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HT
1
9 tháng 10 2021
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
TK
2
12 tháng 4 2017
\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-2.3.\sqrt{6}}+\sqrt{33-2.6\sqrt{6}}\)
\(=\sqrt{3^2-2.3.\sqrt{6}+\sqrt{6^2}}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left(3-\sqrt{6}\right)-\left(2\sqrt{6}-3\right)\)
\(=3-\sqrt{6}-2\sqrt{6}+3\)
\(=6-3\sqrt{6}\)
24 tháng 7 2019
Ko vt lại đề nha bn:
\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-2.3.\sqrt{6}}+\sqrt{33-2.6\sqrt{6}}\)
\(=\sqrt{3^2-2.3.\sqrt{6}+\sqrt{6^2}}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left(3-\sqrt{6}\right)-\left(2\sqrt{6}-3\right)\)
\(=3-\sqrt{6}-2\sqrt{6}+3\)
\(=6-3\sqrt{6}\)
Rất vui vì giúp đc bn !!!
14 tháng 8 2019
\(A=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=1-\sqrt{3}-\sqrt{3}-2\)
\(=-2\sqrt{3}-1\)
\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(4-2\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+4-2\sqrt{3}\)
\(=6-3\sqrt{3}\)
H9
HT.Phong (9A5)
CTVHS
17 tháng 7 2023
1) \(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|\)
\(=2+\sqrt{5}+2-\sqrt{5}\)
\(=4\)
2) \(\sqrt{12-6\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
\(=\sqrt{3^2-2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{3^2+2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(=\left|3-\sqrt{3}\right|+\left|3+\sqrt{3}\right|\)
\(=3-\sqrt{3}+3+\sqrt{3}\)
\(=6\)
VA
1
12 tháng 4 2020
Ta có :
\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-\sqrt{128}}}}\)
Ta có :
\(18-\sqrt{128}=18-8\sqrt{2}=16-2.4.\sqrt{2}+2=\left(4-\sqrt{2}\right)^2\)
Vậy
\(\sqrt{18-\sqrt{128}}=4-\sqrt{2}\)
Thay vào ta có
\(\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-\sqrt{128}}}}\)
\(=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}\)
\(=\sqrt{6-2\sqrt{4+2\sqrt{3}}}\)
Lại có :
\(4+2\sqrt{3}=3+2.1.\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\)
Do đó :
\(\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
Vậy :
\(\sqrt{6-2\sqrt{4+2\sqrt{3}}}=\sqrt{6-2\left(\sqrt{3}+1\right)}\)
\(=\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3-2.1.\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}-1\)
Vậy : \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}=\sqrt{3}-1\)
2
1
Y
18 tháng 6 2023
\(\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
\(=\sqrt{2^2-2.2.\sqrt{3}+\sqrt{3^2}}+\sqrt{3^2+2.3.\sqrt{3}+\sqrt{3^2}}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|+\left|3+\sqrt{3}\right|\)
\(=2-\sqrt{3}+3+\sqrt{3}\)
\(=5\)
28 tháng 5 2022
9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3=6
13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)
\(=\sqrt{5}-2-3+\sqrt{5}=2\sqrt{5}-5\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 1 2022
\(A=\sqrt{\left(9\sqrt{2}+2\sqrt{3}\right)^2}-\sqrt{\left(9\sqrt{2}-\sqrt{3}\right)^2}\)
\(=\left|9\sqrt{2}+2\sqrt{3}\right|-\left|9\sqrt{2}-\sqrt{3}\right|\)
\(=9\sqrt{2}+2\sqrt{3}-9\sqrt{2}+\sqrt{3}=3\sqrt{3}\)
Kiểm tra lại đề bài câu B, chỗ \(\sqrt{2+\sqrt{2+2}}\)
MN
18 tháng 1 2022
30,001x3=3(0,1x)3=0,1x;
\sqrt[3]{-125 a^{12}}=\sqrt[3]{\left(-5 a^{4}\right)^{3}}=-5 a^{4};3−125a12=3(−5a4)3=−5a4;
\sqrt[3]{27 x^{6}}=\sqrt[3]{\left(3 x^{2}\right)^{3}}=3 x^{2};327x6=3(3x2)3=3x2;
\sqrt[3]{-0,343 a^{3}}=\sqrt[3]{(-0,7 a)^{3}}=-0,7 a;3−0,343a3=3(−0,7a)3=−0,7a;
18 tháng 1 2022
Ta rút gọn các biểu thức như sau:
\(\sqrt[3]{0,001x^3}=\sqrt[3]{\left(0,1x\right)^3}=0,1x.\)
\(\sqrt[3]{-125a^{12}}=\sqrt[3]{\left(-5a^4\right)^3}=-5a^4\)
\(\sqrt[3]{27x^6}=\sqrt[3]{\left(3x^2\right)^3}=3x^2\)
\(\sqrt[3]{-0,343a^3}=\sqrt[3]{\left(-0,7a\right)^3}=-0,7a\)
25 tháng 6 2017
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
= \(2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(2-\sqrt{3}+\sqrt{3}-1\) = \(1\)
b) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
= \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
= \(3-\sqrt{6}+2\sqrt{6}-3\) = \(\sqrt{6}\)
c) \(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)
= \(\dfrac{15\sqrt{200}}{\sqrt{10}}-\dfrac{3\sqrt{450}}{\sqrt{10}}+\dfrac{2\sqrt{50}}{\sqrt{10}}\)
= \(15\sqrt{20}-3\sqrt{45}+2\sqrt{5}\)
= \(30\sqrt{5}-9\sqrt{5}+2\sqrt{5}\) = \(23\sqrt{5}\)
\(=\sqrt{\left(2\sqrt{6}-3\right)^2}=\left|2\sqrt{6}-3\right|=2\sqrt{6}-3\)
\(\sqrt{33-12\sqrt{6}=\sqrt{3^2-2.3.2\sqrt{6}+\left(2\sqrt{6}\right)^2}}=\sqrt{\left(3-2\sqrt{6}\right)^2}=\left|3-2\sqrt{6}\right|=2\sqrt{6}-3\)