bài 2 : rút gọn f, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
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
HT
1
9 tháng 10 2021
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
8 tháng 12 2023
Câu 1:
a: \(\dfrac{2}{5}\sqrt{75}-0,5\cdot\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\cdot\sqrt{12}\)
\(=\dfrac{2}{5}\cdot5\sqrt{3}-0,5\cdot4\sqrt{3}+10\sqrt{3}-\dfrac{2}{3}\cdot2\sqrt{3}\)
\(=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}\)
\(=10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)
b: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
\(=\dfrac{\sqrt{3}\cdot3\sqrt{3}-2\sqrt{3}}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{9-6}\)
\(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+3-\sqrt{6}\)
\(=\dfrac{\sqrt{3}}{\sqrt{2}}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}\)
c: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
=\(\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
Bài 2:
a: 
b: Phương trình hoành độ giao điểm là:
\(3x+2=-x-4\)
=>4x=-6
=>x=-3/2
Thay x=-3/2 vào y=-x-4, ta được:
\(y=-\left(-\dfrac{3}{2}\right)-4=\dfrac{3}{2}-4=-\dfrac{5}{2}\)
Vậy: \(A\left(-\dfrac{3}{2};-\dfrac{5}{2}\right)\)
c: Vì (d2)//(d) nên \(\left\{{}\begin{matrix}a=-1\\b\ne-4\end{matrix}\right.\)
Vậy: (d2): y=-x+b
Thay x=-2 và y=5 vào (d2), ta được:
\(b-\left(-2\right)=5\)
=>b+2=5
=>b=5-2=3
Vậy: (d2): y=-x+3
DS
1
2 tháng 4 2015
\(\sqrt{3^2-2.3.\sqrt{6}+6}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2.\sqrt{6}\right)^2}\)
= \(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3-2.\sqrt{6}\right)^2}=\left|3-\sqrt{6}\right|+\left|3-2\sqrt{6}\right|=3-\sqrt{6}-3+2\sqrt{6}=\sqrt{6}\)
4 tháng 9 2016
1. \(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}\)
2. \(\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\\ =3-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}\)
3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(3+\sqrt{6}\right)^2}\\ =2\sqrt{6}-3+3+\sqrt{6}=3\sqrt{6}\)
4 tháng 9 2016
1.\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)
\(=\sqrt{5}+\sqrt{2}-\left(\sqrt{5}-\sqrt{2}\right)=2\sqrt{2}\)
2. \(\sqrt{12-6\sqrt{3}+\sqrt{21-12\sqrt{3}}}=\sqrt{12-6\sqrt{3}+\sqrt{\left(3-2\sqrt{3}\right)^2}}\)
\(=\sqrt{12-6\sqrt{3}+2\sqrt{3}-3}=\sqrt{9-4\sqrt{3}}\)
3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)
\(=2\sqrt{6}-3+3-\sqrt{6}=\sqrt{6}\)
MT
18 tháng 10 2015
\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}=2-\sqrt{3}+\sqrt{3-2\sqrt{3}+1}\left(\text{vì }2>\sqrt{3}\right)\)
\(=2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}=2-\sqrt{3}+\sqrt{3}-1\left(\text{vì }\sqrt{3}>1\right)\)
\(=1\)
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{9-2.3\sqrt{6}+6}+\sqrt{\left(2\sqrt{6}\right)^2-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=3-\sqrt{6}+2\sqrt{6}-3\left(\text{vì }3>\sqrt{6};2\sqrt{6}>3\right)\)
\(=\sqrt{6}\)
TH
18 tháng 10 2015
a/
\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=2-\sqrt{3}+\sqrt{3}-1\)
\(=1\)
b/
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}\)
\(=3-\sqrt{6}+\sqrt{24}-3\)
\(=-\sqrt{6}+2\sqrt{6}=\sqrt{6}\)
tick cho mình nha
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) \(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)
= \(\dfrac{15\sqrt{200}}{\sqrt{10}}-\dfrac{3\sqrt{450}}{\sqrt{10}}+\dfrac{2\sqrt{50}}{\sqrt{10}}\)
= \(15\sqrt{20}-3\sqrt{45}+2\sqrt{5}\)
= \(30\sqrt{5}-9\sqrt{5}+2\sqrt{5}\) = \(23\sqrt{5}\)
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 !!!
19 tháng 7 2018
\(C=\sqrt{15-6\sqrt{6}}+\sqrt{33+12\sqrt{6}}=\sqrt{9-2.3\sqrt{6}+6}+\sqrt{24+2.3.2\sqrt{6}+9}=3-\sqrt{6}+2\sqrt{6}+3=6+\sqrt{6}\) \(D=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\) \(F=\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\)
19 tháng 7 2018
\(C=\sqrt{15-6\sqrt{6}}+\sqrt{33+12\sqrt{6}}=\sqrt{\left(\sqrt{9}-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}+\sqrt{9}\right)^2}=3-\sqrt{6}+2\sqrt{6}+3=6+\sqrt{6}\)
\(D=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\sqrt{2}D=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}-1-\sqrt{3}-1=-2\)
\(\Rightarrow D=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)
\(F=\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=\left(3\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2=27-2=25\)
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{9-2.3\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)