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a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
a: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b:Sửa đề: 2A
2A=2căn x+5
=>(2căn x+2)/căn x=2căn x+5
=>2x+5căn x-2căn x-2=0
=>2x+3căn x-2=0
=>(căn x+2)(2căn x-1)=0
=>x=1/4
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
a: ĐKXĐ: x>=0; x<>1
b: \(A=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
c: x>=0
=>-x<=0
=>-x+1<=1
Dấu = xảy ra khi x=0
a,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
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!
T không làm được à?
dù ko đáp ứng điều kiện nhưng có ai cấm làm đâu
a) \(A=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-x}{\sqrt{x}-1}\right)\left(1+\dfrac{1}{\sqrt{x}}\right)\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left(1+\dfrac{1}{\sqrt{x}}\right)\)
\(=\left(\sqrt{x}+\sqrt{x}\right)\left(1+\dfrac{1}{\sqrt{x}}\right)\)
\(=2\sqrt{x}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=2\left(\sqrt{x}+1\right)=2\sqrt{x}+2\)
b) A = 4 \(\Leftrightarrow\) \(2\sqrt{x}+2=4\)
\(\Leftrightarrow2\sqrt{x}=2\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
c) Để \(\dfrac{3}{A}\) là số nguyên thì \(3⋮A\)
hay A \(\in\) Ư(3) = {1;-1;3;-3}
*\(2\sqrt{x}+2=1\Leftrightarrow2\sqrt{x}=-1\)(loại)
*\(2\sqrt{x}+2=-1\Leftrightarrow2\sqrt{x}=-3\)(loại)
*\(2\sqrt{x}+2=3\Leftrightarrow2\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{\sqrt{2}}\)(nhận)
*\(2\sqrt{x}+2=-3\Leftrightarrow2\sqrt{x}=-5\) (loại)
Vậy x = \(\dfrac{1}{\sqrt{2}}\).