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LM
24 tháng 6 2018
a)\(\sqrt{28-16\sqrt{3}}=\sqrt{12-2.4.2\sqrt{3}+16}=\sqrt{\left(2\sqrt{3}\right)^2-2.4.2\sqrt{3}+4^2}=\sqrt{\left(2\sqrt{3}-4\right)^2}\)\(=\left|2\sqrt{3}-4\right|=4-2\sqrt{3}\)
b) \(\sqrt{29-12\sqrt{5}}=\sqrt{3^2-2.3.2\sqrt{5}+\left(2\sqrt{5}\right)^2}=\sqrt{\left(3-2\sqrt{5}\right)^2}=2\sqrt{5}-3\)
c)\(\sqrt{23-\sqrt{240}}=\sqrt{23-4\sqrt{15}}=\sqrt{\left(2\sqrt{5}\right)^2-2.\sqrt{3}.2\sqrt{5}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2\sqrt{5}-\sqrt{3}\right)^2}=2\sqrt{5}-\sqrt{3}\)
d)\(\sqrt{33-12\sqrt{6}}=\sqrt{\left(2\sqrt{6}\right)^2-2.3.2\sqrt{6}+3^2}=\sqrt{\left(2\sqrt{6}-3\right)^2}=2\sqrt{6}-3\)
22 tháng 7 2020
Trả lời:
a)\(\sqrt{28-16\sqrt{3}}\)
\(=\sqrt{16-16\sqrt{3}+12}\)
\(=\sqrt{\left(4-2\sqrt{3}\right)^2}\)
\(=4-2\sqrt{3}\)
b) \(\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{20-12\sqrt{5}+9}\)
\(=\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(=2\sqrt{5}-3\)
c) \(\sqrt{23-\sqrt{240}}\)
\(=\sqrt{23-4\sqrt{15}}\)
\(=\sqrt{20-4\sqrt{15}+3}\)
\(=\sqrt{\left(2\sqrt{5}-\sqrt{3}\right)^2}\)
\(=2\sqrt{5}-\sqrt{3}\)
d) \(\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{24-12\sqrt{6}+9}\)
\(=\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=2\sqrt{6}-3\)
28 tháng 4 2021
a) Ta có: \(A=\sqrt{23+6\sqrt{10}}-\sqrt{23-6\sqrt{10}}\)
\(=\sqrt{18+2\cdot3\sqrt{2}\cdot\sqrt{5}+5}-\sqrt{18-2\cdot3\sqrt{2}\cdot\sqrt{5}+5}\)
\(=\sqrt{\left(3\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{2}-\sqrt{5}\right)^2}\)
\(=3\sqrt{2}+\sqrt{5}-3\sqrt{2}+\sqrt{5}\)
\(=2\sqrt{5}\)
28 tháng 4 2021
b) Ta có: \(B=\left(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+1\right)\left(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}-1\right)\)
\(=\left(\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}+1\right)\left(\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-1\right)\)
\(=\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)\)
=2-1=2
AH
Akai Haruma
Giáo viên
1 tháng 12 2019
a)
\((2\sqrt{5}-\sqrt{7})(2\sqrt{5}+\sqrt{7})=(2\sqrt{5})^2-(\sqrt{7})^2=13\)
b)
\((\sqrt{5-2\sqrt{6}}+\sqrt{2})\sqrt{3}=(\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2})\sqrt{3}\)
\(=(\sqrt{(\sqrt{3}-\sqrt{2})^2}+\sqrt{2})\sqrt{3}=(\sqrt{3}-\sqrt{2}+\sqrt{2})\sqrt{3}=\sqrt{3}.\sqrt{3}=3\)
c)
\(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}=\sqrt{2^2+3-2.2\sqrt{3}}+\sqrt{2^2+3+2.2\sqrt{3}}\)
\(=\sqrt{(2-\sqrt{3})^2}+\sqrt{(2+\sqrt{3})^2}=2-\sqrt{3}+2+\sqrt{3}=4\)
AH
Akai Haruma
Giáo viên
1 tháng 12 2019
d)
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{3^2+6-2.3\sqrt{6}}+\sqrt{9+24-2\sqrt{9.24}}\)
\(=\sqrt{(3-\sqrt{6})^2}+\sqrt{(\sqrt{24}-3)^2}=3-\sqrt{6}+\sqrt{24}-3\)
\(=\sqrt{6}\)
e)
\(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}=\sqrt{\frac{6+2\sqrt{5}}{2}}+\sqrt{\frac{6-2\sqrt{5}}{2}}\)
\(=\sqrt{\frac{5+1+2\sqrt{5.1}}{2}}+\sqrt{\frac{5+1-2\sqrt{5.1}}{2}}=\sqrt{\frac{(\sqrt{5}+1)^2}{2}}+\sqrt{\frac{(\sqrt{5}-1)^2}{2}}\)
\(=\frac{\sqrt{5}+1}{\sqrt{2}}+\frac{\sqrt{5}-1}{\sqrt{2}}=\sqrt{10}\)
g)
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{20+3-2\sqrt{20.3}}\)
\(=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{20}-\sqrt{3})^2}\)
\(=\sqrt{5}-\sqrt{3}-(\sqrt{20}-\sqrt{3})=\sqrt{5}-\sqrt{20}=-\sqrt{5}\)
BH
12 tháng 8 2019
\(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}=\left(2\sqrt{5}+3\right)-\left(2\sqrt{5}-3\right)=6\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}-\sqrt{3}\right)=-\sqrt{5}\)
\(\sqrt{8-12\sqrt{5}}+\sqrt{48+6\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)+\left(3\sqrt{5}+\sqrt{3}\right)=4\sqrt{5}\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\left(5-2\sqrt{6}\right)+\left(5+2\sqrt{6}\right)=10\)
\(\sqrt{15-6\sqrt{15}}+\sqrt{33-12\sqrt{6}}\) đề này sai ạ
\(\sqrt{16-6\sqrt{7}}+\sqrt{64-24\sqrt{7}}=\left(3-\sqrt{7}\right)+\left(6-2\sqrt{7}\right)=9-3\sqrt{7}\)
\(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\left(3-\sqrt{5}\right)+\left(3+\sqrt{5}\right)=6\)
\(\sqrt{1-6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\left(2\sqrt{2}+5\right)+\left(2\sqrt{2}-5\right)=4\sqrt{2}\)
\(\sqrt{46-6\sqrt{5}}+\sqrt{29-12\sqrt{5}}=\left(3\sqrt{5}-1\right)+\left(2\sqrt{5}-3\right)=5\sqrt{5}-4\)
#Học tốt ạ
Y
5 tháng 7 2019
\(\sqrt{8-2\sqrt{15}}+\sqrt{48+6\sqrt{15}}\\ =\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}+\sqrt{45+2\cdot3\sqrt{5}\cdot\sqrt{3}+3}\\ =\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\\ =\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{5}+\sqrt{3}\right)^2}\\ =\sqrt{5}-\sqrt{3}+3\sqrt{5}+\sqrt{3}=4\sqrt{5}\)
Y
5 tháng 7 2019
\(\sqrt{8-\sqrt{60}}-\sqrt{23-\sqrt{240}}\\ =\sqrt{8-\sqrt{4\cdot15}}-\sqrt{23-\sqrt{4\cdot60}}\\ =\sqrt{8-2\sqrt{15}}-\sqrt{23-2\sqrt{60}}\\ =\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{20-2\cdot\sqrt{20}\cdot\sqrt{3}+3}\\ =\sqrt{5}-\sqrt{3}-\sqrt{\left(\sqrt{20}-\sqrt{3}\right)^2}\\ =\sqrt{5}-\sqrt{3}-\sqrt{20}+\sqrt{3}\\ =\sqrt{5}-2\sqrt{5}=-\sqrt{5}\)
6 tháng 7 2021
\(A=\sqrt{12+\sqrt{12+\sqrt{12}}}+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}< \sqrt{12+\sqrt{12+\sqrt{16}}}+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{9}}}}\)\(=7\)
\(B=\sqrt{14}+\sqrt{11}>\sqrt{13,69}+\sqrt{10,89}=7\)
\(\Rightarrow A< B\)
H
6 tháng 7 2021
Ta có:
\(12< 16\Rightarrow\sqrt{12}< \sqrt{16}=4\\ 6< 9\Rightarrow\sqrt{6}< \sqrt{9}=3\)
\(\Rightarrow A< \sqrt{12+\sqrt{12+4}}+\sqrt{6+\sqrt{6+\sqrt{6+3}}}=\sqrt{12+4}+\sqrt{6+3}=4+3=7\) (1)
Lại có :
\(B=\sqrt{14}+\sqrt{11}\Rightarrow B^2=25+2\sqrt{14.11}=25+2\sqrt{154}>25+2\sqrt{144}=25+2.12=49=7^2\)
Mà B > 0
\(\Rightarrow B>7\) (2)
Từ (1),(2) suy ra A<B
A=....
A^2=46-2√(23^2-
de sai.
23<12√6
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