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

Toán Lớp 8: Tìm x: b)(x+2)(x-3)-x-2=0 c)36x^2-40=0

Home/ Toán Lớp 8: Tìm x: b)(x+2)(x-3)-x-2=0 c)36x^2-40=0

Toán Lớp 8: Tìm x: b)(x+2)(x-3)-x-2=0 c)36x^2-40=0

Tháng Tám 8, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: Tìm x:
b)(x+2)(x-3)-x-2=0
c)36x^2-40=0

Comments ( 2 )

  1. ngoclanquynh
    ngoclanquynh
    8 Tháng Tám, 2022 at 3:14 chiều
    Reply
    b) (x+2)(x-3)-x-2=0
    <=> x^2-3x+2x-6-x-2=0 
    <=> x^2-2x-8=0
    <=> (x-4)(x+2)=0 
    TH1:
    x-4=0
    <=> x=4
    TH2:
    x+2=0
    <=> x=-2
    Vậy x∈{4;-2} 
    c) 36x^2-40=0
    <=>36x^2=40
    <=> x^2=40/36
    <=> x^2=10/9
    <=> x=±\sqrt{\frac{10}{9}}
    Vậy x∈{±\sqrt{\frac{10}{9}}}
     

  2. diepbich93
    diepbich93
    8 Tháng Tám, 2022 at 3:14 chiều
    Reply
    b)
    (x+2)(x-3) – x – 2 = 0
    =>(x+2)(x-3)-(x+2) = 0
    => (x+2)(x-3-1) = 0
    => (x+2)(x-4) = 0
    \(\left[ \begin{array}{l}x+2=0 =>x=-2\\x-4=0 => x = 4\end{array} \right.\)
    => x in {2;4}
    c)
    36^2 – 40 = 0
    Dùng hàng đẳng thức A^2-B^2 = (A-B)(A+B)
    =>(6x)^2 -sqrt(40)^2 = 0
    => (6x-sqrt(40))(6x+sqrt(40)) = 0
    => (6x – 2sqrt(5))(6x+2sqrt(5)) = 0
    => 6x-2sqrt(5) = 0 => x = sqrt(5)/3
    => 6x+2sqrt(5) = 0 => x = (-sqrt(5))/3
    Vậy x in {sqrt(5)/3;(-sqrt(5))/3}
     

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

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