Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài giải:
a) (-2x5 + 3x2 – 4x3) : 2x2 = (- )x5 – 2 + x2 – 2 + (-)x3 – 2 = - x3 + – 2x.
b) (x3 – 2x2y + 3xy2) : (- x) = (x3 : -x) + (-2x2y : -x) + (3xy2 : -x)
= -2x2 + 4xy – 6y2
c)(3x2y2 + 6x2y3 – 12xy) : 3xy = (3x2y2 : 3xy) + (6x2y2 : 3xy) + (-12xy : 3xy)
= xy + 2xy2 – 4.
a) (-2x5+3x2-4x3) : 2x2
= (-2x5:2x2)-(4x3:2x2)+(3x2:2x2)
= -x3-2x+\(\dfrac{3}{2}\)
b) \(\left(x^3-2x^2y+3xy^2\right):\left(-\dfrac{1}{2}x\right)\)
= \(\left(x^3:\dfrac{-1}{2}x\right)+\left(-2x^2y:\dfrac{-1}{2}x\right)+\left(3xy^2:\dfrac{-1}{2}x\right)\)
= \(-2x^2+4xy-6y^2\)
c) \(\left(3x^2y^2+6x^2y^3-12xy\right):3xy\)
= \(\left(6x^2y^3:3xy\right)+\left(3x^2y^2:3xy\right)+\left(-12xy:3xy\right)\)
= \(xy^2+xy-4\)
sau bạn đăng tách ra cho mn cùng giúp nhé
a, \(\left(-2x^5+3x^2-4x^3\right):2x^2=-x^3+\frac{3}{2}-2x\)
b, \(\left(x^3-2x^2y+3xy^2\right):\left(-\frac{1}{2}x\right)=-\frac{x^2}{2}+xy-\frac{3y^2}{2}\)
c, \(\left(3x^2y^2+6x^3y^3-12xy^2\right):3xy=xy+2x^2y^2-4y\)
d, \(\left(4x^3-3x^2y+5xy^2\right):\frac{1}{2}x=2x^2-\frac{3xy}{2}+\frac{5y^2}{2}\)
e, \(\left(18x^3y^5-9x^2y^2+6xy^2\right):3xy^2=6x^2y^3-3x+2\)
f, \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)=x^2+y^2\)
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 \(N=y^2-x^2\)
\(a,3x\left(3x+6\right)=9x^2+18x\)
\(b,-\dfrac{1}{2}xy\left(4x^2+6x\right)\)
\(=-2x^3y-3x^2y\)
\(c,-2x^2y^3\left(\dfrac{1}{2}xy+4y^2\right)\)
\(=-x^3y^4-8x^2y^5\)
\(d,-6x^2\left(\dfrac{1}{3}xy^2-\dfrac{1}{2}y\right)\)
\(=-2x^3y^2+3x^2y\)
#\(Urushi\)
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
Bài 1 . ( 20x4y - 25x2y2 - 3x2 y) : 5x2y
= 5x2y.( 4x2 - 5y - \(\dfrac{3}{5}\)) : 5x2y
= 4x2 - 5y - \(\dfrac{3}{5}\)
Bài 2 . a) ( -2x5 + 3x2 - 4x3) : 2x2
= 2x2.( -x3 + \(\dfrac{3}{2}\) - 2x ) : 2x2
= - x3 - 2x + \(\dfrac{3}{2}\)
b) ( x3 - 2x2y + 3xy2) : ( \(\dfrac{1}{2}x\))
= \(\dfrac{1}{2}x\).( 2x2 - 4xy + 6y2) : ( \(\dfrac{1}{2}x\))
= 2x2 - 4xy + 6y2
c) ( 3x2y2 + 6x2y3 - 12xy ) : 3xy
= 3xy.( xy + 2xy2 - 4 ) : 3xy
= xy + 2xy2 - 4