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

Toán Lớp 7: Cho các đa thức f(x) = 5x^4 + 5x^3 – x – 1 g(x) = x^3(x-1) + 2(x^2 +2) h(x) = 4x^3(x-1) – ( x+3) a) Tính f(x) + g( x) – h(x

Toán Lớp 7: Cho các đa thức f(x) = 5x^4 + 5x^3 – x – 1
g(x) = x^3(x-1) + 2(x^2 +2)
h(x) = 4x^3(x-1) – ( x+3)
a) Tính f(x) + g( x) – h(x)
b) Tính p(x) = f(x) – g(x) + h(x) rồi tính p(2-2)

Comments ( 2 )

  1. Giải đáp+Lời giải và giải thích chi tiết:
    a)
    f(x)+g(x)-h(x)
    =(5x^4+5x^3-x-1)+[x^3(x-1)+2(x^2+2)]-[4x^3(x-1)-(x+3)]
    =5x^4+5x^3-x-1+x^4-x^3+2x^2+4-4x^4+4x^3+x+3
    =(5x^4+x^4-4x^4)+(5x^3-x^3+4x^3)+2x^2+(x-x)+(4+3-1)
    =2x^4+8x^3+2x^2+6
    b)
    p(x)=f(x)-g(x)+h(x)
    =(5x^4+5x^3-x-1)-[x^3(x-1)+2(x^2+2)]+[4x^3(x-1)-(x+3)]
    =5x^4+5x^3-x-1-x^4+x^3-2x^2-4+4x^4-4x^3-x-3
    =(5x^4-x^4+4x^4)+(5x^3+x^3-4x^3)-2x^2-(x+x)-(1+4+3)
    =8x^4+2x^3-2x^2-2x-8
    p(2-2)=p(0)
    Ta có:
    p(x)=8x^4+2x^3-2x^2-2x-8
    ->p(0)=8*0^4+2*0^3-2*0^2-2*0-8=0+0-0-0-8=-8
    Vậy p(2-2)=-8
     

  2. Ta có: 
    f(x) = 5x^4 + 5x^3 – x – 1
    g(x) = x^3(x-1) + 2(x^2 +2)
    = x^4 – x^3 + 2x^2 + 4
    h(x) = 4x^3(x-1) – ( x+3)
    = 4x^4 – 4x^3 – x – 3
    a) 
    f(x) + g( x) – h(x) = 5x^4 + 5x^3 – x – 1 + x^4 – x^3 + 2x^2 + 4 – 4x^4 + 4x^3 + x + 3
    = ( 5x^4 + x^4 – 4x^4) + (5x^3 – x^3 + 4x^3) + 2x^2 + (x – x) + (4 + 3 – 1)
    = 2x^4 + 8x^3 + 2x^2 + 6
    b) 
    p(x) = 5x^4 + 5x^3 – x – 1 – x^4 + x^3 – 2x^2 – 4 + 4x^4 – 4x^3 – x – 3
    = (5x^4 – x^4 + 4x^4) + (5x^3 + x^3 – 4x^3) – 2x^2 – (x+x) – (1+4+3)
    = 8x^4 + 2x^3 – 2x^2 – 2x – 8
    p(2-2) = p(0) = 8. 0^4 + 2 . 0^3 – 2 . 0^2 – 2.0 – 8
    = 8.0+2.0-2.0-2.0 – 8
    = 0 – 8
    = -8
    (Chúc bạn học tốt)

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