\(=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)\)
=3-1=2
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Các câu hỏi tương tự
a)\(\sqrt{\left(2\sqrt{2}-3\right)^2+\sqrt{15}}\)
b)\(\sqrt{\left(\sqrt{10}-3\right)}^2+\sqrt{\left(\sqrt{10}-4\right)^2}\)
c)\(11+6\sqrt{2}=\left(3+\sqrt{2}\right)^2\)
d)\(\sqrt{11}+6\sqrt{2}+\sqrt{11-6\sqrt{2}=6}\)
x^3left(sqrt[3]{5+2sqrt{6}}+sqrt[3]{5-2sqrt{6}}right)^3sqrt[3]{5+2sqrt{6}}^3+3sqrt[3]{left(5+2sqrt{6}right)^2}.sqrt[3]{5-2sqrt{6}}+3sqrt[3]{5+2sqrt{6}}.sqrt[3]{left(5-2sqrt{6}right)^2}+sqrt[3]{5-2sqrt{6}}^35+2sqrt{6}+3sqrt[3]{left(5+2sqrt{6}right)left(5-2sqrt{6}right)}.sqrt[3]{5+2sqrt{6}}+3sqrt[3]{left(5+2sqrt{6}right)left(5-2sqrt{6}right)}.sqrt[3]{5-2sqrt{6}}+5-2sqrt{6}5+5+3sqrt[3]{left(25-4.6right)}.sqrt[3]{5+2sqrt{6}}+3sqrt[3]{left(25-4.6right)}.sqrt[3]{5-2sqrt{6}}10+
3sqrt[3]{5+2sqrt{6}}+3sq...
Đọc tiếp
\(x^3=\left(\sqrt[3]{5+2\sqrt{6}}+\sqrt[3]{5-2\sqrt{6}}\right)^3=\sqrt[3]{5+2\sqrt{6}}^3\)
\(+3\sqrt[3]{\left(5+2\sqrt{6}\right)^2}.\sqrt[3]{5-2\sqrt{6}}+3\sqrt[3]{5+2\sqrt{6}}.\sqrt[3]{\left(5-2\sqrt{6}\right)^2}+\sqrt[3]{5-2\sqrt{6}}^3\)
\(=5+2\sqrt{6}+3\sqrt[3]{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}.\sqrt[3]{5+2\sqrt{6}}\)
\(+3\sqrt[3]{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}.\sqrt[3]{5-2\sqrt{6}}+5-2\sqrt{6}\)
\(=5+5+3\sqrt[3]{\left(25-4.6\right)}.\sqrt[3]{5+2\sqrt{6}}+3\sqrt[3]{\left(25-4.6\right)}.\sqrt[3]{5-2\sqrt{6}}\)
\(=10+ 3\sqrt[3]{5+2\sqrt{6}}+3\sqrt[3]{5-2\sqrt{6}}\)
p/s : có bạn hỏi nên mình đăng , các bạn đừng report nhé
27 tháng 7 2021 lúc 16:36
Bài 2: a) sqrt{left(3-2sqrt{2}right)^2}+sqrt{left(3+2sqrt{2}right)^2} b) sqrt{left(5-2sqrt{6}right)^2}-sqrt{left(5+2sqrt{6}right)^2}c) sqrt{left(sqrt{5}-sqrt{2}right)^2}+sqrt{left(sqrt{5}+sqrt{2}right)^2} d) sqrt{left(sqrt{2}+1right)^2}-sqrt{left(sqrt{2}-5right)^2}e) sqrt{left(2-sqrt{3}right)^2}+sqrt{left(1-sqrt{3}right)^2} f) sqrt{left(3+sqrt{2}right)^2}-sqrt{left(1-sqrt{2}right)^2}Hộ mk vs ạ
Đọc tiếp
Bài 2:
a) \(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\) b) \(\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}\)
c) \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\) d) \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}\)
e) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}\) f) \(\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
Hộ mk vs ạ
19 tháng 8 2021 lúc 10:04
Tính
A= \(\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}\)
B=\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
C=\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
1) frac{3+2sqrt{3}}{sqrt{3}}+frac{2+sqrt{2}}{sqrt{2}+1}-left(sqrt{2}+3right)2) 0.1sqrt{left(-3right)^2}cdotleft[6sqrt{left(-frac{1}{3}right)^2}-sqrt{left(sqrt{3}-2right)^2}right]^23) left(frac{3+2sqrt{3}}{sqrt{3}+2}+frac{2+sqrt{2}}{sqrt{2}+1}right)divleft(1divfrac{1}{sqrt{2}+sqrt{3}}right)4) left(frac{3sqrt{2}+sqrt{6}}{sqrt{12}+2}-frac{sqrt{54}}{3}right)cdotfrac{2}{sqrt{6}}5) sqrt{frac{5+2sqrt{6}}{5-2sqrt{6}}}+sqrt{frac{5-2sqrt{6}}{5+2sqrt{6}}}
Đọc tiếp
1) \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
2) \(0.1\sqrt{\left(-3\right)^2}\cdot\left[6\sqrt{\left(-\frac{1}{3}\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\right]^2\)
3) \(\left(\frac{3+2\sqrt{3}}{\sqrt{3}+2}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\right)\div\left(1\div\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
4) \(\left(\frac{3\sqrt{2}+\sqrt{6}}{\sqrt{12}+2}-\frac{\sqrt{54}}{3}\right)\cdot\frac{2}{\sqrt{6}}\)
5) \(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
1 tháng 9 2023 lúc 9:00
Tính:
c) \(\sqrt{2-\sqrt{3}}.\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{2-\sqrt{3}}.\sqrt{2}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
Làm tiếp ...
a : \(\sqrt{5+2\sqrt{6}}-\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)
b : \(\dfrac{\sqrt{7}-2\sqrt{7}}{2-\sqrt{7}}+\dfrac{6}{\sqrt{7}+1}+\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)
\(a:\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{5+2\sqrt{6}}\)
b : \(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{2}\)
c : \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right).\left(2+\dfrac{5-3\sqrt{5}}{3-\sqrt{5}}\right)\)
d : \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
1> Rut gon
a)\(\sqrt{6-2\sqrt{2}+2\sqrt{3}-2\sqrt{6}}\)
b) \(\left(\sqrt{2}+1\right)\left(\left(\sqrt{2}\right)^2+1\right)\left(\left(\sqrt{2}\right)^4+1\right)\left(\left(\sqrt{2}\right)^8+1\right)\left(\left(\sqrt{2}\right)^{16}+1\right)\)
c)\(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
d) \(\sqrt{3-\frac{4\sqrt{5}}{3}}+\sqrt{3+\frac{4\sqrt{5}}{3}}\)
Rút gọn biểu thức 1) frac{sqrt{5+2sqrt{6}}+sqrt{8+2sqrt{15}}}{sqrt{7+2sqrt{10}}}2) left(2+frac{3+sqrt{3}}{sqrt{3}+1}right)left(2+frac{3-sqrt{3}}{sqrt{3}-1}right):left(sqrt{5}-2right)3) left(frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}-frac{12}{3-sqrt{6}}right).left(sqrt{6}+11right)4) frac{1}{1+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{1}{sqrt{3}+sqrt{4}}+...+frac{1}{sqrt{99}+sqrt{100}}5) frac{1}{1-sqrt{2}}-frac{1}{sqrt{2}-sqrt{3}}+frac{1}{sqrt{3}-sqrt{4}}-...-frac{1}{sqrt{98}-sqrt{99}}+frac{1}{sqrt{99}-s...
Đọc tiếp
Rút gọn biểu thức
1) \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8+2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
2) \(\left(2+\frac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\frac{3-\sqrt{3}}{\sqrt{3}-1}\right):\left(\sqrt{5}-2\right)\)
3) \(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)\)
4) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
5) \(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{98}-\sqrt{99}}+\frac{1}{\sqrt{99}-\sqrt{100}}\)
6) \(\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
7)\(\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{3}{2}}+2\right)\left(\frac{\sqrt{2}+\sqrt{3}}{4\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)\left(24+8\sqrt{6}\right)\left(\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}+\frac{\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)\)