrt{6}}\)

\(=-\frac{3}{\sqrt{6}}=\frac{-\sqrt{3}}{\sqrt{2}}\)

c) Ta có: \(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\)

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

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

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

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

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

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

\(=4\)

d) Ta có: \(\frac{\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)

\(=\frac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)

\(=\frac{\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\cdot\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)

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

\(=\frac{\left|\sqrt{5}-1\right|\cdot\left(5+2\cdot\sqrt{5}\cdot1+1\right)}{2\cdot\left(\sqrt{5}+1\right)\cdot2}\)

\(=\frac{\left(\sqrt{5}-1\right)\cdot\left(\sqrt{5}+1\right)^2}{2\cdot\left(\sqrt{5}+1\right)\cdot2}\)(Vì \(\sqrt{5}>1\))

\(=\frac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}{4}\)

\(=\frac{5-1}{4}=\frac{4}{4}=1\)

e) Ta có: \(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

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

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

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

\(=\frac{2-\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}{-\sqrt{6}}+\frac{2+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{6}}\)

\(=\frac{-2+\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{6}}+\frac{2+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{6}}\)

\(=\frac{\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|}{\sqrt{6}}\)

\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{6}}\)

\(=\frac{2\sqrt{3}}{\sqrt{6}}=\frac{\sqrt{12}}{\sqrt{6}}=\sqrt{2}\)

f) Ta có: \(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)

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

\(=\frac{9-4\sqrt{5}}{2\cdot\left(\sqrt{5}-2\right)}\)

\(=\frac{5-2\cdot\sqrt{5}\cdot2+2}{2\cdot\left(\sqrt{5}-2\right)}\)

\(=\frac{\left(\sqrt{5}-2\right)^2}{2\left(\sqrt{5}-2\right)}\)

\(=\frac{\sqrt{5}-2}{2}\)

21 tháng 7 2019

a) \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)

\(\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(1-\sqrt{5}\right)}+\frac{8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)

\(\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)+8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)

\(\frac{10-2\sqrt{5}+2\sqrt{10}-2\sqrt{2}}{\sqrt{5}+\sqrt{2}-5-\sqrt{10}}\)

\(\frac{2\left(5-\sqrt{5}+\sqrt{10}-\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}-5-\sqrt{10}}\)

= -2

b); c); d) làm tương tự

NV
18 tháng 6 2019

a/ \(\frac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8\left(1+\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)

b/ \(\frac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\frac{-2}{\sqrt{6}}-\frac{1}{\sqrt{6}}=\frac{-3}{\sqrt{6}}=-\frac{\sqrt{6}}{2}\)

c/ \(\frac{\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}+\frac{\sqrt{\left(2+\sqrt{3}\right)^2}}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}=2-\sqrt{3}+2+\sqrt{3}=4\)

d/ \(\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)

\(=\frac{\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\frac{\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)}{8}=\frac{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{4}=1\)

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

\(=\frac{\sqrt{2}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}=\frac{\sqrt{2}}{3+\sqrt{3}}+\frac{\sqrt{2}}{3-\sqrt{3}}=\frac{\sqrt{2}\left(3-\sqrt{3}+3+\sqrt{3}\right)}{6}=\sqrt{2}\)

f/ \(\frac{9+4\sqrt{5}-8\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{9-4\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{\left(\sqrt{5}-2\right)^2}{2\left(\sqrt{5}-2\right)}=\frac{\sqrt{5}-2}{2}\)

a: \(=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{-\sqrt{6}}{3}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{2}\)

b: \(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)

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

d: \(=-\left(\sqrt{5}+\sqrt{2}\right)\cdot\left(\sqrt{5}-\sqrt{2}\right)=-3\)

19 tháng 6 2017

Phần d mình sửa lại đề nha : \(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{4}-4}\)

11 tháng 5 2018

bn xem lại đề câu d đi sao mẫu lại bằng 0 rồi

Bài 2: Thực hiện phép tínha) \(\sqrt{5}-\sqrt{48}+5\sqrt{27}-\sqrt{45}\)b) \(\left(\sqrt{5}+\sqrt{2}\right)\left(3\sqrt{2}-1\right)\)c) \(3\sqrt{50}-2\sqrt{75}-4\frac{\sqrt{54}}{\sqrt{3}}-3\sqrt{\frac{1}{3}}\)d) \(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)e) \(\sqrt{48-2\sqrt{135}}-\sqrt{45}+\sqrt{18}\)f) \(\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\frac{6}{2-\sqrt{10}}-\frac{20}{\sqrt{10}}\)Bài 3: Thực hiện phép...
Đọc tiếp

Bài 2: Thực hiện phép tính

a) \(\sqrt{5}-\sqrt{48}+5\sqrt{27}-\sqrt{45}\)

b) \(\left(\sqrt{5}+\sqrt{2}\right)\left(3\sqrt{2}-1\right)\)

c) \(3\sqrt{50}-2\sqrt{75}-4\frac{\sqrt{54}}{\sqrt{3}}-3\sqrt{\frac{1}{3}}\)

d) \(\sqrt{\left(\sqrt{3}-3\right)^2}+\sqrt{4-2\sqrt{3}}\)

e) \(\sqrt{48-2\sqrt{135}}-\sqrt{45}+\sqrt{18}\)

f) \(\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\frac{6}{2-\sqrt{10}}-\frac{20}{\sqrt{10}}\)

Bài 3: Thực hiện phép tính

a) \(\sqrt{9-4\sqrt{5}}\)

b) \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)

c) \(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)

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

e) \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)

f*) \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

Bài 4: Rút gọn

a) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}\)

b) \(\left(2\sqrt{3}+\sqrt{4}\right)\left(\sqrt{3}-2\right)\)

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

d) \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}+\sqrt{6}\)

e) \(\left(\frac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\frac{4}{1+\sqrt{5}}+4\right)\)

f) \(\frac{1}{5}\sqrt{50}-2\sqrt{96}-\frac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\frac{1}{6}}\)

0
18 tháng 8 2016

a/\(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=\frac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=2\sqrt{5}+\frac{8}{1-\sqrt{5}}\)

\(=\frac{2\sqrt{5}-10+8}{1-\sqrt{5}}=\frac{-2\left(1-\sqrt{5}\right)}{1-\sqrt{5}}=-2\)

b/Đề sai

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

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

d/ \(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}=\frac{9+4\sqrt{5}-8\sqrt{5}}{2\sqrt{5}-4}=\frac{9-4\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{\left(\sqrt{5}-2\right)^2}{2\left(\sqrt{5}-2\right)}=\frac{\sqrt{5}-2}{2}\)

7 tháng 9 2019

22) \(\frac{1}{\sqrt{5}+\sqrt{2}}+\frac{1}{\sqrt{5}-\sqrt{2}}\)

\(=\frac{\left(\sqrt{5}-\sqrt{2}\right)+\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}\)

\(=\frac{2\sqrt{5}}{\sqrt{5^2}-\sqrt{2^2}}\)

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