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

Toán Lớp 8: B1.Rút gọn biểu thức: a) ( x+ 2)(x+3) – ( x – 2) mũ 2 + 3x – 2 b) ( 2x-1) mũ 2 + ( x + 1) mũ 2 – 5x(x +3) c) ( x+3) mũ 2 – 2(x-3)(x+3)

Home/ Toán Lớp 8: B1.Rút gọn biểu thức: a) ( x+ 2)(x+3) – ( x – 2) mũ 2 + 3x – 2 b) ( 2x-1) mũ 2 + ( x + 1) mũ 2 – 5x(x +3) c) ( x+3) mũ 2 – 2(x-3)(x+3)

Toán Lớp 8: B1.Rút gọn biểu thức: a) ( x+ 2)(x+3) – ( x – 2) mũ 2 + 3x – 2 b) ( 2x-1) mũ 2 + ( x + 1) mũ 2 – 5x(x +3) c) ( x+3) mũ 2 – 2(x-3)(x+3)

Tháng Hai 16, 2023 Toán học 2 Comments 0 views

Toán Lớp 8: B1.Rút gọn biểu thức:
a) ( x+ 2)(x+3) – ( x – 2) mũ 2 + 3x – 2
b) ( 2x-1) mũ 2 + ( x + 1) mũ 2 – 5x(x +3)
c) ( x+3) mũ 2 – 2(x-3)(x+3) + ( x -3) mũ 2
d) ( 2x – 3) mũ 2 – 2(4x mũ 2 -9) + ( 2x + 3) mũ 2.
B2. Tìm x, biết:
a) ( x -3) mũ 2 – 16= 0
b) ( x-2) mũ 2 – 4(x mũ 2 – 4)+ 4( x+2) mũ 2 = 0

Comments ( 2 )

  1. lanngocha
    lanngocha
    16 Tháng Hai, 2023 at 11:07 sáng
    Reply
    Giải đáp:
    $\\$
    Bài 1.
    a,
    (x +2) (x+3) – (x-2)^2 + 3x – 2
    = x (x+3) + 2(x+3) – (x^2 – 2x . 2 + 2^2) + 3x-2
    = x^2 + 3x + 2x + 6 – (x^2 – 4x + 4) +3x-2
    = x^2 + 3x + 2x + 6 – x^2 + 4x – 4 + 3x-2
    = (x^2 – x^2) + (3x + 2x + 4x + 3x) + (6 – 4 – 2)
    = 12x
    b,
    (2x-1)^2 + (x+1)^2 – 5x (x+3)
    = [ (2x)^2 – 2 . 2x . 1 + 1^2] + (x^2 + 2x.1 + 1^2) – 5x^2 – 15x
    = [4x^2 – 4x + 1] + (x^2 + 2x+1) – 5x^2- 15x
    = 4x^2 – 4x + 1 + x^2 + 2x+1 – 5x^2-15x
    = (4x^2 + x^2 – 5x^2) + (-4x + 2x – 15x) + (1 + 1)
    = -17x + 2
    c,
    (x+3)^2 – 2 (x-3) (x+3) + (x-3)^2
    = [ (x+3) – (x-3)]^2
    = [x + 3 – x + 3]^2
    = [ (x-x) + (3+3)]^2
    = 6^2
    = 36
    d,
    (2x-3)^2 – 2 (4x^2 – 9) + (2x + 3)^2
    = [ (2x)^2 – 2 . 2x . 3 + 3^2] – 8x^2 + 18 + [ (2x)^2 + 2 . 2x . 3 + 3^2]
    = [4x^2 – 12x + 9] – 8x^2 + 18 + [4x^2 + 12x + 9]
    = 4x^2 – 12x + 9 – 8x^2 + 18 + 4x^2 + 12x + 9
    = (4x^2 – 8x^2 + 4x^2) + (-12x + 12x) + (9 + 18  + 9)
    = 36
    $\\$
    Bài 2
    a,
    (x-3)^2-16=0
    ↔ (x-3)^2=0+16
    ↔(x-3)^2=16
    ↔ \(\left[ \begin{array}{l}(x-3)^2=4^2\\(x-3)^2=(-4)^2\end{array} \right.\) 
    ↔ \(\left[ \begin{array}{l}x-3=4\\x-3=-4\end{array} \right.\) 
    ↔ \(\left[ \begin{array}{l}x=4+3\\x=-4+3\end{array} \right.\) 
    ↔ \(\left[ \begin{array}{l}x=7\\x=-1\end{array} \right.\) 
    Vậy S ∈ {7;-1}
    b,
    (x-2)^2 – 4 (x^2 – 4) + 4 (x+2)^2=0
    ↔ (x^2 – 2x . 2 + 2^2) – 4x^2 + 16 + 4 (x^2 + 2x .2 + 2^2)=0
    ↔ (x^2 – 4x +4) – 4x^2 + 16 + 4 (x^2 + 4x  +4)=0
    ↔ x^2 – 4x + 4 – 4x^2 + 16 + 4x^2 + 16x +16=0
    ↔ (x^2 – 4x^2 + 4x^2) + (-4x + 16x) + (4 + 16 + 16) = 0
    ↔ x^2 + 12x + 36=0
    ↔ x^2 + 2 . 6 . x + 6^2 =0
    ↔ (x+6)^2=0
    ↔x+6=0
    ↔x=0-6
    ↔x=-6
    Vậy S ∈ {-6}

  2. daithanh
    daithanh
    16 Tháng Hai, 2023 at 11:07 sáng
    Reply
    Bài 1:
    a)
    (x+2)(x+3)-(x-2)^2+3x-2
    =x^2+5x+6-(x^2-4x+4)+3x-2
    =x^2+5x+6-x^2+4x-4+3x-2
    =(x^2-x^2)+(5x+4x+3x)+(6-4-2)
    =12x
    b)
    (2x-1)^2+(x+1)^2-5x(x+3)
    =4x^2-4x+1+x^2+2x+1-5x^2-15x
    =(4x^2+x^2-5x^2)+(-4x+2x-15x)+(1+1)
    =-17x+2
    c)
    (x+3)^2-2(x-3)(x+3)+(x-3)^2
    =[(x+3)-(x-3)]^2
    =(x+3-x+3)^2
    =6^2=36
    Áp dụng: (A-B)^2=A^2-2AB+B^2
    d)
    (2x-3)^2-2(4x^2-9)+(2x+3)^2
    =(2x-3)^2-2(2x-3)(2x+3)+(2x+3)^2
    =[(2x-3)-(2x+3)]^2
    =(2x-3-2x-3)^2
    =(-6)^2=36
    Bài 2:
    a)
    (x-3)^2-16=0
    <=>(x-3)^2-4^2=0
    <=>(x-3-4)(x-3+4)=0
    <=>(x-7)(x+1)=0
    <=> \(\left[ \begin{array}{l}x-7=0\\x+1=0\end{array} \right.\) <=> \(\left[ \begin{array}{l}x=7\\x=-1\end{array} \right.\) 
    Vậy S={-1;7}
    b)
    (x-2)^2-4(x^2-4)+4(x+2)^2=0
    <=>x^2-4x+4-4x^2+16+4x^2+16x+16=0
    <=>x^2+12x+36=0
    <=>x^2+2.x.6+6^2=0
    <=>(x+6)^2=0
    <=>x+6=0
    <=>x=-6
    Vậy S={-6}
     

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

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