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1 tháng 8 2018
Bài 1 : \(\sqrt{49-12\sqrt{5}}+\sqrt{49+12\sqrt{5}}\)
\(=\sqrt{45-4\sqrt{45}+4}+\sqrt{45+4\sqrt{45}+4}\)
\(=\sqrt{\left(\sqrt{45}-2\right)^2}+\sqrt{\left(\sqrt{45}+2\right)^2}\)
\(=\sqrt{45}-2+\sqrt{45}+2=2\sqrt{45}\)
Bài 2 : \(\sqrt{29+12\sqrt{5}}+\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{20+6\sqrt{20}+9}+\sqrt{20-6\sqrt{20}+9}\)
\(=\sqrt{\left(\sqrt{20}+3\right)^2}+\sqrt{\left(\sqrt{20}-3\right)^2}\)
\(=\sqrt{20}+3+\sqrt{20}-3=2\sqrt{20}\)
Bài 3 : \(\sqrt{31-12\sqrt{3}}+\sqrt{31+12\sqrt{3}}\)
\(=\sqrt{27-4\sqrt{27}+4}+\sqrt{27+4\sqrt{27}+4}\)
\(=\sqrt{\left(\sqrt{27}-2\right)^2}+\sqrt{\left(\sqrt{27}+2\right)^2}\)
\(=\sqrt{27}-2+\sqrt{27}+2=2\sqrt{27}\)
Chúc bạn học tốt
1 tháng 8 2018
4 , Ta có :
\(\sqrt{39-12\sqrt{3}}-\sqrt{39+12\sqrt{3}}\)
\(=\sqrt{3-2.6.\sqrt{3}+6^2}-\sqrt{3+2.6.\sqrt{3}+6^2}\)
\(=\sqrt{\left(\sqrt{3}-6\right)^2}-\sqrt{\left(\sqrt{3}+6\right)^2}\)
\(=\left|\sqrt{3}-6\right|-\left|\sqrt{3}+6\right|\)
\(=6-\sqrt{3}-\sqrt{3}-6\)
\(=-2\sqrt{3}\)
3 tháng 7 2015
\(\sqrt{31-12\sqrt{3}}+\sqrt{31+12\sqrt{3}}\)
=\(\sqrt{\left(3\sqrt{3}\right)^2-2.3\sqrt{3}.2+4}+\sqrt{\left(3\sqrt{3}\right)^2+2.3\sqrt{3}.2+4}\)
=\(\sqrt{\left(3\sqrt{3}-2\right)^2}+\sqrt{\left(3\sqrt{3}+2\right)^2}\)
=\(3\sqrt{3}-2+3\sqrt{3}+2=6\sqrt{3}\)
13 tháng 9 2021
a, \(\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)
\(=\sqrt{17-2.3.2\sqrt{2}}-\sqrt{17+2.3.2\sqrt{2}}\)
\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)
\(=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(3+2\sqrt{2}\right)^2}=\left|3-2\sqrt{2}\right|-\left|3+2\sqrt{2}\right|\)
\(=3-2\sqrt{2}-3-2\sqrt{2}=-4\sqrt{2}\)
b, \(\sqrt{31-12\sqrt{3}}-\sqrt{31+12\sqrt{3}}\)
\(=\sqrt{31-2.2.3\sqrt{3}}-\sqrt{31+2.2.3\sqrt{3}}\)
\(=\sqrt{\left(3\sqrt{3}-2\right)^2}-\sqrt{\left(3\sqrt{3}+2\right)^2}=\left|3\sqrt{3}-2\right|-\left|3\sqrt{3}+2\right|\)
\(=3\sqrt{3}-2-3\sqrt{3}-2=-4\)
AD
0
AK
6 tháng 7 2018
a )
\(\sqrt{31}+4< \sqrt{36}+4=10\left(1\right)\)
\(6+\sqrt{17}>6+\sqrt{16}=6+4=10\left(2\right)\)
Từ ( 1 ) ; ( 2 )
\(\Rightarrow\sqrt{31}+4< 10< 6+\sqrt{17}\)
\(\Rightarrow\sqrt{31}+4< \sqrt{17}+6\)
b )
\(\sqrt{3}+\sqrt{2}>\sqrt{1}+\sqrt{1}=2\)
c )
\(\sqrt{12+13}=\sqrt{25}=5\left(1\right)\)
\(\sqrt{12}+\sqrt{13}>\sqrt{4}+\sqrt{9}=2+3=5\left(2\right)\)
Từ ( 1 ) ; ( 2 )
\(\Rightarrow\sqrt{12+13}< \sqrt{12}+\sqrt{13}\)
26 tháng 6 2017
a. Ta co: \(2\sqrt{31}=\sqrt{2^2\cdot31}=\sqrt{124}\)
\(10=\sqrt{10}\)
\(Vi\sqrt{10}< \sqrt{124}hay10< 2\sqrt{31}\)
26 tháng 6 2017
c.. \(6+2\sqrt{2}=6+\sqrt{8}=\sqrt{6}+\sqrt{8}=\sqrt{14}\)(1)
\(9=\sqrt{3}\)(2)
Ta co (1)>(2) hay \(\sqrt{14}>\sqrt{3}\) nen \(6+2\sqrt{2}>9\)
24 tháng 6 2019
\(E=\)( ghi đề vào đây )
\(E=\sqrt[3]{4+\frac{5}{3}.\frac{\sqrt{31}}{\sqrt{3}}}+\sqrt[3]{4-\frac{5}{3}.\frac{\sqrt{31}}{3}}\)
\(E=\sqrt[3]{4+\frac{5\sqrt{31}}{3\sqrt{3}}}+\sqrt[3]{4+\frac{5.\sqrt{31}}{3\sqrt{3}}}\)
\(E\approx1\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 6 2019
\(E^3=4+\frac{5}{3}\sqrt{\frac{31}{3}}+4-\frac{5}{3}\sqrt{\frac{31}{3}}+3\sqrt[3]{\left(16-\frac{25}{9}.\frac{31}{3}\right)}\left(\sqrt[3]{4+\frac{5}{3}\sqrt{\frac{31}{3}}}+\sqrt[3]{4-\frac{5}{3}\sqrt{\frac{31}{3}}}\right)\)
\(\Leftrightarrow E^3=8-7E\)
\(\Leftrightarrow E^3+7E-8=0\)
\(\Leftrightarrow\left(E-1\right)\left(E^2+E+8\right)=0\)
\(\Leftrightarrow E=1\)
27 tháng 8 2020
Bài 3:
a) Ta có: \(4+2\sqrt{3}\)
\(=3+2\cdot\sqrt{3}\cdot1+1\)
\(=\left(\sqrt{3}+1\right)^2\)
b) Ta có: \(7+4\sqrt{3}\)
\(=4+2\cdot2\cdot\sqrt{3}+3\)
\(=\left(2+\sqrt{3}\right)^2\)
c) Ta có: \(9+4\sqrt{5}\)
\(=5+2\cdot\sqrt{5}\cdot2+4\)
\(=\left(\sqrt{5}+2\right)^2\)
d) Ta có: \(31+10\sqrt{6}\)
\(=25+2\cdot5\cdot\sqrt{6}+6\)
\(=\left(5+\sqrt{6}\right)^2\)
e) Ta có: \(13+4\sqrt{3}\)
\(=12+2\cdot2\sqrt{3}\cdot1+1\)
\(=\left(2\sqrt{3}+1\right)^2\)
g) Ta có: \(21+12\sqrt{3}\)
\(=12+2\cdot2\sqrt{3}\cdot3+9\)
\(=\left(2\sqrt{3}+3\right)^2\)
h) Ta có: \(29+12\sqrt{5}\)
\(=20+2\cdot2\sqrt{5}\cdot3+3\)
\(=\left(2\sqrt{5}+3\right)^2\)
i) Ta có: \(49+8\sqrt{3}\)
\(=48+2\cdot4\sqrt{3}\cdot1\)
\(=\left(4\sqrt{3}+1\right)^2\)
k) Sửa đề: \(14-6\sqrt{5}\)
Ta có: \(14-6\sqrt{5}\)
\(=9-2\cdot3\cdot\sqrt{5}+5\)
\(=\left(3-\sqrt{5}\right)^2\)
l) Ta có: \(23-8\sqrt{7}\)
\(=16-2\cdot4\cdot\sqrt{7}+7\)
\(=\left(4-\sqrt{7}\right)^2\)
m) Ta có: \(15-4\sqrt{11}\)
\(=11-2\cdot\sqrt{11}\cdot2+4\)
\(=\left(\sqrt{11}-2\right)^2\)
n) Sửa đề: \(28-10\sqrt{3}\)
Ta có: \(28-10\sqrt{3}\)
\(=25-2\cdot5\cdot\sqrt{3}+3\)
\(=\left(5-\sqrt{3}\right)^2\)
o) Ta có: \(17-12\sqrt{2}\)
\(=9-2\cdot3\cdot2\sqrt{2}+8\)
\(=\left(3-2\sqrt{2}\right)^2\)
p) Ta có: \(43-30\sqrt{2}\)
\(=25-2\cdot5\cdot3\sqrt{2}+18\)
\(=\left(5-3\sqrt{2}\right)^2\)
q) Ta có: \(51-10\sqrt{2}\)
\(=50-2\cdot5\sqrt{2}\cdot1\)
\(=\left(5\sqrt{2}-1\right)^2\)
r) Ta có: \(49-12\sqrt{5}\)
\(=45-2\cdot3\sqrt{5}\cdot2+4\)
\(=\left(3\sqrt{5}-2\right)^2\)
21 tháng 6 2018
\(1)\) Ta có :
\(\left(\sqrt{3\sqrt{2}}\right)^4=\left[\left(\sqrt{3\sqrt{2}}\right)^2\right]^2=\left(3\sqrt{2}\right)^2=9.2=18\)
\(\left(\sqrt{2\sqrt{3}}\right)^4=\left[\left(\sqrt{2\sqrt{3}}\right)^2\right]^2=\left(2\sqrt{3}\right)^2=4.3=12\)
Vì \(18>12\) nên \(\left(\sqrt{3\sqrt{2}}\right)^4>\left(\sqrt{2\sqrt{3}}\right)^4\)
\(\Rightarrow\)\(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Vậy \(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Chúc bạn học tốt ~
\(\sqrt{31-12\sqrt{3}}-\sqrt{31+12\sqrt{3}}\)
\(=\sqrt{\left(3\sqrt{3}\right)^2-2.3\sqrt{3}.2+2^2}-\sqrt{\left(3\sqrt{3}\right)^2+2.3\sqrt{3}.2+2^2}\)
\(=\sqrt{\left(3\sqrt{3}-2\right)^2}-\sqrt{\left(3\sqrt{3}+2\right)^2}=\left|3\sqrt{3}-2\right|-\left|3\sqrt{3}+2\right|\)
\(=3\sqrt{3}-2-3\sqrt{3}-2=-4\)
a yêu e