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a)\(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}\)
\(\frac{-5}{2\left(2+y\right)}+\frac{y-2}{y\left(2+y\right)}\)
\(\frac{-5y}{2y\left(2+y\right)}+\frac{2y-4}{2y\left(2+y\right)}\)
\(\frac{-5y+2y-4}{2y\left(2+y\right)}\)
\(\frac{-3y-4}{2y\left(2+y\right)}\)
b)\(\frac{x-1}{x^2-2xy}+\frac{3}{2xy-x^2}\)
\(\frac{x-1}{x\left(x-2y\right)}+\frac{3}{x\left(2y-x\right)}\)
\(\frac{x-1}{x\left(x-2y\right)}+\frac{-3}{x\left(x-2y\right)}\)
\(\frac{x-1-3}{x\left(x-2y\right)}\)
\(\frac{x-4}{x\left(x-2\right)}\)
1. \(x^3-x^2+x-1=(x^3-x^2)+(x-1)\)
\(=x^2(x-1)+(x-1)=(x^2+1)(x-1)\)
2. \(6x^2y-2xy^2+3x-y=2xy(3x-y)+(3x-y)\)
\(=(3x-y)(2xy+1)\)
3. \(4x^2+1\) thì còn cái gì để phân tích hả bạn? Hay ý bạn là \(4x^4+1\)?
\(4x^4+1=(2x^2)^2+1=(2x^2)^2+1+4x^2-4x^2\)
\(=(2x^2+1)^2-(2x)^2=(2x^2+1-2x)(2x^2+1+2x)\)
4. \(x^2-9x+8=(x^2-x)-(8x-8)\)
\(=x(x-1)-8(x-1)=(x-1)(x-8)\)
5. \(x^3-2x^2y+3xy^2=x(x^2-2xy+3y^2)\)
6. \(x^2-6x+y-y^2\) (sai đề)
7. \(x^2-xy-2x+2y=(x^2-xy)-(2x-2y)\)
\(=x(x-y)-2(x-y)=(x-y)(x-2)\)
1, \(x^2+4x-2xy-4y+y^2=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)=\left(x-y\right)^2+4\left(x-y\right)=\left(x-y\right)\left(x-y+4\right)\)
2, \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
3, \(2x^2+4x+2-2y^2=2\left(x^2-y^2\right)+2\left(2x+1\right)=2\left(x^2+2x+1-y^2\right)=2\left[\left(x+1\right)^2-y^2\right]=2\left(x+1-y\right)\left(x+1+y\right)\)
4, \(x^4-2x^2=x^4-2x^2+1-1=\left(x^2-1\right)^2-1=\left(x^2-1-1\right)\left(x^2-1+1\right)=\left(x^2-2\right)x^2\)
5, \(x^3+2x^2y+xy^2-9x=x\left(x^2+2xy+y^2-9\right)=x\left[\left(x+y\right)^2-3^2\right]=x\left(x+y-3\right)\left(x+y+3\right)\)
6, \(x^3-\frac{1}{4}x=x\left(x^2-\frac{1}{4}\right)=x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\)
7, \(2x-2y-x^2+2xy-y^2=\left(2x-2y\right)-\left(x^2-2xy+y^2\right)=2\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(2-x+y\right)\)
8, \(\left(2x+3\right)^2-\left(x+1\right)^2=\left(2x+3+x+1\right)\left(2x+3-x-1\right)=\left(3x+4\right)\left(x+2\right)\)
Lời giải:
a) \(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}=\frac{-5}{2(y+2)}+\frac{y-2}{y(y+2)}=\frac{-5y}{2y(y+2)}+\frac{2(y-2)}{2y(y+2)}\)
\(=\frac{-5y+2(y-2)}{2y(y+2)}=\frac{-(3y+4)}{2y(y+2)}\)
b)
\(\frac{x-1}{x^2-2xy}+\frac{3}{2xy-x^2}=\frac{x-1}{x^2-2xy}-\frac{3}{x^2-2xy}=\frac{x-1-3}{x^2-2xy}=\frac{x-4}{x(x-2y)}\)
2xy+x2y4 +1
=2x3 y5 +1
=3x3y5