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a: \(A=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\left(\dfrac{x\left(x^2-4x+4\right)+4x^2}{2\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3-4x^2+4x+4x^2}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)
b: Để A là số nguyên thì x+1 chia hết cho 2x
=>2x+2 chia hết cho 2x
=>2 chia hết cho 2x
=>2x=2
=>x=1(nhận)
a: \(\left[\dfrac{1}{2}x^2\left(2x-1\right)^m-\dfrac{1}{2}x^{m+2}\right]:\dfrac{1}{2}x^2=0\)
\(\Leftrightarrow\left(2x-1\right)^m-x^m=0\)
\(\Leftrightarrow\left(2x-1\right)^m=x^m\)
=>2x-1=x
=>x=1
b: \(\left(2x-3\right)^8=\left(2x-3\right)^6\)
\(\Leftrightarrow\left(2x-3\right)^6\cdot\left(2x-4\right)\left(2x-2\right)=0\)
hay \(x\in\left\{\dfrac{3}{2};2;1\right\}\)
c: \(\Leftrightarrow4x^2-4x+1+y^2-\dfrac{2}{3}y+\dfrac{1}{9}+\dfrac{6}{9}=0\)
\(\Leftrightarrow\left(2x-1\right)^2+\left(y-\dfrac{1}{3}\right)^2+\dfrac{6}{9}=0\)(vô lý)
a: \(M=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\left(\dfrac{x\left(x-2\right)^2+4x^2}{2\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3-4x^2+4x+4x^2}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)
b: Thay x=1/2 vào M, ta được:
\(M=\left(\dfrac{1}{2}+1\right):\left(2\cdot\dfrac{1}{2}\right)=\dfrac{3}{2}\)
Câu a :
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)
\(A=\left(\dfrac{x^2-2x}{2x^2+8x}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{\left(x^2-2x\right)\left(x-2\right)+2.2x^2}{2\left(x-2\right)\left(x^2+4\right)}\right)\left(\dfrac{x^2-x-2}{x^2}\right)\)
\(=\dfrac{x}{2\left(x-2\right)}\times\dfrac{\left(x+1\right)\left(x-2\right)}{x^2}\)
\(=\dfrac{x+1}{2x}\)
Câu b : Dễ rồi
b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{-3x^2-2x+1}{3x}\right)\cdot\dfrac{x}{x+1}\)
\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)
\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)
c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)
\(=\left(\dfrac{2}{x\left(x+1\right)^2}+\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}\right)\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{2x+x^2+1}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}\cdot\dfrac{x}{x-1}=\dfrac{x}{x-1}\)
a) Ta có: \(P=\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{4\left(2-x\right)+x^2\left(2-x\right)}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(2-x\right)\left(x^2+4\right)}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\left(\dfrac{\left(x^2-2x\right)\left(x-2\right)}{2\left(x-2\right)\left(x^2+4\right)}+\dfrac{4x^2}{2\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\dfrac{x^3-x^2-2x^2+4x+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)\)
\(=\dfrac{x^3+x^2+4x}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{x\left(x^2+x+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{\left(x^2+x+4\right)\left(x+1\right)}{2x\left(x^2+4\right)}\)
\(=\left[\dfrac{2x-3}{\left(2x-5\right)\left(2x-1\right)}-\dfrac{3}{2x-1}-\dfrac{2\left(x-4\right)}{\left(x-4\right)\left(2x-5\right)}\right].\dfrac{2x\left(2x+3\right)-\left(2x+3\right)}{-2x\left(4x-7\right)-3\left(4x-7\right)}+1\)
\(=\left[\dfrac{2x-3-6x+15-4x+2}{\left(2x-5\right)}\right].\dfrac{2\left(x+\dfrac{3}{2}\right)}{\left(-2x-3\right)\left(4x-7\right)}+1\)
\(=\dfrac{-2\left(4x-7\right)}{2x-5}.\dfrac{2\left(x+\dfrac{3}{2}\right)}{\left(-2x-3\right)\left(4x-7\right)}+1\)
\(=\dfrac{1}{2x-5}.2+1\)
\(=\dfrac{2+2x-5}{2x-5}\)
\(=\dfrac{-3+2x}{2x-5}\)
a: \(M=\left[\dfrac{x^2-2x+1}{x^2+x+1}+\dfrac{2x^2-4x-1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right]\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^2+1}{2}\)
a: ĐKXĐ: x<>2; x<>0
b: \(M=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3-2x^2-2x^2+4x}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x}{2}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)
c: M>=-3
=>(x+1+6x)/2x>=0
=>(7x+1)/x>=0
=>x>0 hoặc x<=-1/7