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a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(M=\frac{\sqrt{x}}{\sqrt{x}-x}-\frac{\sqrt{x}+2}{1-x}\)
\(\Leftrightarrow M=\frac{1}{1-\sqrt{x}}-\frac{\sqrt{x}+2}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
\(\Leftrightarrow M=\frac{1+\sqrt{x}-\sqrt{x}-2}{1-x}\)
\(\Leftrightarrow M=\frac{-1}{1-x}\)
\(\Leftrightarrow M=\frac{1}{x-1}\)
b) Để M nhận giá trị nguyên
\(\Leftrightarrow\frac{1}{x-1}\inℤ\)
\(\Leftrightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{0;2\right\}\)
Mà \(x>0\)
Vậy để M nguyên \(\Leftrightarrow x=2\)
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
Vậy ...
a) \(P=\dfrac{x-1+4\left(\sqrt{x}+1\right)+1}{x-1}.\dfrac{x-1}{x+2\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}+4}{x+2\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
b) \(P=\dfrac{\sqrt{x}+2}{\sqrt{x}}=1+\dfrac{2}{\sqrt{x}}\in Z\)
Do \(\sqrt{x}>0\)
\(\Rightarrow\sqrt{x}\inƯ\left(2\right)=\left\{1;2\right\}\)
\(\Rightarrow x\in\left\{1;4\right\}\)
a/ \(P=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(x-1\right)+x-1}\right]:\left[\frac{1}{\sqrt{x}-1}-\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\left[\frac{1}{\sqrt{x}+1}-\frac{2}{\left(\sqrt{x}+1\right)^2}\right]:\left[\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(=\frac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b/ Ta có: \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Để \(P\in Z\) thì \(\left(\sqrt{x}+1\right)\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
+ Với \(\sqrt{x}+1=1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
+ Với \(\sqrt{x}+1=-1\Rightarrow\sqrt{x}=-2\left(vn\right)\)
+ Với \(\sqrt{x}+1=2\Rightarrow\sqrt{x}=1\Rightarrow x=1\)(loại)
+ Với \(\sqrt{x}+1=-2\Rightarrow\sqrt{x}=-3\left(vn\right)\)
Vậy x = 0 thì P nguyên
a) \(P=\left(\frac{1}{\sqrt{x}+1}-\frac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2}{x-1}\right)\)
\(=\frac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}:\frac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\frac{x-1}{\sqrt{x}-1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Để P nguyên thì \(\sqrt{x}+1\in\left\{1;2\right\}\Leftrightarrow x\in\left\{0\right\}\) (Vì x khác 1 - điều kiện)
c) \(\sqrt{x}+1\ge1\Leftrightarrow\frac{2}{\sqrt{x}+1}\le\frac{1}{2}\Leftrightarrow1-\frac{2}{\sqrt{x}+1}\ge\frac{1}{2}\)
\(\Rightarrow P\ge\frac{1}{2}\). Dấu đẳng thức xảy ra khi x = 0
Vậy Min P = 1/2 <=> x = 0
a: Sửa đề: \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right):\dfrac{2}{x^2-2x+1}\)
\(=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)^2}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{\sqrt{x}-1}\cdot\dfrac{1}{2}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)
b: Để P>0 thì \(-\dfrac{\sqrt{x}}{\sqrt{x}-1}>0\)
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-1}< 0\)
=>\(\sqrt{x}< 1\)
=>\(0< =x< 1\)
c: Thay \(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\) vào P, ta được:
\(P=\dfrac{-\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2-\sqrt{3}\right)^2}-1}\)
\(=\dfrac{-\left(2-\sqrt{3}\right)}{2-\sqrt{3}-1}=\dfrac{-2+\sqrt{3}}{1-\sqrt{3}}=\dfrac{2-\sqrt{3}}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{3}-1}{2}\)
a: \(P=\left(\dfrac{2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)
hay \(x\in\left\{0;4;9\right\}\)
\(B=\left[\dfrac{\sqrt{x-2}}{\left(\sqrt{x}-1\right)^2}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)
\(=\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)
\(=\left[\dfrac{x+\sqrt{x}-2\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)
\(=\left[\dfrac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)
\(=\dfrac{-2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\dfrac{2x}{x-1}\)
b/
\(B=-\dfrac{2\left(x-1\right)+2}{x-1}=-2+\dfrac{2}{x-1}\)
Để B nguyên
\(x-1=\left\{-1;-2;1;2\right\}\Rightarrow x=\left[0;-1;2;3\right]\)