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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Phân tích các đa thức sau thành nhân tử R(x)= (3x+2)(3x-5)(x-1)(9x+10)+24x^2 K(x)= (6x+5)^2(3x+2)(x+1)-35

Tháng Mười Hai 13, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: Phân tích các đa thức sau thành nhân tử
R(x)= (3x+2)(3x-5)(x-1)(9x+10)+24x^2
K(x)= (6x+5)^2(3x+2)(x+1)-35

Comments ( 2 )

  1. ngoclandiep
    ngoclandiep
    13 Tháng Mười Hai, 2022 at 10:58 sáng
    Reply
    $R(x)= (3x+2)(3x-5)(x-1)(9x+10)+24x^2$
    $=(9x^2−9x−10)(9x^2+x-10)+24x^2$
    $=(9x^2−9x−10)(9x^2+x-10)+24x^2$
    $=[(9x^2-4x-10)-5x][(9x^2-4x-10)+5x]+24x^2$
    $=(9x^2-4x-10)^2-25x^2+24x^2$
    $=(9x^2-4x-10)^2-x^2$
    $=(9x^2-4x-10+x)(9x^2-4x-10-x)$
    $=(9x^2-3x-10)(9x^2-5x-10)$
    $K(x)= (6x+5)^2(3x+2)(x+1)-35$
    $=(36x^2+60x+25)(3x^2+5x+2)-35$
    Đặt $y=3x^2+5x+2(1)$
    $⇒36x^2+60x+24+1=12(3x^2+5x+2)+1=12y+1$
    $⇒K(x)=(12y+1).y-35$
    $=12y^2+y-35$
    $=12y^2-20y+21y-35$
    $=4y(3y-5)+7(3y-5)$
    $=(4y+7)(3y-5)(2)$
    Thay $(1)$ vào $(2)$
    $⇒[4(3x^2+5x+2)+7][3(3x^2+5x+2)-5]$
    $=(12x^2+20x+15)(9x^2+15x+1)$

  2. hauthanhyen98
    hauthanhyen98
    13 Tháng Mười Hai, 2022 at 10:58 sáng
    Reply
    R(x)=(3x+2)(3x-5)(x-1)(9x+10)+24x^2
    R(x)=[(3x+2)(3x-5)][(x-1)(9x+10)]+24x^2
    R(x)=(9x^2-15x+6x-10)(9x^2+10x-9x-10)+24x^2
    R(x)=(9x^2-9x-10)(9x^2+x-10)+24x^2
    R(x)=(9x^2-4x-10-5x)(9x^2-4x-10+5x)+24x^2
    R(x)=(9x^2-4x-10)^2-25x^2+24x^2
    R(x)=(9x^2-4x-10)^2-x^2
    R(x)=(9x^2-4x-10-x)(9x^2-4x-10+x)
    R(x)=(9x^2-5x-10)(9x^2-3x-10)
    ——————————————
    K(x)=(6x+5)^2(3x+2)(x+1)-35
    K(x)=(36x^2+60x+25)(3x^2+5x+2)-35
    Đặt 3x^2+5x+2=a -> 36x^2+60x+25=12a+1
    K(x)=a(12a+1)-35
    K(x)=12a^2+a-35
    K(x)=12a^2-20a+21a-35
    K(x)=4a(3a-5)+7(3a-5)
    K(x)=(3a-5)(4a+7)
    Thay a=3x^2+5x+2 vào K(x) thì:
    K(x)=[3(3x^2+5x+2)-5][4(3x^2+5x+2)+7]
    K(x)=(9x^2+15x+6-5)(12x^2+20x+8+7)
    K(x)=(9x^2+15x+1)(12x^2+20x+15)

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222-9+11+12:2*14+14 = ? ( )

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