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

Toán Lớp 8: Tìm x 1)7x^2 + 2x=0 2)x(x-1) – x^2 +2x =5 3)x^2 – 2x – 15=0 4)x(x-1) + 2x -2=0 5)(3x-1)^2 – (x+5)^2=0 6) (x+1)^3 = 4x + 4 7) 2x^2 (x-1)

Home/ Toán Lớp 8: Tìm x 1)7x^2 + 2x=0 2)x(x-1) – x^2 +2x =5 3)x^2 – 2x – 15=0 4)x(x-1) + 2x -2=0 5)(3x-1)^2 – (x+5)^2=0 6) (x+1)^3 = 4x + 4 7) 2x^2 (x-1)

Toán Lớp 8: Tìm x 1)7x^2 + 2x=0 2)x(x-1) – x^2 +2x =5 3)x^2 – 2x – 15=0 4)x(x-1) + 2x -2=0 5)(3x-1)^2 – (x+5)^2=0 6) (x+1)^3 = 4x + 4 7) 2x^2 (x-1)

Tháng Năm 23, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: Tìm x
1)7x^2 + 2x=0
2)x(x-1) – x^2 +2x =5
3)x^2 – 2x – 15=0
4)x(x-1) + 2x -2=0
5)(3x-1)^2 – (x+5)^2=0
6) (x+1)^3 = 4x + 4
7) 2x^2 (x-1) – x=-x^2
8) 6(1-x) + x^2 -2x + 1=0
9) x^3 + 27 = (x+3)(x+9)
mình cần mấy bạn làm chính xác nhé 9h tối nay các bạn phải làm xong nhá

Comments ( 1 )

  1. ngocbichchau
    ngocbichchau
    23 Tháng Năm, 2022 at 1:52 chiều
    Reply
    1) 7x² + 2x=0
    => x (7x + 2) = 0
    => \(\left[ \begin{array}{l}x=0\\7x+2=0\end{array} \right.\) 
    => \(\left[ \begin{array}{l}x=0\\7x=-2\end{array} \right.\) 
    => \(\left[ \begin{array}{l}x=0\\x=\frac{-2}{7}\end{array} \right.\) 
    Vậy x ∈ {0; $\frac{-2}{7}$}
    2) x (x – 1) – x² + 2x = 5
    => x² – x – x² + 2x = 5
    => x = 5
    Vậy x = 5
    3) x² – 2x – 15 = 0
    => x² – 2x + 1 – 16 = 0
    => (x – 1)² – 4² = 0
    => (x – 1 + 4) (x – 1 – 4) = 0
    => (x + 3) (x – 5) = 0
    => \(\left[ \begin{array}{l}x+3=0\\x-5=0\end{array} \right.\) 
    => \(\left[ \begin{array}{l}x=-3\\x=5\end{array} \right.\) 
    Vậy x ∈ {-3; 5}
    4) x (x – 1) + 2x – 2 = 0
    => x (x – 1) + 2 (x – 1) = 0
    => (x + 2) (x – 1) = 0
    => \(\left[ \begin{array}{l}x+2=0\\x-1=0\end{array} \right.\) 
    => \(\left[ \begin{array}{l}x=-2\\x=1\end{array} \right.\) 
    Vậy x ∈ {-2; 1}
    5) (3x – 1)² – (x + 5)² = 0
    => (3x – 1 – x – 5) (3x – 1 + x + 5) = 0
    => (2x – 6) (4x + 4) = 0
    => \(\left[ \begin{array}{l}2x-6=0\\4x+4=0\end{array} \right.\) 
    => \(\left[ \begin{array}{l}2x=6\\4x=-4\end{array} \right.\) 
    => \(\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.\) 
    Vậy x ∈ {3; -1}
    6) (x + 1)³ = 4x + 4
    => (x + 1)³ – 4x – 4 = 0
    => (x + 1)³ – 4 (x + 1) = 0
    => (x + 1) [(x + 1)² – 4] = 0
    => (x + 1) (x + 1 – 2) (x + 1 + 2) = 0
    => (x + 1) (x – 1) (x + 3) = 0
    – Trường hợp 1: x + 1 = 0 = > x = -1
    – Trường hợp 2: x – 1 = 0 = > x = 1
    – Trường hợp 3: x + 3 = 0 = > x = -3
    Vậy x ∈ {±1; -3}
    7) 2x² (x – 1) – x = -x²
    => 2x² (x – 1) – x + x² = 0
    => 2x² (x – 1) – x (1 – x) = 0
    => 2x² (x – 1) + x (x – 1) = 0
    => (2x² + x) (x – 1) = 0
    => x (2x + 1) (x – 1) = 0
    – Trường hợp 1: x = 0
    – Trường hợp 2: 2x + 1 = 0 = > 2x = -1 => x = -1/2
    – Trường hợp 3: x – 1 = 0 = > x = 1
    Vậy x ∈ {0; -1/2; 1}
    8) 6 (1 – x) + x² – 2x + 1 = 0
    => 6 (1 – x) + (x – 1)² = 0
    => 6 (1 – x) + (1 – x)² = 0
    => (1 – x) (6 + 1 – x) = 0
    => (1 – x) (7 – x) = 0
    => \(\left[ \begin{array}{l}1-x=0\\7-x=0\end{array} \right.\) 
    => \(\left[ \begin{array}{l}x=1\\x=7\end{array} \right.\) 
    Vậy x ∈ {1; 7}
    9) x³ + 27 = (x + 3) (x + 9)
    => (x + 3) (x² – 3x + 9) – (x + 3) (x + 9) = 0
    => (x  + 3) (x² – 3x + 9 – x – 9) = 0
    => (x + 3) (x² – 4x) = 0
    => (x + 3) x (x – 4) = 0
    – Trường hợp 1: x + 3 = 0 = > x = -3
    – Trường hợp 2: x = 0
    – Trường hợp 3: x – 4 = 0 = > x = 4
    Vậy x ∈ {-3; 0; 4}

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

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