bài 6 a)(5x-2y)( x^2-xy+1) b)(x-1)(x+1)(x+2) c) 29/07/2021 Bởi Kaylee bài 6 a)(5x-2y)( x^2-xy+1) b)(x-1)(x+1)(x+2) c)
Giải thích các bước giải: a/ $(5x-2y)(x^2-xy+1)$ $=5x(x^2-xy+1)-2y(x^2-xy+1)$ $=5x^3-5x^2y+2xy^2+5x-2y$ b/ $(x-1)(x+1)(x+2)$ $=(x^2-1)(x+2)$ $=x^3+2x^2-x-2$ c/ $\dfrac{1}{2}x^2y^2(2x-y)(2x+y)$ $=\dfrac{1}{2}x^2y^2(4x^2-y^2)$ $=2x^4y^2-\dfrac{1}{2}x^2y^4$ Chúc bạn học tốt !!! Bình luận
a)(5x-2y)( x^2-xy+1)== 5x.( x²- xy + 1) – 2y.(x² – xy + 1)= 5x³ – 5x²y + 5x – 2x²y + 2xy² – 2y= 5x³ – 7x²y + 5x + 2xy² – 2y b) (x – 1).(x + 1).(x + 2)= (x² – 1)(x + 2)= x³+2x²-x-2 c)$\frac{1}{2}$x²y² (2x+y)(2x-y) =$\frac{1}{2}$ x²y²(4x²-y²) =2$x^{4}$y²-$\frac{1}{2}$ x²$y^{4}$ Cho mình ctlhn nha~ Bình luận
Giải thích các bước giải:
a/ $(5x-2y)(x^2-xy+1)$
$=5x(x^2-xy+1)-2y(x^2-xy+1)$
$=5x^3-5x^2y+2xy^2+5x-2y$
b/ $(x-1)(x+1)(x+2)$
$=(x^2-1)(x+2)$
$=x^3+2x^2-x-2$
c/ $\dfrac{1}{2}x^2y^2(2x-y)(2x+y)$
$=\dfrac{1}{2}x^2y^2(4x^2-y^2)$
$=2x^4y^2-\dfrac{1}{2}x^2y^4$
Chúc bạn học tốt !!!
a)(5x-2y)( x^2-xy+1)== 5x.( x²- xy + 1) – 2y.(x² – xy + 1)
= 5x³ – 5x²y + 5x – 2x²y + 2xy² – 2y
= 5x³ – 7x²y + 5x + 2xy² – 2y
b) (x – 1).(x + 1).(x + 2)
= (x² – 1)(x + 2)
= x³+2x²-x-2
c)$\frac{1}{2}$x²y² (2x+y)(2x-y)
=$\frac{1}{2}$ x²y²(4x²-y²)
=2$x^{4}$y²-$\frac{1}{2}$ x²$y^{4}$
Cho mình ctlhn nha~