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19 tháng 10 2019

b)\(\frac{3\sqrt{2}}{\sqrt{3}+1}\)

\(=\frac{3\sqrt{2}\left(\sqrt{3}-1\right)}{(\sqrt{3}+1)\left(\sqrt{3}-1\right)}\)

\(=\frac{3\left(\sqrt{6}-\sqrt{2}\right)}{3-1}\)

\(=\frac{3\left(\sqrt{6}-\sqrt{2}\right)}{2}\)

19 tháng 10 2019

a)\(\left(\sqrt{3}+1\right)^2+\left(1-\sqrt{3}\right)^2\)

\(=3+2\sqrt{3}+1+1-2\sqrt{3}+3\)

\(=8\)

b)\(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)

\(=\sqrt{28.7}-2\sqrt{3.7}+\sqrt{7}.\sqrt{7}+\sqrt{84}\)

\(=\sqrt{196}-2\sqrt{21}+7+\sqrt{4.21}\)

\(=\sqrt{14^2}-2\sqrt{21}+7+2\sqrt{21}\)

\(=14-2\sqrt{21}+7+2\sqrt{21}\)

\(=21\)

12 tháng 10 2020

a) \(\left(\frac{\sqrt{9}}{2}+\frac{\sqrt{1}}{2}-\sqrt{2}\right)\sqrt{2}\)

\(=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-2\)

\(=\frac{4\sqrt{2}}{2}-2=2\sqrt{2}-2\)

b) \(\left(\frac{\sqrt{8}}{3}-\sqrt{24}+\frac{\sqrt{50}}{3}\right)\sqrt{6}\)

\(=\frac{4\sqrt{3}}{3}-12+\frac{10\sqrt{3}}{3}\)

\(=\frac{14\sqrt{3}}{3}-12\)

c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{1}\right)\) (đã sửa đề)

\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\sqrt{2}\)

\(=\left(3-1\right)\sqrt{2}\)

\(=2\sqrt{2}\)

d) \(\left(3\sqrt{2}+1\right)\left(\sqrt{3\sqrt{2}-1}\right)\)

\(=\sqrt{3\sqrt{2}+1}\cdot\left(\sqrt{3\sqrt{2}+1}\cdot\sqrt{3\sqrt{2}-1}\right)\)

\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{18-1}\)

\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{17}\)

...

19 tháng 6 2017

2\(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)

\(14-\sqrt{84}+7-\sqrt{84}\)

= 21

19 tháng 6 2017

Câu 1 = 15

Câu 2 = 21

Nha!

K VÀ KB NHA !

7 tháng 6 2019

Thêm câu này hộ tớ nx nhé !
e) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0.4}\right)\)

14 tháng 7 2019

\(a,\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{12}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}-\frac{6\sqrt{6}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}}{2}-\frac{4\sqrt{6}}{2}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{\sqrt{6}-4\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{-3\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)

\(=-\frac{3}{2}\)

a: Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}-11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}-11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}-11\right)\)

\(=127-22\sqrt{6}\)

b: Ta có: \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)

=-1+5

=4

27 tháng 7 2021

a) \(-0,8\sqrt{\left(-0,125\right)^2}=-0,8.\left|-0,125\right|=-0.8.0,125=-\dfrac{1}{10}\)

b) \(\sqrt{\left(-2\right)^6}=\sqrt{\left(\left(-2\right)^3\right)^2}=\left|\left(-2\right)^3\right|=8\)

c) \(\sqrt{\left(\sqrt{3}-2\right)^2}=\left|\sqrt{3}-2\right|=2-\sqrt{3}\)

d) \(\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)

27 tháng 7 2021

Thanks ạ hihi

30 tháng 6 2021

a) \(\text{2}\sqrt{\text{18}}-9\sqrt{50}+3\sqrt{8}\)

\(\text{6}\sqrt{\text{2}}-45\sqrt{2}+6\sqrt{2}\)

\(-33\sqrt{2}\)

30 tháng 6 2021

b) = \(7-2.\sqrt{7}.\sqrt{3}+3+7.2\sqrt{21}\)

\(10-2\sqrt{21}+14\sqrt{21}\)

\(10+12\sqrt{21}\)

bài 1: 

a: Ta có: \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)

\(=6\sqrt{2}-45\sqrt{2}+6\sqrt{2}\)

\(=-33\sqrt{2}\)

b: Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)

\(=10-2\sqrt{21}+14\sqrt{21}\)

\(=12\sqrt{21}+10\)

Bài 2: 

a: Ta có: \(\sqrt{\left(2x+3\right)^2}=8\)

\(\Leftrightarrow\left|2x+3\right|=8\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)

b: Ta có: \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)

\(\Leftrightarrow4\sqrt{x}=8\)

hay x=4

c: Ta có: \(\sqrt{9x-9}+1=13\)

\(\Leftrightarrow3\sqrt{x-1}=12\)

\(\Leftrightarrow x-1=16\)

hay x=17