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

Toán Lớp 8: tìm x. a) 4x(x-7)=x^2 – 49 b) x^2+3x = 28. phân tích đa thức thành nhân tử: a) ( 2x +1 ) (x-4)+2x-8 b) 5x^4 – 20x^3 + 20x^2

Toán Lớp 8: tìm x.
a) 4x(x-7)=x^2 – 49
b) x^2+3x = 28.
phân tích đa thức thành nhân tử:
a) ( 2x +1 ) (x-4)+2x-8
b) 5x^4 – 20x^3 + 20x^2

Comments ( 2 )

  1. 1.
    a)
    4x(x-7)=x^2 – 49
    ⇔4x(x-7)-(x^2 – 49)=0
    ⇔4x(x-7)-(x-7)(x+7)=0
    ⇔(x-7)(4x-x-7)=0
    ⇔(x-7)(3x-7)=0
    TH1:x-7=0⇔x=7
    TH2:3x-7=0⇔x=7/3
    Vậy x∈{7; 7/3}
    b) x^2+3x-28=0
    ⇔x^2-4x+7x-28=0
    ⇔x(x-4)+7(x-4)=0
    ⇔(x+7)(x-4)=0
    TH1:x+7=0⇔x=-7
    TH2:x-4=0⇔x=4
    Vậy x∈{-7; 4}
    2.
    a) (2x +1 )(x-4)+2x-8
    =(2x+1)(x-4)+2(x-4)
    =(x-4)(2x+1+2)
    =(x-4)(2x+3)
    b) 5x^4 – 20x^3 + 20x^2
    =5x^2(x^2-4x+4)
    =5x^2(x-2)^2
    =5x^2(x-2)(x-2)

  2. Câu 1:
    a)
    4x (x-7) = x^2 – 49
    => 4x (x-7) – (x^2 – 49) = 0
    => 4x (x-7) – (x^2-7^2)=0
    =>4x(x-7) – (x-7)(x+7)=0
    => (x-7) [ 4x – (x+7)] =0
    => (x-7)(4x – x – 7)=0
    =>(x-7)(3x-7)=0
    =>x-7=0 hoặc 3x-7=0
    +)x-7=0=>x=7
    +)3x-7=0=>3x=7=>x=7/3
    Vậy x \in {7;7/3}
    b)
    x^2 + 3x = 28
    => x^2 + 3x – 28 =0
    => x^2 – 4x + 7x – 28 =0
    => x (x-4) + 7(x-4)=0
    =>(x+7)(x-4)=0
    =>x+7=0 hoặc x-4=0
    +)x+7=0=>x=-7
    +)x-4=0=>x=4
    Vậy x \in {4;-7}
    Câu 2:
    a)
    (2x+1)(x-4) + 2x -8
    = (2x+1)(x-4) + 2(x-4)
    = (x-4)(2x+1+2)
    =  (x-4)(2x+3)
    b)
    5x^4 – 20x^3 + 20x^2
    = 5x^2 (x^2 – 4x  +4)
    = 5x^2 (x-2)^2
     

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

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