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a: Ta có: \(\frac{8x\cdot\sqrt{x}-1}{2x-\sqrt{x}}-\frac{8x\cdot\sqrt{x}+1}{2x+\sqrt{x}}\)
\(=\frac{\left(2\sqrt{x}\right)^3-1}{\sqrt{x}\left(2\sqrt{x}-1\right)}-\frac{\left(2\sqrt{x}\right)^3+1}{\sqrt{x}\left(2\sqrt{x}+1\right)}\)
\(=\frac{\left(2\sqrt{x}-1\right)\left(4x+2\sqrt{x}+1\right)}{\sqrt{x}\left(2\sqrt{x}-1\right)}-\frac{\left(2\sqrt{x}+1\right)\left(4x-2\sqrt{x}+1\right)}{\sqrt{x}\left(2\sqrt{x}+1\right)}\)
\(=\frac{4x+2\sqrt{x}+1}{\sqrt{x}}-\frac{4x-2\sqrt{x}+1}{\sqrt{x}}=\frac{4\sqrt{x}}{\sqrt{x}}=4\)
Ta có: \(A=\left(\frac{8x\cdot\sqrt{x}-1}{2x-\sqrt{x}}-\frac{8x\cdot\sqrt{x}+1}{2x+\sqrt{x}}\right):\frac{2x+1}{2x-1}\)
\(=4\cdot\frac{2x-1}{2x+1}=\frac{8x-4}{2x+1}\)
b: Để A là số chính phương thì đầu tiên A phải là số tự nhiên
A là số tự nhiên khi \(\begin{cases}8x-4\vdots2x+1\\ \frac{8x-4}{2x+1}\ge0\end{cases}\Rightarrow\begin{cases}8x+4-8\vdots2x+1\\ \frac{2x-1}{2x+1}\ge0\end{cases}\)
=>\(\begin{cases}-8\vdots2x+1\\ \left[\begin{array}{l}x\ge\frac12\\ x<-\frac12\end{array}\right.\end{cases}\Rightarrow\begin{cases}2x+1\in\left\lbrace1;-1;2;-2;4;-4;8;-8\right\rbrace\\ \left[\begin{array}{l}x\ge\frac12\\ x<-\frac12\end{array}\right.\end{cases}\)
=>\(\begin{cases}2x\in\left\lbrace0;-2;1;-3;3;-5;7;-9\right\rbrace\\ \left[\begin{array}{l}x\ge\frac12\\ x<-\frac12\end{array}\right.\end{cases}\)
=>\(\begin{cases}x\in\left\lbrace0;-1;\frac12;-\frac32;\frac32;-\frac52;\frac72;-\frac92\right\rbrace\\ \left[\begin{array}{l}x\ge\frac12\\ x<-\frac12\end{array}\right.\end{cases}\)
=>x∈{-1;1/2;-3/2;3/2;-5/2;7/2;-9/2}
Kết hợp ĐKXĐ, ta được: x\(\in\left\lbrace\frac32;\frac72\right\rbrace\)
TH1: \(x=\frac32\)
=>2x=3
=>2X+1=4; 2x-1=2
\(A=\frac{8x-4}{2x+1}=4\cdot\frac{2x-1}{2x+1}=4\cdot\frac24=2\) không là số chính phương
=>Loại
TH2: \(x=\frac72\)
=>2x=7
=>2x+1=8; 2x-1=6
\(A=4\cdot\frac{2x-1}{2x+1}=4\cdot\frac68=\frac{24}{8}=3\) không là số chính phương
=>Loại
Vậy: x∈∅
Bài 1:
a: \(A=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{9x-1}\right):\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+5\sqrt{x}+1}{9x-1}:\dfrac{3}{3\sqrt{x}+1}\)
\(=\dfrac{3x+3\sqrt{x}}{9x-1}\cdot\dfrac{3\sqrt{x}+1}{3}=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
b: \(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{1}\cdot\dfrac{\sqrt{x}-1}{2}\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
a, Ta có : \(A=\left(\frac{x-\sqrt{x}+2}{x-1}-\frac{1}{\sqrt{x}-1}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{x-\sqrt{x}+2-\left(\sqrt{x}+1\right)}{x-1}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{x-2\sqrt{x}+1}{x-1}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{\left(\sqrt{x}-1\right)^2}{x-1}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\frac{x+2\sqrt{x}}{2x-2\sqrt{x}}\)
=> \(A=\frac{\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)}\frac{\left(x+2\sqrt{x}\right)}{\left(2x-2\sqrt{x}\right)}\)
=> \(A=\frac{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(2x-2\sqrt{x}\right)}\)
=> \(A=\frac{\left(\sqrt{x}-1\right)\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)2\sqrt{x}\left(\sqrt{x}-1\right)}\)
=> \(A=\frac{\sqrt{x}+2}{2\sqrt{x}+2}\)
b, Ta có : \(A=\frac{\sqrt{x}+1+1}{2\left(\sqrt{x}+1\right)}=\frac{1}{2}+\frac{1}{2\left(\sqrt{x}+1\right)}\)
- Ta thấy : \(\sqrt{x}+1>0\)
=> \(\frac{1}{2\left(\sqrt{x}+1\right)}>0\)
=> \(\frac{1}{2\left(\sqrt{x}+1\right)}+\frac{1}{2}>\frac{1}{2}\)
=> \(A>\frac{1}{2}\) ( đpcm )