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a,Ta có \(x=4-2\sqrt{3}=\sqrt{3}^2-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)do \(\sqrt{3}-1>0\)
\(\Rightarrow A=\frac{1}{\sqrt{3}-1-1}=\frac{1}{\sqrt{3}-2}\)
b, Với \(x\ge0;x\ne1\)
\(B=\left(\frac{-3\sqrt{x}}{x\sqrt{x}-1}-\frac{1}{1-\sqrt{x}}\right):\left(1-\frac{x+2}{1+\sqrt{x}+x}\right)\)
\(=\left(\frac{-3\sqrt{x}+x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x+\sqrt{x}+1-x-2}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}.\frac{x+\sqrt{x}+1}{\sqrt{x}-1}=1\)
Vậy biểu thức ko phụ thuộc biến x
c, Ta có : \(\frac{2A}{B}\)hay \(\frac{2}{\sqrt{x}-1}\)để biểu thức nhận giá trị nguyên
thì \(\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| \(\sqrt{x}-1\) | 1 | -1 | 2 | -2 |
| \(\sqrt{x}\) | 2 | 0 | 3 | -1 |
| x | 4 | 0 | 9 | vô lí |
1: Thay x=16 vào A, ta được:
\(A=\dfrac{6-2\cdot4}{4-5}=\dfrac{-2}{-1}=2\)
a: Thay x=36 vào B, ta được:
\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)
\(B=\left[\dfrac{\sqrt{x-2}}{\left(\sqrt{x}-1\right)^2}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)
\(=\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)
\(=\left[\dfrac{x+\sqrt{x}-2\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)
\(=\left[\dfrac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)
\(=\dfrac{-2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\dfrac{2x}{x-1}\)
b/
\(B=-\dfrac{2\left(x-1\right)+2}{x-1}=-2+\dfrac{2}{x-1}\)
Để B nguyên
\(x-1=\left\{-1;-2;1;2\right\}\Rightarrow x=\left[0;-1;2;3\right]\)
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(đk:x\ge0,x\ne1\right)\)
\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)
Để A nguyên thì: \(x+\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Mà \(x+\sqrt{x}+1=\left(x+\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
\(\Rightarrow x+\sqrt{x}+1\in\left\{1;2\right\}\)
+ Với \(x+\sqrt{x}+1=1\)
\(\Leftrightarrow\sqrt[]{x}\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow x=0\left(tm\right)\left(do.\sqrt{x}+1\ge1>0\right)\)
+ Với \(x+\sqrt{x}+1=2\)
\(\Leftrightarrow\left(x+\sqrt{x}+\dfrac{1}{4}\right)=\dfrac{5}{4}\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\\\sqrt{x}+\dfrac{1}{2}=-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{5}-1}{2}\\\sqrt{x}=-\dfrac{\sqrt{5}+1}{2}\left(VLý\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{3-\sqrt{5}}{2}\left(tm\right)\)
Vậy \(S=\left\{1;\dfrac{3-\sqrt{5}}{2}\right\}\)
\(P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=x-\sqrt{x}+1\)
\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^3+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=4\end{matrix}\right.\) \(\Rightarrow a+b=7\)