Toán Lớp 8: bài 1 : Tính
a, (5x-2y)( $x^{2}$ – xy +1)
b,( x-1 )( x+ 1)( x+ 2 )
c, $\frac{1}{2}$ $x^{2}$ ( 2x + y )( 2x-y)
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 8: bài 1 : Tính
a, (5x-2y)( $x^{2}$ – xy +1)
b,( x-1 )( x+ 1)( x+ 2 )
c, $\frac{1}{2}$ $x^{2}$ ( 2x + y )( 2x-y)
Comments ( 2 )
a,$\text{a, (5x-2y)(}$$x^{2}$ $\text{-xy+1)}$
$\text{=5x.(}$$x^{2}$ $\text{-xy+1)}$$\text{-2y.(}$$x^{2}$ $\text{-xy+1)}$
= $\text{5}$$x^{3}$ $\text{- 5}$$x^{2}$ $\text{y +5x – $\text{2}$}$$x^{2}$ $\text{y +2x}$$y^{2}$ $\text{ – 2y}$
=$\text{5}$$x^{3}$ $\text{+(5}$$x^{2}$ $\text{y – 2}$$x^{2}$ $\text{y)+ 5x +2x}$$y^{2}$ $\text{ – 2y}$
=$\text{5}$$x^{3}$ $\text{-7}$$x^{2}$ $\text{y + 5x + 2x}$$y^{2}$ $\text{-2y}$
b, $\text{(x – 1)( x+ 1)(x + 2)}$
=$\left[\begin{matrix} x.(x+1)-1.(x+1)\ \end{matrix}]\right.$ \text{.(x+2)}
\text{=(}$x^{2}$ \text{. x . x -1).(x+2)}
$\text{=(}$$x^{2}$ $\text{-1)}$$\text{. ( x + 2 )}$
$\text{=}$$x^{2}$ $\text{(x+2)-1.(x+2)}$
$\text{=}$$x^{3}$ $\text{+2}$$x^{3}$ $\text{- x – 2}$
c,$\frac{1}{2}$ $x^{2}$ (2x+y)(2x-y)
=($\frac{1}{2}$ $x^{2}$ $y^{2}$ .2x + $\frac{1}{2}$ $x^{2}$ $y^{2}$ . y )(2x – y)
=$x^{3}$ $y^{2}$ .(2x-y) + $\frac{1}{2}$ $x^{2}$ $y^{3}$ .(2x-y)
=2$x^{4}$ $y^{2}$ -$x^{3}$$y^{3}$ + $x^{3}$$y^{3}$ – $\frac{1}{2}$ $x^{2}$ $y^{4}$
=2$x^{4}$$y^{2}$ -$\frac{1}{2}$ $x^{2}$ $y^{4}$
a) (5x-2y)(x^2-xy+1)
=5x*x^2-5x*xy+5x-2y*x^2-2y*(-xy)-2y
=5x^3-5x^2y+5x-2x^2y+2xy^2-2y
=5x^3-7x^2y+2xy^2+5x-2y
b) (x-1)(x+1)(x+2)
=(x^2-1)(x+2)
=x^2*x+x^2*2-x-2
=x^3+2x^2-x-2
c) 1/2x^2(2x+y)(2x-y)
=1/2x^2[(2x)^2-y^2]
=1/2x^2*4x^2-1/2x^2*y^2
=2x^4-1/2x^2y^2