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

Toán Lớp 8: Bài 1: Tìm x: (x – 5) (-x + 4) – (x – 1) (x + 3) = -2x^2 Bài 2: Chứng minh rằng: (5a – 3) (3b – 5) – (3a – 5) (5b – 3) chia hết cho 16

Toán Lớp 8: Bài 1: Tìm x:
(x – 5) (-x + 4) – (x – 1) (x + 3) = -2x^2
Bài 2: Chứng minh rằng:
(5a – 3) (3b – 5) – (3a – 5) (5b – 3) chia hết cho 16
( Giải thích rõ giúp mik với ạ)

Comments ( 2 )

  1. Bài 1:
    (x-5)(-x+4)-(x-1)(x+3)=-2x^2
    \to (-x^2+9x-20)-(x^2+2x-3)+2x^2=0
    \to -x^2+9x-20-x^2-2x+3+2x^2=0
    \to (-x^2-x^2+2x^2)+(9x-2x)+(-20+3)=0
    \to 7x-17=0
    \to 7x=17
    \to x=17/7
    Vậy x=17/7
    Bài 2:
    (5a-3)(3b-5)-(3a-5)(5b-3)
    =[5a(3b-5)-3(3b-5)]-[3a(5b-3)-5(5b-3)]
    =[5a.3b-5a.5-3.3b+3.5]-[3a.5b-3a.3-5.5b+5.3]
    =[15ab-25a-9b+15]-[15ab-9a-25b+15]
    =15ab-25a-9b+15-15ab+9a+25b-15
    =(15ab-15ab)+(25b-9b)+(9a-25a)+(15-15)
    =16b-16a
    =16(b-a)\vdots16 (đpcm)

  2. Bài 1:
    (x-5)(-x+4)-(x-1)(x+3)=-2x^2
    <=>-x^2+4x+5x-20-x^2-3x+x+3=-2x^2
    <=>(-x^2-x^2)+(4x+5x-3x+x)-20+3=-2x^2
    <=>-2x^2+7x-17=-2x^2
    <=>7x-17=-2x^2+2x^2
    <=>7x-17=0
    <=>7x=17
    <=>x=17/7
    Vậy x=17/7
    Bài 2:
    (5a-3) (3b – 5) – (3a – 5) (5b – 3)
    =15ab-25a-9b+15-15ab+9a+25b-15
    =(15ab-15ab)+(9a-25a)+(25b-9b)+15-15
    =-16a+16b
    =16(-a+b) $\vdots16$ AAa;b
    ->(5a-3) (3b – 5) – (3a – 5) (5b – 3) $\vdots16(đpcm)$

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

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