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

Toán Lớp 8: Tìm $a,b,c$ sao cho: a) $ax(x^2+b)+2x(1+x)=3x^3+cx^2+5x$ b) $x(ax^2-bx+1)+ã(c+x)=x^3+2x^2+3x$ c) $-3x^2(2ax^2-bx+c)=12x^4-6x^2-3x^2$

Toán Lớp 8: Tìm $a,b,c$ sao cho:
a) $ax(x^2+b)+2x(1+x)=3x^3+cx^2+5x$
b) $x(ax^2-bx+1)+ã(c+x)=x^3+2x^2+3x$
c) $-3x^2(2ax^2-bx+c)=12x^4-6x^2-3x^2$

Comments ( 1 )

  1. a) ax(x^2+b)+2x(1+x)=3x^3+cx^2+5x
    <=>ax^3+abx+2x+2x^2=3x^3+cx^2+5x
    <=> ax^3+2x^2+(ab+2)x=3x^3+cx^2+5x
    =>{(a=3),(2=c),(ab+2=5):}<=>{(a=3),(c=2),(3b=3):}<=>{(a=3),(c=2),(b=1):}
    Vậy (a;b;c)=(3;1;2)
    b) x(ax^2-bx+1)+ax(c+x)=x^3+2x^2+3x
    <=> ax^3-bx^2-x+acx+x^2=x^3+2x^2+3x
    <=> ax^3+(-b+1)x^2+(ac-1)x=x^3+2x^2+3x
    =>{(a=1),(-b+1=2),(ac-1=3):}<=>{(a=1),(b=-1),(c=4):}
    Vậy (a;b;c)=(1;-1;4)
    c) -3x^2(2ax^2-bx+c)=12x^4-6x^3-3x^2
    <=> -6ax^4+3bx^3-3cx^2=12x^4-6x^3-3x^2
    =>{(-6a=12),(3b=-6),(-3c=-3):}<=>{(a=-2),(b=-2),(c=1):}
    Vậy (a;b;c)=(-2;-2;1)

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

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