Rút gọn : C= \(1-\sqrt{x-2\sqrt{x-1}}+\sqrt{x-1}\)với \(x\ge1\)
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22 tháng 8 2021
a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\dfrac{x\sqrt{x}+y\sqrt{y}-\left(x-y\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\dfrac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}+x\sqrt{y}+y\sqrt{x}-y\sqrt{y}}{\sqrt{x}+\sqrt{y}}=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\left|\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)( do \(x\ge1\))
22 tháng 8 2021
a: Ta có: \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
\(=\sqrt{xy}\)
b: Ta có: \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
\(=\dfrac{ \left|\sqrt{x}-1\right|}{\left|\sqrt{x}+1\right|}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
MN
7
25 tháng 8 2016
Gọi cái đó là A
A2 = x + \(2\sqrt{x-1}\) + x - \(2\sqrt{x-1}\)+
\(2\sqrt{\left(x+\sqrt{x-1}\right)\left(x-\sqrt{x-1}\right)}\)
= 2x + 2
DN
5 tháng 6 2018
\(P=\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}+\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)
\(P=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(P=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
\(\Rightarrow P=\sqrt{x-1}+1+\sqrt{x-1}-1
\left(x\ge2\right)\) hoặc \(P=\sqrt{x-1}+1-\sqrt{x-1}+1\left(1\le x\le2\right)\)
\(\Rightarrow P=2\sqrt{x-1}
\left(x\ge2\right)\) hoặc \(P=2
\left(1\le x\le2\right)\)
6 tháng 8 2016
Xét : \(\sqrt{4x-3+4\sqrt{x-1}}=\sqrt{4\left(x-1\right)+4\sqrt{x-1}+1}=\sqrt{\left(2\sqrt{x-1}+1\right)^2}=2\sqrt{x-1}+1\)
Khi đó : \(A=\left(\sqrt{x-1}-1\right)^2+2\sqrt{x-1}=x-1-2\sqrt{x-1}+1+2\sqrt{x-1}+1=x+1\)
29 tháng 10 2023
a: Khi x=25 thì \(A=\dfrac{5-2}{5-1}=\dfrac{3}{4}\)
b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-4}{1-x}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}+\sqrt{x}-4}{x-1}=\dfrac{x-4}{x-1}\)
c: \(P=\dfrac{A}{B}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}:\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
P<1/2
=>P-1/2<0
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{1}{2}< 0\)
=>\(\dfrac{2\sqrt{x}+2-\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}< 0\)
=>\(\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}< 0\)
=>\(x\in\varnothing\)
13 tháng 6 2018
a, \(M=\sqrt{x^2-4x+4}-\sqrt{x^2+4x+4}\) (ĐK : \(\forall x\in R\))
\(=\sqrt{\left(x-2\right)^2}-\sqrt{\left(x+2\right)^2}\)
* Nếu x\(\ge2\Rightarrow M=x-2-x-2=-4\)
*Nếu x<2 => M=2-x-x-2=-2x
b,Để M=2\(\ne-4\)
=>M=-2x
=>-2x=-4
=>x=2
__________________________________________________________________________________________
P=\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
* Nếu \(x\ge2\Rightarrow P=\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\)
* Nếu x<2 =>P=\(\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
VẬY.......
Tk nha!
14 tháng 5 2021
Ta có:
\(A=x-\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{1}{\sqrt{x}+\sqrt{x-1}}\right)\)
\(A=x-\frac{\sqrt{x}+\sqrt{x-1}-\sqrt{x}+\sqrt{x-1}}{\left(\sqrt{x}-\sqrt{x-1}\right)\left(\sqrt{x}+\sqrt{x-1}\right)}\)
\(A=x-\frac{2\sqrt{x-1}}{x-x+1}\)
\(A=x-2\sqrt{x-1}\)
\(A=\left(x-1\right)-2\sqrt{x-1}+1\)
\(A=\left(\sqrt{x-1}-1\right)^2\ge0\left(\forall x\ge1\right)\)
=> đpcm
H
1 tháng 10 2021
\(\sqrt{x+2\sqrt{x+1}}\)
\(\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)
\(\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(\left|\sqrt{x-1}+1\right|\)
6 tháng 6 2019
\(\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}\right)^2-2\cdot\sqrt{x-1}\cdot1+1^2}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|\)
\(C=1-\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1}\)
\(=1-\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{x-1}\)
\(=1-\left|\sqrt{x-1}-1\right|+\sqrt{x-1}\)
+) Nếu \(\sqrt{x-1}-1\ge0\Leftrightarrow x\ge2\) thì \(C=2\)
+) Nếu \(\sqrt{x-1}-1< 0\Leftrightarrow x< 2\Rightarrow1\le x< 2\) thì \(C=2\sqrt{x-1}\)