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a: \(=-xy\cdot x^2-2xy\cdot xy+3\cdot xy\)
\(=-x^3y-2x^2y^2+3xy\)
b: \(=\left(5x-x+4y-4y\right)\left(y-5x\right)\)
\(=4x\left(y-5x\right)=-20x^2+4xy\)
c: \(=\left(4x^2-2y\right)\left(5x^3-4y^2\right)\)
\(=4x^2\cdot5x^3-4x^2\cdot4y^2-2y\cdot5x^3+2y\cdot4y^2\)
\(=20x^5-16x^2y^2-10x^3y+8y^3\)
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Lời giải:
a.
$A=20x^3-10x^2+5x-(20x^3-10x^2-4x)$
$=9x=9.15=135$
b.
$B=(5x^2-20xy)-(4y^2-20xy)=5x^2-4y^2$
$=5(\frac{-1}{5})^2-4(\frac{-1}{2})^2=\frac{-4}{5}$
c.
$C=(6x^2y^2-6xy^3)-(8x^3-8x^2y^2)-(5x^2y^2-5xy^3)$
$=-8x^3+9x^2y^2-xy^3$
$=(-2x)^3+(3xy)^2-xy^3$
$=(-2.\frac{1}{2})^3+(3.\frac{1}{2}.2)^2-\frac{1}{2}.2^3$
$=(-1)^3+3^2-4=4$
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\(C=\left(x^2+\dfrac{y^2}{4}+4-xy+4x-2y\right)+\dfrac{3}{4}\left(y^2-4y+4\right)+1011\)
\(=\left(x-\dfrac{y}{2}+2\right)^2+\dfrac{3}{4}\left(y-2\right)^2+1011\ge1011\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(-1;2\right)\)
a) Ta có: \(B=x^2+4y^2+4x-4y\)
\(=\left(x^2+4x+4\right)+\left(4y^2-4y+1\right)-5\)
\(=\left(x+2\right)^2+\left(2y-1\right)^2-5\ge-5\forall x,y\)
Dấu '=' xảy ra khi \(\left(x,y\right)=\left(-2;\dfrac{1}{2}\right)\)
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a: 2x^2y-50xy=2xy(x-25)
b: 5x^2-10x=5x(x-2)
c: 5x^3-5x=5x(x^2-1)=5x(x-1)(x+1)
d: \(x^2-xy+x=x\left(x-y+1\right)\)
e: x(x-y)-2(y-x)
=x(x-y)+2(x-y)
=(x-y)(x+2)
f: 4x^2-4xy-8y^2
=4(x^2-xy-2y^2)
=4(x^2-2xy+xy-2y^2)
=4[x(x-2y)+y(x-2y)]
=4(x-2y)(x+y)
f1: x^2ỹ-y^2+y
=(x-y)(x+y)+(x+y)
=(x+y)(x-y+1)
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a: \(x^2\left(x-3\right)-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
b: \(2a\left(x+y\right)+x+y=\left(x+y\right)\left(2a+1\right)\)
c: \(6x^2-12x-7x+14\)
\(=6x\left(x-2\right)-7\left(x-2\right)\)
\(=\left(x-2\right)\left(6x-7\right)\)
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a) `x^4+2x^3-4x-4`
`=(x^4-4)+(2x^3-4x)`
`=(x^2-2)(x^2+2)+2x(x^2-2)`
`=(x^2-2)(x^2+2+2x)`
b) `x^3-4x^2+12x-27`
`=(x^3-27)-(4x^2-12x)`
`=(x-3)(x^2+3x+9)-4x(x-3)`
`=(x-3)(x^2+3x+9-4x)`
`=(x-3)(x^2-x+9)`
c) `xy-4y-5x+20`
`=y(x-4)-5(x-4)`
`=(y-5)(x-4)`
a) Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^4-4\right)+2x^3-4x\)
\(=\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
b) Ta có: \(x^3-4x^2+12x-27\)
\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\cdot\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
c) Ta có: \(xy-4y-5x+20\)
\(=y\left(x-4\right)-5\left(x-4\right)\)
\(=\left(x-4\right)\left(y-5\right)\)
\(=x\left(x+4\right)+y\left(x+4\right)=\left(x+y\right)\left(x+4\right)\)
x^2+4x+xy+4y
=x(x+4)+y(x+4)
=(x+4)(x+y)