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a,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
ĐKXĐ : \(x\ne0,x\ne\pm2\)
Câu a :
\(A=\left(\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}+\dfrac{1}{x+2}\right).\left(\dfrac{2}{x}-1\right)\)
\(=\dfrac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}.\left(\dfrac{2}{x}-1\right)\)
\(=\dfrac{4x}{\left(x-2\right)\left(x+2\right)}\times\dfrac{2-x}{x}\)
\(=-\dfrac{4}{x+2}\)
Câu b :
Ta có : \(2x^2+x=0\Leftrightarrow x\left(2x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Thay \(x=0\) vào A ta được \(-\dfrac{4}{0+2}=-2\)
Thay \(x=-\dfrac{1}{2}\) vào A ta được \(-\dfrac{4}{-\dfrac{1}{2}+2}=-\dfrac{8}{3}\)
Câu c :
Để \(A=\dfrac{1}{2}\) thì \(-\dfrac{4}{x+2}=\dfrac{1}{2}\)
\(\Leftrightarrow x+2=-8\Leftrightarrow x=-10\)
Câu d :
Để A nguyên dương thì \(-4⋮x+2\)
Xét :
\(Ư\left(-4\right)=-4;-2;-1;1;2;4\)
\(\left\{{}\begin{matrix}x+2=-4\\x+2=-2\\x+2=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-6\left(N\right)\\x=-4\left(N\right)\\x=-3\left(N\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+2=1\\x+2=2\\x+2=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\left(N\right)\\x=0\left(L\right)\\x=2\left(L\right)\end{matrix}\right.\)
Vậy có 4 giá trị của x thì A nguyên : \(\left\{{}\begin{matrix}x=-6\\x=-4\\x=-3\\x=-1\end{matrix}\right.\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-3\end{cases}}\)
b) \(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{\left(x-2\right)\left(x+2\right)}{6\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{x+4}{6}\)
c) Để P = 0
\(\Leftrightarrow\frac{x+4}{6}=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Để P = 1
\(\Leftrightarrow\frac{x+4}{6}=1\)
\(\Leftrightarrow x+4=6\)
\(\Leftrightarrow x=2\)
d) Để P > 0
\(\Leftrightarrow\frac{x+4}{6}>0\)
\(\Leftrightarrow x+4>0\)(Vì 6>0)
\(\Leftrightarrow x>-4\)
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
1: ĐKXĐ: \(x\notin\left\{0;-1;1\right\}\)
2: \(P=\left(\dfrac{x^2-2x+1}{x^2+x+1}+\dfrac{2x^2-4x-1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right)\cdot\dfrac{x^2+1}{x+1}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{x+1}\)
\(=\dfrac{2x^3-x^2-x-3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{x+1}\)
Để P=0 thì \(2x^3-x^2-x-3=0\)
=>x=3/2
a ) \(D=\left(\dfrac{1}{1-x}+\dfrac{1}{1+x}\right):\left(\dfrac{1}{1-x}-\dfrac{1}{1+x}\right)+\dfrac{1}{x+1}\)
\(=\left(\dfrac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}\right):\left(\dfrac{1+x-1+x}{\left(1-x\right)\left(1+x\right)}\right)+\dfrac{1}{x+1}\)
\(=\dfrac{2}{\left(1-x\right)\left(1+x\right)}:\dfrac{2x}{\left(1-x\right)\left(1+x\right)}+\dfrac{1}{x+1}\)
\(=\dfrac{2}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(1-x\right)\left(1+x\right)}{2x}+\dfrac{1}{x+1}\)
\(=\dfrac{1}{x}+\dfrac{1}{x+1}\)
\(=\dfrac{x+1+x}{x\left(x+1\right)}=\dfrac{2x+1}{x\left(x+1\right)}\)
b ) Khi \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Thay 0,1 vào biểu thức D
Khi \(x=0\), ta có :
\(\dfrac{2.0+1}{0.\left(0+1\right)}\) ( ko được )
Khi \(x=1,\) ta có :
\(\dfrac{2.1+1}{1.\left(1+1\right)}=\dfrac{3}{2}\)
c ) Khi \(D=\dfrac{3}{2}\)
Ta có : \(\dfrac{2x+1}{x\left(x+1\right)}=\dfrac{3}{2}\)
\(\Leftrightarrow4x+2=3x^2+3x\)
\(\Leftrightarrow-3x^2+x+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy ...........