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

Toán Lớp 8: Tìm x, biết: a) (x+1)(x+2)- x^2 = 17 b) (x+1)^3+ 3x(2-x) – 9(x-1) = 18 c) x^2(x-5) – 9x + 45 = 0

Home/ Toán Lớp 8: Tìm x, biết: a) (x+1)(x+2)- x^2 = 17 b) (x+1)^3+ 3x(2-x) – 9(x-1) = 18 c) x^2(x-5) – 9x + 45 = 0

Toán Lớp 8: Tìm x, biết: a) (x+1)(x+2)- x^2 = 17 b) (x+1)^3+ 3x(2-x) – 9(x-1) = 18 c) x^2(x-5) – 9x + 45 = 0

Tháng Chín 26, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: Tìm x, biết:
a) (x+1)(x+2)- x^2 = 17
b) (x+1)^3+ 3x(2-x) – 9(x-1) = 18
c) x^2(x-5) – 9x + 45 = 0

Comments ( 2 )

  1. tuanh
    tuanh
    26 Tháng Chín, 2022 at 7:47 chiều
    Reply
    a) (x+1)(x+2)-x^2=17
    ⇔x^2+2x+x+2-x^2-17=0
    ⇔3x-15=0
    ⇔3x=15
    ⇔x=5
    Vậy S={5}
    b) (x+1)^3+3x(2-x)-9(x-1)=18
    ⇔x^3+3x^2+3x+1+6x-3x^2-9x+9-18=0
    ⇔x^3-8=0
    ⇔x^3=8
    ⇔x^3=2^3
    ⇔x=2
    Vậy S={2}
    c) x^2(x-5)-9x+45=0
    ⇔x^2(x-5)-(9x-45)=0
    ⇔x^2(x-5)-9(x-5)=0
    ⇔(x-5)(x^2-9)=0
    ⇔(x-5)(x-3)(x+3)=0
    ⇔\(\left[ \begin{array}{l}x-5=0\\x-3=0\\x+3=0\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=5\\x=3\\x=-3\end{array} \right.\) 
    Vậy S={5,3,-3}
     

  2. thuminhthong
    thuminhthong
    26 Tháng Chín, 2022 at 7:47 chiều
    Reply
    a)
    (x+1)(x+2)-x^2=17
    <=>x^2+2x+x+2-x^2-17=0
    <=>3x-15=0
    <=>3x=15
    <=>x=5
    Vậy S={5}
    b)
    (x+1)^3+3x(2-x)-9(x-1)=18
    <=>x^3+3x^2+3x+1+6x-3x^2-9x+9-18=0
    <=>x^3-8=0
    <=>x^3=8
    <=>x=2
    Vậy S={2}
    c)
    x^2(x-5)-9x+45=0
    <=>x^2(x-5)-9(x-5)=0
    <=>(x-5)(x^2-9)=0
    <=>(x-5)(x-3)(x+3)=0
    <=>\(\left[ \begin{array}{l}x-5=0\\x-3=0\\x+3=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=5\\x=3\\x=-3\end{array} \right.\) 
    Vậy S={5;3;-3} 

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

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