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a: ĐKXĐ: x>=0; x<>1
b \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{1}{\sqrt{x}+2}\)
c: Khi x=9-4 căn 5 thì \(A=\dfrac{1}{\sqrt{5}-2+2}=\dfrac{\sqrt{5}}{5}\)
d: căn x+2>=2
=>A<=1/2
Dấu = xảy ra khi x=0
a: ĐKXĐ: x>=0; x<>1
\(A=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\)
\(=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}=\dfrac{x+\sqrt{x}}{x-1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b: Khi x=9/4 thì A=3/2:1/2=3/2*2=3
a: ĐKXĐ: a>=0; a<>1
b: \(A=\dfrac{a+\sqrt{a}-2-a+\sqrt{a}+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot\dfrac{\sqrt{a}+1}{1}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}-1}\)
c: Để A là số nguyên thì \(2\sqrt{a}-2+2⋮\sqrt{a}-1\)
=>\(\sqrt{a}-1\in\left\{1;-1;2\right\}\)
=>\(a\in\left\{4;0;9\right\}\)
a) Ta có: \(M=\left(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\right)\cdot\dfrac{x+3\sqrt{x}}{7-\sqrt{x}}\)
\(=\left(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)
\(=\dfrac{x-9-\left(x-2\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)
\(=\dfrac{x-9-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{-\left(\sqrt{x}-7\right)}\)
\(=\dfrac{\sqrt{x}-7}{\sqrt{x}-2}\cdot\dfrac{-1}{\sqrt{x}-7}\)
\(=\dfrac{-1}{\sqrt{x}-2}\)(1)
b) Ta có: \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(loại\right)\end{matrix}\right.\)
Thay x=0 vào biểu thức (1), ta được:
\(M=\dfrac{-1}{\sqrt{0}-2}=\dfrac{-1}{-2}=\dfrac{1}{2}\)
Vậy: Khi \(x^2-4x=0\) thì \(M=\dfrac{1}{2}\)
Nhìn mãi mới hiểu cái đề bài @-@
`a)đk:` $\begin{cases}\sqrt{x^2-2x} \ge 0\\x+\sqrt{x^2-2x} \ne 0\\x-\sqrt{x^2-2x} ne 0\\\end{cases}$
`<=>` $\begin{cases}x \ge 2\,or\,x<0\\x \ne 0\end{cases}$
`b)A=(x+sqrt{x^2-2x})/(x-sqrt{x^2-2x})-(x-sqrt{x^2-2x})/(x+sqrt{x^2+2x})`
`=((x+sqrt{x^2-2x})^2-(x-sqrt{x^2-2x})^2)/((x+sqrt{x^2-2x})(x-sqrt{x^2-2x}))`
`=(x^2+x^2-2x+2sqrt{x^2-2x}-x^2-x^2+2x+2sqrt{x^2-2x})/(x^2-x^2+2x)`
`=(4sqrt{x^2-2x})/(2x)`
`=(2sqrt{x^2-2x})/x`
`c)A<2`
`<=>2sqrt{x^2-2x}<2x`
`<=>sqrt{x^2-2x}<x(x>=2)`(BP 2 vế thì x>=2)
`<=>x^2-2x<x^2`
`<=>2x>0`
`<=>x>0`
`<=>x>=2`
Vậy `x>=2` thì `A<2`.
\(a,ĐK:x\ne\pm1;x\ne0\\ M=\dfrac{1-x+2x}{\left(1+x\right)\left(1-x\right)}:\dfrac{1-x}{x}\\ M=\dfrac{x+1}{\left(x+1\right)\left(1-x\right)}\cdot\dfrac{x}{1-x}=\dfrac{x}{\left(1-x\right)^2}\\ b,ĐK:x\ge0;x\ne4\\ N=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ N=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
Tất cả đều phải tìm điều kiện
Câu 5: B
Câu 3:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne4\end{matrix}\right.\)
b: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right):\dfrac{2\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\cdot\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)
c: Để P>4 thì \(\sqrt{x}>4\)
=>x>16
a: ĐKXĐ: a>=0; a<>4
b: \(M=\dfrac{a\sqrt{a}-a\sqrt{a}+2a-a-2\sqrt{a}}{a-4}=\dfrac{a-2\sqrt{a}}{a-4}=\dfrac{\sqrt{a}}{\sqrt{a}+2}\)
c: Khi a=9 thì \(M=\dfrac{3}{3+2}=\dfrac{3}{5}\)