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a: \(=\dfrac{27a^6b^3\cdot a^2b^6}{a^8b^8}=27b\)
b: \(=3y^2-5x^2y^3-2y^2+3x^2y^3\)
\(=y^2-2x^2y^3\)
c: \(=6x-y+2x^2+3y-2x^2+x\)
\(=7x+2y\)
d: \(=x-y+2y^2-6xy+\dfrac{10x^2}{y}\)
a: \(=3y^2-5x^2y^3-2y^2+3x^2y^3=y^2-2x^2y^3\)
b: \(=6x-y+2x^2+3y^2-2x^2+x=7x-y+3y^2\)
c: \(=x-y+4y^2-6xy+\dfrac{10x^2}{y}\)
\(a.\left(9x^2y^3-15x^4y^4\right):3x^2y-\left(2-3x^2y\right)y^2\)
\(=3y^2-5x^2y^3-2y^2+3x^2y^3\)
\(=y^2-2x^2y^3\)
\(b.\left(6x^2-xy\right):x+\left(2x^3y+3xy^2\right):xy-\left(2x-1\right)x\)
\(=6x-y+2x^2+3y-2+x\)
\(=2x^2+7x+2y-2\)
\(c.\left(x^2-xy\right):x+\left(6x^2y^5-9x^3y^4+15x^4y^3\right):\dfrac{3}{2}x^2y^3\)
\(=x-y+4y^2-6xy+10x^2\)
Bài 1:
a) \(\dfrac{15xy}{10x^2y}\)
= \(\dfrac{3.5xy}{2.5xyx}\)
= \(\dfrac{3}{2x}\)
d) \(\dfrac{6x\left(x+5\right)^3}{2x^2\left(x+5\right)}\)
= \(\dfrac{3.2x\left(x+5\right)\left(x+5\right)^2}{x.2x\left(x+5\right)}\)
= \(\dfrac{3\left(x+5\right)^2}{x}\)
$P=\dfrac{2}{x+3}+\dfrac{1}{x-3}+\dfrac{9-x}{9-x^2}$
$=\dfrac{2(x-3)+(x+3)}{(x+3)(x-3)}+\dfrac{9-x}{(3-x)(3+x)}$
$=\dfrac{3x-3}{x^2-9}+\dfrac{9-x}{9-x^2}$
$=\dfrac{3x-3}{x^2-9}-\dfrac{9-x}{x^2-9}$
$=\dfrac{3x-3-9+x}{x^2-9}$
$=\dfrac{4x-12}{x^2-9}$
$=\dfrac{4(x-3)}{(x-3)(x+3)}$
$=\dfrac{4}{x+3}$
Vậy $P=\dfrac{4}{x+3}$.
$(2x^4-3x^3-3x^2+6x-1):(x^2-2)$
$=2x^2-3x+1+\dfrac{4x+1}{x^2-2}$
Vậy thương là $2x^2-3x+1$, dư $4x+1$.
b)$(15x^4y^6-12x^3y^4-18x^2y^3):(-6x^2y^2)$
$=\dfrac{15x^4y^6}{-6x^2y^2}+\dfrac{-12x^3y^4}{-6x^2y^2}+\dfrac{-18x^2y^3}{-6x^2y^2}$
$=-\dfrac52x^2y^4+2xy^2+3y$
Vậy kết quả là $-\dfrac52x^2y^4+2xy^2+3y$.