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1,\(\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x}{x\left(2x+6\right)}+\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x+x-6}{x\left(2x+6\right)}\)=\(\frac{4x-6}{x\left(2x+6\right)}=\frac{2\left(2x-3\right)}{x\left(2x+6\right)}\)
a: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)
\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)
\(=x^2-2x+3\)
b: \(=\dfrac{x^5-3x^4+5x^3-x^2+3x-5}{x^2-3x+5}=x^2-1\)
c: \(=\dfrac{2x^4-5x^3+2x^2+2x-1}{x^2-x-1}\)
\(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)
\(=2x^2-3x+1\)
f)
$\frac{3x^2-2x}{x^2-1}.\frac{1-x^4}{(2-3x)^3}$
$=\frac{2x-3x^2}{x^2-1}.\frac{x^4-1}{(2-3x)^3}=\frac{x(2-3x)(x^2-1)(x^2+1)}{(x^2-1)(2-3x)^3}$
$=\frac{x(x^2+1)}{(2-3x)^2}$
g)
$\frac{5xy}{2x-3}:\frac{15xy^3}{12-8x}=\frac{5xy}{2x-3}.\frac{12-8x}{15xy^3}$
$=\frac{5xy}{2x-3}.\frac{-4(2x-3)}{15xy^3}=\frac{-4}{3y^2}$
h)
$\frac{x^2+2x}{3x^2-6x+3}:\frac{2x+4}{5x-5}=\frac{x(x+2)}{3(x-1)^2}:\frac{2(x+2)}{5(x-1)}$
$=\frac{x(x+2)}{3(x-1)^2}.\frac{5(x-1)}{2(x+2)}$
$=\frac{5x}{6(x-1)}$
d)
$\frac{x+8}{x^2-16}-\frac{2}{x^2+4x}=\frac{x+8}{(x-4)(x+4)}-\frac{2}{x(x+4)}$
$=\frac{x(x+8)}{x(x-4)(x+4)}-\frac{2(x-4)}{x(x+4)(x-4)}$
$=\frac{x^2+8x-2(x-4)}{x(x+4)(x-4)}=\frac{x^2+6x+8}{x(x+4)(x-4)}$
$=\frac{(x+2)(x+4)}{x(x+4)(x-4)}=\frac{x+2}{x(x-4)}$
e)
$\frac{x^2-49}{2x+1}.\frac{3}{7-x}=\frac{(x-7)(x+7)}{2x+1}.\frac{-3}{x-7}$
$=\frac{-3(x+7)}{2x+1}$
câu c:x^4-2x^3-x^2+x^3-2x^2-x+5x^2-10x-5=x^2(x^2-2x-1)+x(x^2-2x-1)+5(x^2-2x-1)=(x^2-2x-1)(x^2+x+5)
\(1,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2x+6}-\frac{x-6}{x\left(2x-6\right)}=\frac{3x-x+6}{x\left(2x-6\right)}=\frac{2x+6}{x\left(2x-6\right)}\)
\(2,\frac{1}{1-x}+\frac{2x}{x^2-1}=\frac{-1\left(x+1\right)+2x}{x^2-1}=\frac{x-1}{x^2-1}=\frac{1}{x+1}\)
\(3,\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
\(4,\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}=\frac{-5}{2}\)
\(5,\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2x\left(x+4\right)}\)
\(6,\frac{12x}{5y^3}.\frac{15y^4}{8x^3}=\frac{9y}{2x^2}\)
\(1.\)
\(x^2-2x+1-xy-y=\left(x-1\right)^2-y\left(x-1\right)=\left(x-1\right)\left(x-1-y\right)\)
\(2.\)
\(x^3-4x^2+4x-2x+2=x\left(x^2-4x+4\right)-2\left(x-1\right)=x\left(x-2\right)^2-2\left(x-1\right)\)
\(3.\)
\(10x-25-x^2+4y^2=4y^2-\left(x^2-10x+25\right)=4y^2-\left(x-5\right)^2=\left(2y+x-5\right)\left(2y-x+5\right)\)
\(4.\)
\(4x^2-2x+2xy-y=2x\left(2x-1\right)+y\left(2x-1\right)=\left(2x-1\right)\left(2x+y\right)\)
\(5.\)
\(4x\left(x-3\right)^2-3x^2+9x=4x\left(x-3\right)^2-3x\left(x-3\right)=\left(x-3\right)\left(4x^2-12x-3x\right)\)