Rút gọn biểu thức
A= \(\sqrt{\frac{2+\sqrt{3}}{2}}-\sqrt{\frac{2-\sqrt{3}}{2}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
6 tháng 7 2021
a, Ta có : \(A=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(\Rightarrow A^2=2-\sqrt{3}+2+\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=4-2\sqrt{4-3}=4-2=2\)
\(\Rightarrow A=-\sqrt{2}\)
b, Ta có : \(B=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(\Rightarrow B\sqrt{2}=\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\)
\(=\sqrt{5+2\sqrt{5}+1}+\sqrt{9-2.3\sqrt{5}+5}-2\)
\(=\sqrt{5}+1+3-\sqrt{5}-2=2\)
\(\Rightarrow B=\sqrt{2}\)
MN
1
5 tháng 1 2022
\(a,=5\sqrt{2}-3\sqrt{2}+\sqrt{2}=3\sqrt{2}\\ b,=\dfrac{\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}=\dfrac{2\sqrt{3}}{3-2}=2\sqrt{3}\)
AD
2
Y
1 tháng 7 2023
\(A=2\sqrt{27}-\sqrt{\left(1-\sqrt{3}\right)^2}+\dfrac{1}{2-\sqrt{3}}\\ =2.3\sqrt{3}-\left|1-\sqrt{3}\right|+\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\\ =6\sqrt{3}-\left(-1+\sqrt{3}\right)+\dfrac{2+\sqrt{3}}{2^2-\sqrt{3^2}}\\ =6\sqrt{3}+1-\sqrt{3}+2+\sqrt{3}\\ =6\sqrt{3}+3\)
24 tháng 6 2021
Xét \(\sqrt{2}.A=\sqrt{\dfrac{4+2\sqrt{3}}{2}}-\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)
= \(\sqrt{\dfrac{\left(1+\sqrt{3}\right)^2}{2}}-\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{2}}\)
= \(\dfrac{1+\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}\)
<=> A = 1
12 tháng 10 2018
\(\frac{\sqrt{2-\sqrt{3}}}{2}:\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right).\)
\(=\frac{2\sqrt{2-\sqrt{3}}}{4}:\left(\frac{2\sqrt{2+\sqrt{3}}}{4}-\frac{2}{\sqrt{6}}+\frac{2\sqrt{2+\sqrt{3}}}{4\sqrt{3}}\right)\)
\(=\frac{\sqrt{4-2\sqrt{3}}}{4}:\left(\frac{\sqrt{4+2\sqrt{3}}}{4}-\frac{2}{\sqrt{6}}+\frac{\sqrt{4+2\sqrt{3}}}{4\sqrt{3}}\right)\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{4}:\left[\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{4}-\frac{2}{\sqrt{6}}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{4\sqrt{3}}\right]\)
\(=\frac{\sqrt{3}-1}{4}:\left[\frac{\sqrt{6}\left(\sqrt{3}+1\right)}{4\sqrt{6}}-\frac{2.4}{4\sqrt{6}}+\frac{\sqrt{2}\left(\sqrt{3}+1\right)}{4\sqrt{6}}\right]\)
\(=\frac{\sqrt{3}-1}{4}:\frac{\sqrt{18}+\sqrt{6}-8+\sqrt{6}+\sqrt{2}}{4\sqrt{6}}\)
\(=\frac{\sqrt{3}-1}{4}.\frac{4\sqrt{6}}{\sqrt{2}\left(\sqrt{9}+2\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{2}\left(\sqrt{3}+1\right)^2}=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)^2}\)............
VD
18 tháng 2 2021
\(A=a+2\sqrt{a}-3\sqrt{a}-6-a-2\sqrt{a}-1+3\sqrt{a}\)
\(A=-7\)
18 tháng 2 2021
Ta có: \(A=\left(\sqrt{a}+2\right)\left(\sqrt{a}-3\right)-\left(\sqrt{a}+1\right)^2+\sqrt{9a}\)
\(=a-3\sqrt{a}+2\sqrt{a}-6-a-2\sqrt{a}-1+3\sqrt{a}\)
\(=-7\)
21 tháng 6 2023
a: \(A=6\sqrt{3}+10\sqrt{3}-12\sqrt{3}=4\sqrt{3}\)
b: \(B=7\sqrt{3}+5\sqrt{3}-12\sqrt{3}=0\)
c: \(=12\sqrt{2}-6+3\left(9-4\sqrt{2}\right)=12\sqrt{2}-6+27-12\sqrt{2}=21\)
d: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)
5 tháng 7 2021
a. \(\sqrt{48}-2\sqrt{32}-\sqrt{75}+3\sqrt{50}\) = \(4\sqrt{3}-2.4\sqrt{2}-5\sqrt{3}+3.5\sqrt{2}\)
= \(4\sqrt{3}-8\sqrt{2}-5\sqrt{3}+15\sqrt{2}\) = \(-\sqrt{3}+7\sqrt{2}\)
b. \(\sqrt{20}-15\sqrt{\dfrac{1}{5}}+\sqrt{\left(1-\sqrt{5}\right)^2}\) = \(2\sqrt{5}-3.5.\sqrt{\dfrac{1}{5}}+\left|1-\sqrt{5}\right|\)
= \(2\sqrt{5}-3\sqrt{25.\dfrac{1}{5}}+\sqrt{5}-1\) = \(2\sqrt{5}-3\sqrt{5}+\sqrt{5}-1\) = \(-1\)
c. \(\dfrac{3}{3+2\sqrt{3}}+\dfrac{3}{3-2\sqrt{3}}\) = \(\dfrac{3\left(3-2\sqrt{3}\right)+3\left(3+2\sqrt{3}\right)}{\left(3+2\sqrt{3}\right)\left(3-2\sqrt{3}\right)}\)
= \(\dfrac{9-6\sqrt{3}+9+6\sqrt{3}}{\left(3+2\sqrt{3}\right)\left(3-2\sqrt{3}\right)}\) = \(\dfrac{18}{9-12}=\dfrac{18}{-3}=-6\)
24 tháng 6 2017
A=\(\sqrt{2}\), cái kết quả này bấm máy tính là ra được, quan trọng là phải làm thế nào để ra
\(A=\sqrt{\frac{2+\sqrt{3}}{2}}-\sqrt{\frac{2-\sqrt{3}}{2}}\)
\(A=\sqrt{\frac{4+2\sqrt{3}}{4}}-\sqrt{\frac{4-2\sqrt{3}}{4}}\)
\(A=\sqrt{\frac{3+2\sqrt{3}+1}{4}}-\sqrt{\frac{3-2\sqrt{3}+1}{4}}\)
\(A=\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{4}}-\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{4}}\)
\(A=\left|\frac{\sqrt{3}+1}{2}\right|-\left|\frac{\sqrt{3}-1}{2}\right|\)
\(A=\frac{\sqrt{3}+1}{2}-\frac{\sqrt{3}-1}{2}\)
\(A=1\)
A=\(\sqrt{\frac{4+2\sqrt{3}}{4}}-\sqrt{\frac{4-2\sqrt{3}}{4}}=\)\(\sqrt{\frac{3+2\sqrt{3}+1}{4}}-\sqrt{\frac{3-2\sqrt{3+1}}{4}}\)=\(\frac{\sqrt{3}+1}{2}-\frac{\sqrt{3}-1}{2}=\frac{\sqrt{3}}{2}\)