\(\sqrt{\left(2\sqrt{6}-4\right)^2}+\sqrt{15-6\sqrt{6}}\)

b) 

K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

29 tháng 5

Bài 1:

a: \(\sqrt{\left(2\sqrt6-4\right)^2}+\sqrt{15-6\sqrt6}\)

\(=\left|2\sqrt6-4\right|+\sqrt{\left(3-\sqrt6\right)^2}\)

\(=2\sqrt6-4+3-\sqrt6=-1+\sqrt6\)

b: \(\sqrt{\left(3-2\sqrt2\right)^2}+\sqrt{19+2\cdot\sqrt{18}}\)

\(=\left|3-2\sqrt2\right|+\sqrt{\left(3\sqrt2+1\right)^2}\)

\(=3-2\sqrt2+3\sqrt2+1=4+\sqrt2\)

c: \(\sqrt{9+4\sqrt5}-\sqrt{\left(1-\sqrt5\right)^2}\)

\(=\sqrt{\left(\sqrt5+2\right)^2}-\left|1-\sqrt5\right|\)

\(=\sqrt5+2-\left(\sqrt5-1\right)=3\)

Bài 2:

a: \(\frac{1}{\sqrt7+3}+\frac{1}{\sqrt7-3}=\frac{\sqrt7-3+\sqrt7+3}{\left(\sqrt7-3\right)\left(\sqrt7+3\right)}=\frac{2\sqrt7}{7-9}=-\sqrt7\)

b: \(\frac{3}{\sqrt2-1}+\frac{\sqrt6+\sqrt2}{\sqrt3+1}\)

\(=\frac{3\left(\sqrt2+1\right)}{\left(\sqrt2-1\right)\left(\sqrt2+1\right)}+\frac{\sqrt2\left(\sqrt3+1\right)}{\sqrt3+1}\)

\(=3\left(\sqrt2+1\right)+\sqrt2=4\sqrt2+3\)

c: \(\frac{1}{7+4\sqrt3}+\frac{1}{7-4\sqrt3}=\frac{7-4\sqrt3+7+4\sqrt3}{\left(7+4\sqrt3\right)\left(7-4\sqrt3\right)}=\frac{14}{49-48}=14\)

Rút gọn biểu thức: 1) \(\sqrt{12}+5\sqrt{3}-\sqrt{48}\) 2) \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\) 3) \(2\sqrt{32}+4\sqrt{8}-5\sqrt{18}\) 4) \(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\) 5) \(\sqrt{12}+\sqrt{75}-\sqrt{27}\) 6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\) 7) \(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\) 8) \(\left(\sqrt{2}+2\right)\sqrt{2}-2\sqrt{2}\) 9) \(\dfrac{1}{\sqrt{5}-1}-\dfrac{1}{\sqrt{5}+}\) 10) \(\dfrac{1}{\sqrt{5}-2}+\dfrac{1}{\sqrt{5}+2}\) 11)...
Đọc tiếp

Rút gọn biểu thức:

1) \(\sqrt{12}+5\sqrt{3}-\sqrt{48}\)

2) \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)

3) \(2\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)

4) \(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)

5) \(\sqrt{12}+\sqrt{75}-\sqrt{27}\)

6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)

7) \(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)

8) \(\left(\sqrt{2}+2\right)\sqrt{2}-2\sqrt{2}\)

9) \(\dfrac{1}{\sqrt{5}-1}-\dfrac{1}{\sqrt{5}+}\)

10) \(\dfrac{1}{\sqrt{5}-2}+\dfrac{1}{\sqrt{5}+2}\)

11) \(\dfrac{2}{4-3\sqrt{2}}-\dfrac{2}{4+3\sqrt{2}}\)

12) \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\)

13) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

14) \(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)

15) \(\left(\sqrt{6}-\sqrt{5}\right)^2-\sqrt{120}\)

16) \(\left(2\sqrt{3}-3\sqrt{2}\right)^2+2\sqrt{6}+3\sqrt{24}\)

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

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

19) \(\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)

20) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)\)

4
3 tháng 1 2019

1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)

2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)

3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2} \)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)

4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)

5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)

6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)

7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)

8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2


4 tháng 1 2019
https://i.imgur.com/pmexRQv.jpg
14 tháng 11 2018

Đề không khó, mỗi tội dài

14 tháng 11 2018

vậy thì bn làm hộ mik vs , mik cần gấp

16 tháng 10 2018

2]\(\sqrt{3}\)+1+\(\sqrt{4-4\sqrt{3}+3}\)=\(\sqrt{3}+1+\sqrt{\left(2-\sqrt{3}\right)^2}=\sqrt{3}+1+2-\sqrt{3}=3\)

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

19 tháng 10 2022

1: \(=2\sqrt{7}-12\sqrt{7}+15\sqrt{7}+27\sqrt{7}=32\sqrt{7}\)

3: \(=\sqrt{5}-2-\sqrt{14+6\sqrt{5}}\)

\(=\sqrt{5}-2-3-\sqrt{5}=-5\)

4: \(=2\sqrt{3}+3+4-2\sqrt{3}=7\)

5: \(=3-\sqrt{2}+3+\sqrt{2}+4-3=7\)

6: \(=\sqrt{\dfrac{6+2\sqrt{5}}{4}}+\sqrt{\dfrac{14-6\sqrt{5}}{4}}\)

\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}}{2}=\dfrac{4}{2}=2\)

8: \(=\sqrt{5}-1+\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{4}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{4}}\)

\(=\sqrt{5}-1+\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}\)

\(=\dfrac{2\sqrt{5}-2+3-\sqrt{5}-3-\sqrt{5}}{2}=\dfrac{-2}{2}=-1\)

4 tháng 8 2022

a: \(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+\dfrac{9}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)

\(=60-20\sqrt{18}+\dfrac{45}{2}\sqrt{12}\)

\(=60-60\sqrt{2}+45\sqrt{3}\)

b: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)

\(=\dfrac{2\sqrt{5}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}=\dfrac{2\sqrt{5}+3}{9+6\sqrt{2}}\)

 

19 tháng 5 2022

a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)

b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)

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

\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)

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

16 tháng 12 2022

a: \(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=12\sqrt{2x}\)

b: \(=6-4\sqrt{3}+4\sqrt{3}-8=-2\)

c: \(=\sqrt{2}+1+2-\sqrt{2}=3\)

d: \(=\dfrac{1}{\sqrt{2}}\cdot\left(\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}+2\sqrt{3}\right)=0\)

f: \(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}=-6\sqrt{6}\)

21 tháng 7 2022

a: \(=\left(-\sqrt{5}-\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)

\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

=-2

b: \(=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}\)

c: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)

\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)