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Đk:x \(\ge0\)
+) x không là số chính phương
=> \(\sqrt{x}\) là số vô tỉ (loại)
+) x là số chính phương
\(A=3+\dfrac{\sqrt{x}-5}{2\sqrt{x}+1}\)
Để A nhận giá trị nguyên dương
\(\Rightarrow\left(\sqrt{x}-5\right)⋮\left(2\sqrt{x}+1\right)\)
\(\Leftrightarrow\left(2\sqrt{x}-10\right)⋮\left(2\sqrt{x}+1\right)\)
\(\Leftrightarrow-11⋮\left(2\sqrt{x}+1\right)\)
\(\Rightarrow\left(2\sqrt{x}+1\right)\inƯ\left(11\right)=\left\{1;11\right\}\left(2\sqrt{x}+1>0\right)\)
| \(2\sqrt{x}+1\) | 1 | 11 |
| \(\sqrt{x}\) | 0 | 5 |
| \(x\) | 0 | 25 |
Thay vào => x=25
\(a,\)
\(=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3}{3\sqrt{x}+1}\right)\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+3x}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
Vậy \(P=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
\(b,\)Thay \(P=\dfrac{6}{5}\) vào pt, ta có :
\(\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}=\dfrac{6}{5}\)
\(\Leftrightarrow5\left(3\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)
\(\Leftrightarrow15\sqrt{x}+5-18\sqrt{x}+6=0\)
\(\Leftrightarrow-3\sqrt{x}+11=0\)
\(\Leftrightarrow-3\sqrt{x}=-11\)
\(\Leftrightarrow\sqrt{x}=\dfrac{11}{3}\)
\(\Leftrightarrow x=\left(\dfrac{11}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{121}{9}\)
Vậy \(x=\dfrac{121}{9}\) thì \(P=\dfrac{6}{5}\)
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(đk:x\ge0,x\ne1\right)\)
\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)
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 \(S=\left\{1;\dfrac{3-\sqrt{5}}{2}\right\}\)
a)ĐK: x khác 1; x>0
A=\(\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)+\(\frac{2\sqrt{x}}{x-1}\)-\(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)=\(\frac{\sqrt{x}-1+2x-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)=\(\frac{2}{\sqrt{x}}\)
b) Để A nhận giá trị nguyên thì \(\sqrt{x}\in\)Ước của 2=>\(\sqrt{x}=2;\sqrt{x}=-2\)(loại)=>x=4
a) A= \(\dfrac{\sqrt{x}}{\sqrt{x-2}}-\dfrac{4}{x-2\sqrt{x}}=\dfrac{\sqrt{x}\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\sqrt{x}}=\dfrac{x+2\sqrt{x}}{x}\)
b) Ta có x >0 nên \(\sqrt{x}\) >0
<=> \(2\sqrt{x}\) > 0
<=> \(x+2\sqrt{x}\) > x
<=> \(\dfrac{x+2\sqrt{x}}{x}\) > \(\dfrac{x}{x}\)
hay A > 1
c)
Ta có: \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
Để A nguyên thì \(\sqrt{x}⋮\sqrt{x}-2\)
\(\Leftrightarrow2⋮\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}-2\in\left\{-2;-1;1;2\right\}\)
Vậy: Có 4 giá trị nguyên của x thỏa mãn yêu cầu đề bài
