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a: \(N=\dfrac{3x^5-4x^4+6x^3}{-2x^2}=-\dfrac{3}{2}x^3+2x^2-3x\)
b: \(N=\dfrac{\left(6x^4y^5-3x^3y^4+\dfrac{1}{2}x^4y^3z\right)}{-\dfrac{1}{3}x^2y^3}=-18x^2y^2+9xy-\dfrac{3}{2}x^2z\)
c: \(\Leftrightarrow N\cdot\left(y-x\right)=\left(x-y\right)^3\)
\(\Leftrightarrow N=\dfrac{\left(x-y\right)^3}{y-x}=-\left(y-x\right)^2\)
d: \(\Leftrightarrow N\cdot\left(y^2-x^2\right)=\left(y^2-x^2\right)^2\)
hay \(N=y^2-x^2\)
mik ko bít
I don't now
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1)\(x^2-y^2+2x-4y-10=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)-\left(y^2+4y+4\right)-7=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)
\(\Leftrightarrow\left(x+1-y-2\right)\left(x+1+y+2\right)=7\)
\(\Leftrightarrow\left(x-y-1\right)\left(x+y+3\right)=7\)
Xét ước
2) \(x^2+2y^2+3xy+3x+3y=15\)
\(\Leftrightarrow x^2+y^2+2xy+y^2+xy+3x+3y=15\)
\(\Leftrightarrow\left(x+y\right)^2+y\left(x+y\right)+3\left(x+y\right)=15\)
\(\Leftrightarrow\left(x+y\right)\left(x+2y+3\right)=15\)
Xét ước
\(2x^2+3xy-2y^2=7\)
\(\Leftrightarrow2x^2+4xy-2y^2-xy=7\)
\(\Leftrightarrow2x\left(x+2y\right)-y\left(x+2y\right)=7\)
\(\Leftrightarrow\left(2x-y\right)\left(x+2y\right)=7\)
Xét ước
a) x^2 - 5xy +4y^2= x^2 -xy -4xy+4y^2= (x^2-xy) - (4xy - 4y^2)= x(x-y)-4y(x-y)=(x-y)*(x - 4y)
b) x^2 -y^4+9y -x(9+y-y^3= x^2-y^4 +9y-9x-xy+xy^3= (x^2-xy)-(9x-9y)+(xy^3-y^4)=x(x-y)-9(x-y)+y^3(x-y)=(x-y)*(y^3+x-9)
d) 2u^2+2v^2-5uv=(2u^2-4uv)+(2v^2-uv)=2u(u-2v)+v(2v-u)= 2u(u-2v)-v(u-2v)=(u-2v)*(2u-v)
`a, -xy(x^2+xy-y^2)`
`= -x^3y - x^2y^2 + xy^3`.
`b, 5x^2y(2y^2-xy)`
`= 10x^2y^3 - 5x^3y^2`.
`c, (-2x^3 - 1/4y - 4y^2).8xy^2`.
`= -16x^4y^2 - 2xy^3 - 32xy^4`.
`d, (2x^3 - 3xy + 12x)(-1/6xy)`
`= -2/3x^4y + 1/2x^2y^2 - 2x^2y`.
Bài 2:
a: \(3\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)\)
\(=3\left(x^3-1\right)+x^3-3x^2+3x-1-4x\left(x^2-1\right)\)
\(=3x^3-3+x^3-3x^2+3x-1-4x^3+4x\)
\(=-3x^2+7x-4\)
\(=-3\cdot\left(-1\right)^2+7\cdot\left(-1\right)-4\)
=-3-4-7=-14
b: \(=27x^3y^3-8-3xy\left(9x^2y^2+6xy+1\right)\)
\(=27x^3y^3-8-27x^3y^3-18x^2y^2-3xy\)
\(=-18x^2y^2-3xy-8\)
\(=-18\cdot\left[\left(-2010\right)\cdot\left(-\dfrac{1}{2010}\right)\right]^2-3\cdot\left(-2010\right)\cdot\dfrac{-1}{2010}-8\)
\(=-18-3-8=-29\)
a) \(3x^2-3xy-5x+5y\)
\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) \(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left[x^2-\left(y+1\right)^2\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
c) \(x^2+1+2x-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
f) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
a: =3x(x-y)-5(x-y)
=(x-y)(3x-5)
b: \(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
d:
Sửa đề: x^2+4x-2xy-4y+y^2
=x^2-2xy+y^2+4x-4y
=(x-y)^2+4(x-y)
=(x-y)(x-y+4)
e: =x(x^2-2x+1)
=x(x-1)^2
f: =2(x^2+2x+1-y^2)
=2[(x+1)^2-y^2]
=2(x+1+y)(x+1-y)
\(1.5x\left(x^2+2x-1\right)-3x^2\left(x-2\right)=5x^3+10x^2-5x-3x^3+6x^2\)
\(=2x^3+16x^2-5x\)
\(=\left(2x^3-x\right)+\left(16x^2-4x\right)\)
\(=x\left(2x^2-1\right)+4x\left(4x-1\right)\left(ĐCCM\right)\)
\(M=\left(x^2-3xy+2y^2\right).\left(x-2y\right)-4y^2.\left(2x-y\right)\)
\(\Rightarrow M=-4y^3+8xy^2-5x^2y+x^3-8xy^2+4y^3\)
\(\Rightarrow M=x^3-5x^2y\)
Ta thay \(x=1,y=-3598\) vào M:
\(M=1^3-5.1^2.\left(-3598\right)=17991\)