Toán Lớp 8: tính a)(4x-1)(2x^2-x-1) b)(4x^3+8x^2-2x):2x c)(6x^3-7x^2-16x+12):(2x+3)

Question

Toán Lớp 8:  tính a)(4x-1)(2x^2-x-1) b)(4x^3+8x^2-2x):2x c)(6x^3-7x^2-16x+12):(2x+3)

tonhitruc · Answer

Giải đáp:   a) 8x^3-6x^2-3x+1   b) 2x^2+4x-1   c) 3x^2-8x+4   Lời giải và giải thích chi tiết:   a) (4x-1)(2x^2-x-1)   =4x*2x^2-4x*x-4x-2x^2-x*(-1)-1*(-1)   =8x^3-4x^2-4x-2x^2+x+1   =8x^3+(-4x^2-2x^2)+(x-4x)+1   =8x^3-6x^2-3x+1   --   b) (4x^3+8x^2-2x):2x   =2x(2x^2+4x-1):2x   =2x^2+4x-1   --   c) (6x^3-7x^2-16x+12):(2x+3)   =(6x^3+9x^2-16x^2-24x+8x+12):(2x+3)   =[(6x^3+9x^2)+(-16x^2-24x)+(8x+12)]:(2x+3)   =[3x^2(2x+3)-8x(2x+3)+4(2x+3)]:(2x+3)   =(2x+3)(3x^2-8x+4):(2x+3)   =3x^2-8x+4

lanngocha · Answer

a)     $(4x&minus;1)(2x^2&minus;x&minus;1)$     $=8x^3&minus;4x^2&minus;4x&minus;2x^2+x+1$     $=8x^3&minus;6x62&minus;3x+1$     b)     $(4x^3+8x^2-2x):2$   $=2x^3+4x^2-x$     c)      text{Ta c&oacute;:}     $6x^3&minus;7x^2&minus;16x+12$     $=(6x^3+9x^2)&minus;(16x^2+24x)+(8x+12)$     $=3x^2.(2x+3)&minus;8x.(2x+3)+4.(2x+3)$     $=(2x+3)(3x^2&minus;8x+4)$     $&rArr; (6x^3&minus;7x^2&minus;16x+12):(2x+3)$     $=[(2x+3)(3x^2&minus;8x+4)]:(2x+3)$     $=3x^2&minus;8x+4$