Bài 1:
a: \(x^3-5x^2+8x-4\)
\(=x^3-x^2-4x^2+4x+4x-4\)
\(=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\cdot\left(x-2\right)^2\)
b: \(A⋮B\)
=>\(10x^2-7x-5⋮2x-3\)
=>\(10x^2-15x+8x-12+7⋮2x-3\)
=>\(7⋮2x-3\)
=>\(2x-3\in\left\{1;-1;7;-7\right\}\)
=>\(2x\in\left\{4;2;10;-4\right\}\)
=>\(x\in\left\{2;1;5;-2\right\}\)
c: x+y=1
=>y=1-x
\(\dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{2\left(x-y\right)}{x^2y^2+3}\)
\(=\dfrac{x}{\left(1-x\right)^3-1}-\dfrac{1-x}{x^3-1}+\dfrac{2\left(x-1+x\right)}{x^2\left(1-x\right)^2+3}\)
\(=\dfrac{x}{\left(1-x-1\right)\left[\left(1-x\right)^2+1\left(1-x\right)+1\right]}+\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2\left(2x-1\right)}{\left[x\left(1-x\right)\right]^2+3}\)
\(=\dfrac{x}{-x\cdot\left(x^2-2x+1+1-x+1\right)}+\dfrac{1}{x^2+x+1}+\dfrac{2\left(2x-1\right)}{\left(x^2-x\right)^2+3}\)
\(=\dfrac{-1}{x^2-3x+3}+\dfrac{1}{x^2+x+1}+\dfrac{2\left(2x-1\right)}{\left(x^2-x\right)^2+3}\)
\(=\dfrac{-x^2-x-1+x^2-3x+3}{\left(x^2-3x+3\right)\left(x^2+x+1\right)}+\dfrac{2\left(2x-1\right)}{\left(x^2-x\right)^2+3}\)
\(=\dfrac{-4x+2}{x^4+x^3+x^2-3x^3-3x^2-3x+3x^2+3x+3}+\dfrac{4x-2}{\left(x^2-x\right)^2+3}\)
\(=\dfrac{-4x+2}{x^4-2x^3+x^2+3}+\dfrac{4x-2}{x^4-2x^3+x^2+3}=0\)
Bài 2:
a: \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
=>\(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
=>\(\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)
mà \(x^2+x+6=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>=\dfrac{23}{4}>0\forall x\)
nên \(x^2+x-2=0\)
=>(x+2)(x-1)=0
=>\(\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
b: \(\dfrac{x+1}{2008}+\dfrac{x+2}{2007}+\dfrac{x+3}{2006}=\dfrac{x+4}{2005}+\dfrac{x+5}{2004}+\dfrac{x+6}{2003}\)
=>\(\left(\dfrac{x+1}{2008}+1\right)+\left(\dfrac{x+2}{2007}+1\right)+\left(\dfrac{x+3}{2006}+1\right)=\left(\dfrac{x+4}{2005}+1\right)+\left(\dfrac{x+5}{2004}+1\right)+\left(\dfrac{x+6}{2003}+1\right)\)
=>\(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}=\dfrac{x+2009}{2005}+\dfrac{x+2009}{2004}+\dfrac{x+2009}{2003}\)
=>\(\left(x+2009\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}-\dfrac{1}{2004}-\dfrac{1}{2003}\right)=0\)
=>x+2009=0
=>x=-2009
