\(\frac{2}{5}\)x2y + xy2 - 3xy + \(\frac{1}{3}\)xy2 - 3xy - \(\frac{1}{2}\)x2y
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a: \(x^4+x^3+x+1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+1\right)=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
b: \(x^4-x^3-x^2+1\)
\(=x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
c: \(x^2y+xy^2-x-y\)
=xy(x+y)-(x+y)
=(x+y)(xy-1)
d: \(a^2x+a^2y-7x-7y\)
\(=a^2\left(x+y\right)-7\left(x+y\right)\)
\(=\left(x+y\right)\left(a^2-7\right)\)
e: \(a\cdot x^2+ay-bx^2-by\)
\(=a\left(x^2+y\right)-b\left(x^2+y\right)\)
\(=\left(x^2+y\right)\left(a-b\right)\)
g: \(12x^2-3xy+8xz-2yz\)
=3x(4x-y)+2z(4x-y)
=(4x-y)(3x+2z)
a) \(2x-72x^3=2x\left(1-36x^2\right)=2x\left(1-6x\right)\left(1+6x\right)\)
f) \(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)
Ta có A + 2B = (x2y - xy2 + 3x2) + 2(x2y + xy2 - 2x2 - 1)
= x2y - xy2 + 3x2 + 2x2y + 2xy2 - 4x2 - 2
= 3x2y + xy2 - x2 - 2. Chọn C
a: \(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\)
\(=x^5y-\dfrac{1}{5}x^3y^3-x^2y\)
b: \(\left(\dfrac{1}{2}x-5\right)\left(x^2-2x+3\right)\)
\(=\dfrac{1}{2}x^3-x^2+\dfrac{3}{2}x-5x^2+10x-15\)
\(=\dfrac{1}{3}x^3-6x^2+\dfrac{23}{2}x-15\)
`a)(x-1)(x^2+x+1)`
`=x^3+x^2+x-x^2-x-1`
`=x^3-1`
`b)(x^3+x^2y+xy^2+y^3)(x-y)`
`=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4`
`=x^4-y^4`
a) VT`=(x-1)(x^2+x+1)`
`=x^3 +x^2 +x -x^2-x-1 `
`=x^3-1=` VP.
b) VT `=(x^3+x^2y+xy^2+y^3)(x-y)`
`=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4`
`=x^4-y^4=` VP.
a/ \(P+Q=\left(x^2y+x^3-xy^2+3\right)+\left(x^3+xy^2-xy-6\right)\)
\(=x^2y+x^3-xy^2+3+x^3+xy^2-xy-6\)
\(=\left(x^3+x^3\right)+\left(xy^2-xy^2\right)+\left(3-6\right)+x^2y-xy\)
\(=2x^3+x^2y-xy-3\)
b/ \(M+N=\left(x^2y+0,5xy^3-7,5x^3y^2+x^3\right)+\)
\(\left(3xy^3-x^2y+5,5x^3y^2\right)\)
\(=x^2y+0,5xy^3-7,5x^3y^2+x^3+3xy^3-x^2y+5,5x^3y^2\)
\(=\left(x^2y-x^2y\right)+\left(0,5xy^3+3xy^3\right)+\left(5,5x^3y^2-7,5x^3y^2\right)+x^3\)
\(=3,5xy^3-2x^3y^2+x^3\)
\(A=4x^2y+\dfrac{14}{15}xy^2-2xy-\dfrac{2}{3}\) bậc : 3
\(B=2xy^2z-1\) bậc :4
+ Thu gọn :
\(A=4x^2y+\dfrac{14}{15}xy^2-2xy-\dfrac{2}{3}\)
\(B=2xy^2z-1\)
+ Bậc
Đa thức \(A\) có 4 hạng tử :
\(4x^2y\) có bậc \(3\)
\(\dfrac{14}{15}xy^2\) có bậc \(3\)
\(-2xy\) có bậc \(2\)
\(-\dfrac{2}{3}\) có bậc \(0\)
Đa thức \(B\) có \(2\) hạng tử :
\(2xy^2z\) có bậc \(4\)
\(-1\) có bậc \(0\)
Làm lại nha
\(\dfrac{2}{5}x^2y+xy^2-3xy+\dfrac{1}{3}xy^2-3xy-\dfrac{1}{2}x^2y\)
\(=\left(\dfrac{2}{5}x^2y+\dfrac{1}{3}x^2y\right)+\left(xy^2+\dfrac{1}{3}xy^2\right)+\left(-3xy^2-3xy^2\right)\)
\(=-\dfrac{1}{10}x^2y+\dfrac{4}{3}xy^2-6xy\)
\(\dfrac{2}{5}x^2y+xy^2-3xy+\dfrac{1}{3}xy^2-3xy-\dfrac{1}{2}x^2y\)
\(=\left(\dfrac{2}{5}x^2y-\dfrac{1}{2}x^2y\right)+\left(xy^2+\dfrac{1}{3}xy^2\right)+\left(3xy-3xy\right)\)
\(=-\dfrac{1}{10}x^2y+\dfrac{4}{3}xy^2\)