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`a, 20x^3y^5 : 5x^2y^2`
`= (20:5)x^(3-2) . y^(5-2)`
`= 4xy^3`
`b, 18x^3y^5 : (3(-x^3)y^2)`
`= -(18:3)y^(5-3)`
`= -6y^2`
a: \(\left(4x^5-8x^3\right):\left(-2x^3\right)\)
\(=-4x^5:2x^3+8x^3:2x^3\)
\(=-2x^2+4\)
b: \(\left(9x^3-12x^2+3x\right):\left(-3x\right)\)
\(=-\frac{9x^3}{3x}+\frac{12x^2}{3x}-\frac{3x}{3x}\)
\(=-3x^2+4x-1\)
c: \(\left(xy^2+4x^2y^3-3x^2y^4\right):\left(-\frac12x^2y^3\right)\)
\(=-xy^2:\frac12x^2y^3-4x^2y^3:\frac12x^2y^3+3x^2y^4:\frac12x^2y^3\)
\(=-\frac{2}{xy}-8+6y\)
d: \(\left\lbrack2\left(x-y\right)^3-7\left(y-x\right)^2-\left(y-x\right)\right\rbrack:\left(x-y\right)\)
\(=\left\lbrack2\left(x-y\right)^3-7\left(x-y\right)^2+\left(x-y\right)\right\rbrack:\left(x-y\right)\)
\(=\frac{2\left(x-y\right)^3}{x-y}-\frac{7\left(x-y\right)^2}{x-y}+\frac{\left(x-y\right)}{x-y}\)
\(=2\left(x-y\right)^2-7\left(x-y\right)+1\)
BÀi 3:
a: \(y\times12,35+y\times12,65=875\)
=>\(y\times\left(12,35+12,65\right)=875\)
=>\(y\times25=875\)
=>y=875:25
=>y=35
b: \(y+y\times2,7+y\times6,3=120\)
=>\(y\times\left(1+2,7+6,3\right)=120\)
=>\(y\times10=120\)
=>y=120:10
=>y=12
a) (x^2+2xy+y^2) : (x+y)
=(x+y)2:(x+y)
=x+y
b) (125x^3+1) : (5x+1)
=(5x+1)(25x2-5x+1):(5x+1)
=25x2-5x+1
c) (x^2-2xy+y^2) : (y-x)
=(x-y)2:(y-x)
=-(x-y)2:(x-y)
=-(x-y)
=-x+y
a)\(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4=x^4-y^4\)
b) \(x\left(3x-18\right)-3\left(x-4\right)\left(x-2\right)+8=3x^2-18x-3x^2+18x-24+8=-16\)
Ta có: \(\dfrac{y}{x-y}-\dfrac{x^3-xy^2}{x^2+y^2}\cdot\left(\dfrac{x}{x^2-2xy+y^2}-\dfrac{y}{x^2-y^2}\right)\)
\(=\dfrac{y}{x-y}-\dfrac{x\left(x^2-y^2\right)}{x^2+y^2}\cdot\left(\dfrac{x\left(x+y\right)}{\left(x-y\right)^2\cdot\left(x+y\right)}-\dfrac{y\cdot\left(x-y\right)}{\left(x-y\right)^2\cdot\left(x+y\right)}\right)\)
\(=\dfrac{y}{x-y}-\dfrac{x\left(x-y\right)\left(x+y\right)}{x^2+y^2}\cdot\dfrac{x^2+xy-xy+y^2}{\left(x-y\right)^2\left(x+y\right)}\)
\(=\dfrac{y}{x-y}-\dfrac{x\cdot\left(x^2+y^2\right)}{\left(x^2+y^2\right)\cdot\left(x-y\right)}\)
\(=\dfrac{y}{x-y}-\dfrac{x}{x-y}\)
\(=\dfrac{y-x}{x-y}=\dfrac{-\left(x-y\right)}{x-y}=-1\)
Chọn đáp án B