K
Khách

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

28 tháng 8 2018

a. 9+4\(\sqrt{5}\)=(\(\sqrt{5}\)​+2)2

VT: 9+4\(\sqrt{5}\)=2\(^2\)+2.2.\(\sqrt{5}\)+​​(\(\sqrt{5}\))\(^2\)=(2+\(\sqrt{5}\))\(^2\)=VP

b. \(\sqrt{23+8\sqrt{7}}\)-\(\sqrt{7}\)=4

\(\Leftrightarrow\)\(\sqrt{4^2+2.4\sqrt{7}+\left(\sqrt{7}\right)^2}\)-\(\sqrt{7}\)=4

\(\Leftrightarrow\)\(\sqrt{4+\sqrt{7}}^2\)-\(\sqrt{7}\)=4

\(\Leftrightarrow\)4+\(\sqrt{7}\)-\(\sqrt{7}\)=4

\(\Leftrightarrow\)4=4

\(\Rightarrow\)VT=VP
\(\sqrt{5}\)\(\sqrt{5}\)

28 tháng 8 2018

Cái dòng \(\sqrt{5}\)\(\sqrt{5}\) máy mình bị lỗi nên đánh thừa thông cảm nha.

25 tháng 8 2017

a) \(9+4\sqrt{5}=4+4\sqrt{5}+5=2^2+2\cdot2\sqrt{5}+\left(\sqrt{5}\right)^2=\left(\sqrt{5}+2\right)^2\left(ĐPCM\right)\)

21 tháng 9 2017

a) \(9+4\sqrt{5}=\left(\sqrt{5}\right)^2+2.\sqrt{5}.2+2^2=\left(\sqrt{5}+2\right)^2\left(đpcm\right)\)

b)\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\left(đpcm\right)\)

c)\(\left(4-\sqrt{7}\right)^2=16-8\sqrt{7}+7=23-8\sqrt{7}\left(đpcm\right)\)

d)\(\sqrt{23+8\sqrt{7}}-\sqrt{7}=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}=4+\sqrt{7}-\sqrt{7}=4\left(đpcm\right)\)

23 tháng 6 2016

a) \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}=\left|\sqrt{5}-2\right|-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\)

b) \(\left(4-\sqrt{7}\right)^2=4^2-2.4.\sqrt{7}+\sqrt{7}^2=16-8\sqrt{7}+7=23-8\sqrt{7}\)

c)  \(\sqrt{23+8\sqrt{7}}=\sqrt{\left(4+\sqrt{7}\right)^2}=\left|4+\sqrt{7}\right|=\sqrt{7}+4\)

Câu 8:

a)

Ta có: \(VT=\sqrt{4-2\sqrt{3}}-\sqrt{3}\)

\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}-\sqrt{3}\)

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

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

Ta có: 3>1

\(\Leftrightarrow\sqrt{3}>\sqrt{1}\)

\(\Leftrightarrow\sqrt{3}>1\)

\(\Leftrightarrow\sqrt{3}-1>0\)

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

Từ (1) và (2) suy ra \(VT=\sqrt{3}-1-\sqrt{3}=-1=VP\)(đpcm)

b) Ta có: \(VP=\left(\sqrt{5}+2\right)^2\)

\(=\left(\sqrt{5}\right)^2+2\cdot\sqrt{5}\cdot2+2^2\)

\(=5+4\sqrt{5}+4\)

\(=9+4\sqrt{5}=VT\)(đpcm)

c) Ta có: \(VT=\sqrt{9+4\sqrt{5}}-\sqrt{5}\)

\(=\sqrt{4+2\cdot2\cdot\sqrt{5}+5}-\sqrt{5}\)

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

\(=\left|2+\sqrt{5}\right|-\sqrt{5}\)

\(=2+\sqrt{5}-\sqrt{5}=2=VP\)(đpcm)

d) Ta có: \(VT=\sqrt{23+8\sqrt{7}}-\sqrt{7}\)

\(=\sqrt{16+2\cdot4\cdot\sqrt{7}+7}-\sqrt{7}\)

\(=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)

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

\(=4+\sqrt{7}-\sqrt{7}\)

\(=4=VP\)(đpcm)

13 tháng 7 2020

em cảm ơn ạ yeu

2 tháng 6 2017
  1. \(\sqrt{\sqrt{5}^2-2.2\sqrt{5}+4}-\sqrt{5}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\left(dpcm\right)\)
  2. \(\sqrt{23+8\sqrt{7}}-\sqrt{7}=\sqrt{\sqrt{7}^2+2.4\sqrt{7}+16}-\sqrt{7}\)\(=\sqrt{\left(\sqrt{7}+4\right)^2}-\sqrt{7}=\sqrt{7}+4-\sqrt{7}=4\left(DPCM\right)\)
29 tháng 8 2016

a) \(\left(\sqrt{5}+2\right)^2=\sqrt{5}^2+4\sqrt{5}+4=5+4\sqrt{5}+4=9+4\sqrt{5}\left(dpcm\right)\)

29 tháng 8 2016

cậu ơi làm câu 2 lun đi 

31 tháng 5 2018

1)d) \(\sqrt{23+8\sqrt{7}}-\sqrt{7}\)

\(=\sqrt{4^2+2.4.\sqrt{7}+\sqrt{7^2}}-\sqrt{7}\)

\(=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)

\(=4+\sqrt{7}-\sqrt{7}\)

\(=4\)

28 tháng 8 2018

a) \(\sqrt{9-4\sqrt{5}}+\sqrt{5}\)

=\(\sqrt{\left(\sqrt{2}\right)^2-2.2\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{5}\)

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

=\(\left|\sqrt{2}-\sqrt{5}\right|+\sqrt{5}\)

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

=\(\sqrt{2}\)

NV
17 tháng 6 2019

\(a\sqrt{b}-b\sqrt{a}=\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)

\(7\sqrt{7}+3\sqrt{3}=\left(\sqrt{7}+\sqrt{3}\right)\left(7-\sqrt{21}+3\right)=\left(\sqrt{7}+\sqrt{3}\right)\left(10-\sqrt{21}\right)\)

\(a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)\)

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

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

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

\(\left(\sqrt{5}+2\right)^2-8\sqrt{5}=\left(\sqrt{5}-2\right)^2\)

2 cái trên đều áp dụng HĐT \(\left(a+b\right)^2-4ab=\left(a-b\right)^2\)

\(5\sqrt{2}-2\sqrt{5}=\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)\)

13 tháng 5 2018

a)\(\sqrt{13-4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{12-2.2\sqrt{3}.1+1}+\sqrt{4-2.2.\sqrt{3}+3}\)

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

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

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

b)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

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

\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\left(\sqrt{5}+1\right)+\left(\sqrt{5}-1\right)=2\sqrt{5}\)

c)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)

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

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

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

d)\(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{4+2.2\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}\)

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

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

e)\(\sqrt{9+4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}=\sqrt{5}+2\)

f)\(\sqrt{23+8\sqrt{7}}=\sqrt{16+2.4.\sqrt{7}+7}=\sqrt{\left(4+\sqrt{7}\right)^2}=4+\sqrt{7}\)

11 tháng 9 2019

undefined