K
Khách

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

13 tháng 7 2016

\(\left(\frac{4\sqrt{a}}{\sqrt{a}+2}+\frac{8a}{4-a}\right):\left(\frac{\sqrt{a}-1}{a-2\sqrt{a}}-\frac{2}{\sqrt{a}}\right)\) (ĐKXĐ : \(a>0;a\ne4;a\ne9\))

\(=\left[\frac{4\sqrt{a}\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}-\frac{8a}{a-4}\right]:\left[\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-2\right)}-\frac{2\left(\sqrt{a}-2\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}\right]\)

\(=\frac{4a-8\sqrt{a}-8a}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}:\frac{\sqrt{a}-1-2\sqrt{a}+4}{\sqrt{a}\left(\sqrt{a}-2\right)}\)

\(=\frac{-4\sqrt{a}\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}:\frac{-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{-4\sqrt{a}}{\sqrt{a}-2}.\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{3-\sqrt{a}}=-\frac{4a}{3-\sqrt{a}}\)

\(=\dfrac{a\sqrt{a}-3-2\left(a-6\sqrt{a}+9\right)-a-4\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{a-1}{a+8}\)

\(=\dfrac{a\sqrt{a}-a-4\sqrt{a}-6-2a+12\sqrt{a}-18}{\left(\sqrt{a}-3\right)}\cdot\dfrac{\sqrt{a}-1}{a+8}\)

\(=\dfrac{a\sqrt{a}-3a+8\sqrt{a}-24}{\left(\sqrt{a}-3\right)}\cdot\dfrac{\sqrt{a}-1}{a+8}=\sqrt{a}-1\)

20 tháng 7 2016

A=căn[(x-5)2]/x-5=|x-5|/x-5

Nếu x>=5 thì A=1

Nếu x<5 thì A=-1

31 tháng 10 2018

\(a,\left(\sqrt{50}+\sqrt{48}-\sqrt{72}\right)2\sqrt{3}\)

\(=\left(5\sqrt{2}+4\sqrt{3}-6\sqrt{2}\right)2\sqrt{3}\)

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

\(=24-2\sqrt{6}\)

18 tháng 6 2015

Đặt B  = \(\left(\frac{\left(a+3\sqrt{a}+1\right)\left(\sqrt{a}+1\right)-\left(a+\sqrt{a}\right)\left(\sqrt{a}+2\right)}{\left(\text{\sqrt{a}+2}\right)\left(a-1\right)}\right)\)  ($\sqrt{ a}$ + 2 là căn a )

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

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

Dặt \(C=\frac{1}{\sqrt{a}+1}-\frac{1}{\sqrt{a}-1}=\frac{\sqrt{a}-1-\sqrt{a}-1}{a-1}=-\frac{2}{a-1}\)

A = B : C = \(\frac{\sqrt{a}+1}{a-1}:-\frac{2}{a-1}=\frac{\sqrt{a}+1}{a-1}\cdot\frac{a-1}{-2}=-\frac{\left(\sqrt{a}+1\right)}{2}\)

17 tháng 6 2015

Rut gon lai là (-1-căn a) / 2 

b,a = 3

15 tháng 11 2022

Câu 3: 

a: =>|2x-1|=4

=>2x-1=4 hoặc 2x-1=-4

=>x=-3/2 hoặc x=5/2

b: \(\Leftrightarrow2\sqrt{x+1}+3\sqrt{x+1}-2\sqrt{x+1}=5\)

=>3căn x+1=5

=>x+1=25/9

=>x=16/9

3 tháng 9 2016

\(A=\left(\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right)\left(ĐK:x\ge0;\ne1\right)\)

\(=\left[\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}+2}\right]\)

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

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

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

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

31 tháng 10 2018

Rút gọn bt:

Câu 1: a, \(\left(\sqrt{50}+\sqrt{48}-\sqrt{72}\right)2\sqrt{3}\)

b, \(\sqrt{25a}+2\sqrt{45a}-3\sqrt{80a}+2\sqrt{16a}\left(a\ge0\right)\)ư

Câu 2: Cho bt: P =\(\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)

a, Tìm ĐKXĐ . Rút gọn P 

B, Tìm x nguyên để P có gt nguyên

c, Tìm GTNN của P với a >1

Câu 3: Giair các pt 

a, \(\sqrt{\left(2x-1\right)^2}=4\)

b, \(\sqrt{4x+4}+\sqrt{9x+9}-8\sqrt{\frac{x+1}{16}}=5\)