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a) ĐKXĐ: \(x>0;x\ne4\)
\(Q=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)
b) Để biểu thức \(Q\) có giá trị âm thì \(\dfrac{3\sqrt{x}}{\sqrt{x}-2}< 0\)
\(\Rightarrow\sqrt{x}-2< 0\) (vì \(3\sqrt{x}>0\forall x>0;x\ne4\))
\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\)
Kết hợp với điều kiện xác định của \(x\), ta được: \(0< x< 4\)
\(\text{#}\mathit{Toru}\)
a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\left(x+\sqrt{x}\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\left(x+\sqrt{x}\right)\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\cdot\left(\sqrt{x}+1\right)\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\)
\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}}{x-1}=\dfrac{2x}{x-1}\)
b: Để Q là số nguyên thì \(2x⋮x-1\)
=>\(2x-2+2⋮x-1\)
=>\(2⋮x-1\)
=>\(x-1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{2;0;3;-1\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;2;3\right\}\)
a) Ta có: \(Q=\left(\dfrac{x-1}{\sqrt{x}-1}-\dfrac{x\sqrt{x}-1}{x-1}\right):\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}-\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x-2\sqrt{x}+1+\sqrt{x}}{\sqrt{x}+1}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\right):\left(\dfrac{x-\sqrt{x}+1}{\sqrt{x}+1}\right)^2\)
\(=\dfrac{x+2\sqrt{x}+1-x-\sqrt{x}-1}{\sqrt{x}+1}:\dfrac{\left(x-\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{\left(x-\sqrt{x}+1\right)^2}\)
\(=\dfrac{x+\sqrt{x}}{\left(x-\sqrt{x}+1\right)^2}\)
\(a,A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(x>0;x\ne1\right)\\ A=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(b,\dfrac{P}{A}\left(x-1\right)=0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x=0\left(\sqrt{x}+1>0\right)\)
a) \(A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(đk:x>0,x\ne1\right)\)
\(=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b) \(\dfrac{P}{A}\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow x=0\)( do \(\sqrt{x}+1\ge1>0\))(không thỏa đk)
Vậy \(S=\varnothing\)
a: \(Q=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
a: \(Q=\left(\dfrac{4\sqrt{x}}{x+2\sqrt{x}}+\dfrac{8\sqrt{x}}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left(\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\left(\dfrac{4}{\sqrt{x}+2}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\dfrac{4\left(\sqrt{x}-2\right)-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\)
\(=\dfrac{-4\sqrt{x}-8}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{-\left(\sqrt{x}-3\right)}=\dfrac{4\sqrt{x}}{\sqrt{x}-3}\)
b: Q<4
=>Q-4<0
=>\(\dfrac{4\sqrt{x}}{\sqrt{x}-3}-4< 0\)
=>\(\dfrac{4\sqrt{x}-4\sqrt{x}+12}{\sqrt{x}-3}< 0\)
=>\(\dfrac{12}{\sqrt{x}-3}< 0\)
=>\(\sqrt{x}-3< 0\)
=>0<=x<9
Kết hợp ĐKXĐ, ta được: 0<x<9 và x<>4
\(a,Q=\left(\dfrac{4\sqrt{x}}{x+2\sqrt{x}}+\dfrac{8\sqrt{x}}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\\ =\left(\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2}{\sqrt{x}}\right)\\ =\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{4x-8\sqrt{x}-8x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\dfrac{-4x-8\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\\ =\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\\ =\dfrac{-4\sqrt{x}}{3-\sqrt{x}}\)
`b,` Để `Q<4` ta có :
\(\dfrac{-4\sqrt{x}}{3-\sqrt{x}}< 4\\ \Leftrightarrow\dfrac{-4\sqrt{x}}{3-\sqrt{x}}-4< 0\\ \Leftrightarrow\dfrac{-4\sqrt{x}-4\left(3-\sqrt{x}\right)}{3-\sqrt{x}}< 0\\ \Leftrightarrow-4\sqrt{x}-12+4\sqrt{x}< 0\\ \Leftrightarrow-12< 0\left(luon.dung\right)\)
a) \(ĐK:x>0,x\ne1\)\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2}{x-1}\)
b) \(P=\dfrac{2}{x-1}\in Z\)
\(\Rightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Kết hợp với đk
\(\Rightarrow x\in\left\{0;2;3\right\}\)
`P=(sqrtx/(sqrtx-1)+sqrtx/(x-1)):(2/x-(2-x)/(xsqrtx+x))`
`đk:x>0,x ne 1`
`P=((x+sqrtx+sqrtx)/(x-1)):(2/x+(x-2)/(x(sqrtx+1)))`
`=(x+2sqrtx)/(x-1):((2sqrtx+2+x-2)/(x(sqrtx+1)))`
`=(x+2sqrtx)/(x-1):(x+2sqrtx)/(x(sqrtx+1))`
`=(x+2sqrtx)/(x-1)*(x(sqrtx+1))/(x+2sqrtx)`
`=(x(sqrtx+1))/((sqrtx-1)(sqrtx+1))`
`=x/(sqrtx-1)`
`b)P>2`
`<=>x/(sqrtx-1)-2>0`
`<=>(x-2sqrtx+2)/(sqrtx-1)>0`
`<=>((sqrtx-1)^2+1)/(sqrtx-1)>0`
`<=>sqrtx-1>0`
`<=>x>1`
a) đk: x>0;x khác 1;0
P = \(\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\left[\dfrac{2}{x}-\dfrac{2-x}{x\left(\sqrt{x}+1\right)}\right]\)
= \(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{x+2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x}{\sqrt{x}-1}\)
b) Để P > 2
<=> \(\dfrac{x}{\sqrt{x}-1}-2>0\)
<=> \(\dfrac{x-2\sqrt{x}+2}{\sqrt{x}-1}>0\)
<=> \(\dfrac{\left(\sqrt{x}-1\right)^2+1}{\sqrt{x}-1}>0\)
<=> \(\sqrt{x}-1>0\)
<=> x > 1
a, x > 0 ; x khác 1
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\dfrac{1}{\sqrt{x}-1}\)
\(=\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{1}{\sqrt{x}-1}=\dfrac{x-2}{\sqrt{x}}\)
b, Ta có : \(P=\dfrac{x-2}{\sqrt{x}}=1\Rightarrow x-2=\sqrt{x}\)
\(\Leftrightarrow x-\sqrt{x}-2=0\Leftrightarrow\left(\sqrt{x}+1>0\right)\left(\sqrt{x}-2\right)=0\Leftrightarrow x=4\)(tm)
a: \(P=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}=\dfrac{x-2}{\sqrt{x}}\)
b: Để P=1 thì \(x-\sqrt{x}-2=0\)
hay x=4
Kết luận giùm mình nha ^^