NA
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
AH
Akai Haruma
Giáo viên
10 tháng 9 2020
Bài 1:
Xét tử số:
\(\sqrt{14+6\sqrt{5}}-\sqrt{14-6\sqrt{5}}=\sqrt{3^2+5+2.3\sqrt{5}}-\sqrt{3^2+5-2.3\sqrt{5}}\)
\(=\sqrt{(3+\sqrt{5})^2}-\sqrt{(3-\sqrt{5})^2}=3+\sqrt{5}-(3-\sqrt{5})=2\sqrt{5}\)
Xét mẫu số:
\(\sqrt{(\sqrt{5}+1)\sqrt{6-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{5+1-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{(\sqrt{5}-1)^2}}\)
\(=\sqrt{(\sqrt{5}+1)(\sqrt{5}-1)}=\sqrt{4}=2\)
Do đó: $A=\frac{2\sqrt{5}}{2}=\sqrt{5}$
S
3 tháng 7 2019
\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(-\sqrt{7}-\sqrt{5}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\sqrt{5}-\sqrt{7}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{7}+\sqrt{5}\right)^2}=\frac{2}{12+2\sqrt{35}}\)
S
3 tháng 7 2019
\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+3\right)}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{8-2\sqrt{15}}{2}+\frac{8+2\sqrt{15}}{2}-\frac{\left(\sqrt{5}+1\right)^2}{4}=8-\frac{6+2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}\)
30 tháng 11 2019
a/ \(A=\sqrt{3^2+2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(A=\sqrt{\left(\sqrt{5}+3\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(A=\sqrt{5}+3+3-\sqrt{5}\text{ vì }\left(3-\sqrt{5}>0\right)\)
\(A=6\)
b/\(B^2=\left(2-\sqrt{3}\right)\left(8+2\sqrt{12}\right)\)
\(B^2=16+4\sqrt{12}-8\sqrt{3}-6\sqrt{36}\)
\(B^2=16-6\sqrt{36}+4\sqrt{12}-8\sqrt{3}\)
\(B^2=4\)
\(\Rightarrow B=2\)
c/ \(C=\frac{2\left(\sqrt{3}+1\right)-\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)'
\(C=\frac{2\sqrt{3}+2-2\sqrt{3}}{2}\)
\(C=1\)
S
31 tháng 8 2019
\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=\frac{32-6-2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}=\frac{14-\sqrt{5}}{2}\) \(\left(\frac{9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2-\left(\frac{9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2=\left(\frac{9-2\sqrt{14}-9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)\left(\frac{9-2\sqrt{14}+9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)=\frac{-72\sqrt{14}}{\sqrt{7}-\sqrt{2}}\)
NV
16 tháng 7 2016
a/ Bạn ghi nhầm đề rồi
c/ \(2\sqrt{18\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{18}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{\sqrt{48}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{4\sqrt{3}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-6\sqrt{5}.\sqrt{\sqrt{3}}\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-4\sqrt{5}\right)\)\(=2\sqrt{2\sqrt{3}}\left(3-2\sqrt{10}\right)\)
f/ \(\sqrt{2}.\sqrt{2+\sqrt{3}}-2\left(\sqrt{3}-1\right)=\sqrt{4+2\sqrt{3}}-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-2\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{3}+1-2\sqrt{3}+2=3-\sqrt{3}=\sqrt{3}\left(\sqrt{3}-1\right)\)
g/ \(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-2\sqrt{3}+2007\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-2\sqrt{3}+2007\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+2007\)
\(=2007\)
13 tháng 8 2020
a) Ta có: \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)
\(=\sqrt{3}\left(2+\sqrt{16}-\sqrt{25}-\sqrt{81}\right)\)
\(=\sqrt{3}\left(2+4-5-9\right)\)
\(=-8\sqrt{3}\)
b) Ta có: \(\left(\frac{\sqrt{7}-\sqrt{14}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}+\sqrt{5}}\)
\(=\left(\frac{\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right)\cdot\left(\sqrt{7}+\sqrt{5}\right)\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)\)
\(=7-5=2\)
c) Ta có: \(\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\cdot\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)
\(=\left(\sqrt{3}+1\right)\cdot\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\cdot\left|\sqrt{3}-1\right|\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)(Vì \(\sqrt{3}>1\))
\(=3-1=2\)
d) Ta có: \(5\sqrt{2}+\sqrt{18}-\sqrt{98}-\sqrt{288}\)
\(=\sqrt{2}\cdot\left(5+\sqrt{9}-\sqrt{49}-\sqrt{144}\right)\)
\(=\sqrt{2}\cdot\left(5+3-7-12\right)\)
\(=-11\sqrt{2}\)
e) Ta có: \(\left(\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{3}+\sqrt{5}}\)
\(=\left(\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right)\cdot\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)\)
\(=3-5=-2\)
g) Ta có: \(\left(\sqrt{3}-1\right)\cdot\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\)
\(=\left(\sqrt{3}-1\right)\cdot\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\cdot\left|\sqrt{3}+1\right|\)
\(=\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)\)(Vì \(\sqrt{3}>1>0\))
\(=3-1=2\)
7 tháng 6 2019
Thêm câu này hộ tớ nx nhé !
e) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0.4}\right)\)
14 tháng 7 2019
\(a,\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{12}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}-\frac{6\sqrt{6}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-\frac{4\sqrt{6}}{2}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{\sqrt{6}-4\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{-3\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)
\(=-\frac{3}{2}\)
\(\left(\sqrt{3}+1\right)\sqrt{\frac{14-6\sqrt{3}}{5+\sqrt{3}}=A}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{25-3}}\)
\(=\left(\sqrt{3+1}\right)\sqrt{\frac{70-14\sqrt{3}-30\sqrt{3}+18}{22}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\frac{88-44\sqrt{3}}{22}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{3-2\sqrt{3}+1}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3+1}\right)\left(\sqrt{3}-1\right)\)
\(=3-1\)
\(=2\)
\(A=2\)
Bài này là rút gọn biểu thức nhé