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

Toán Lớp 8: 13. a) 2x^2 – 7x^2 – 4 = 0 b) 4x^4 + 3x^2 – 1 = 0 14. a) (x + 1)^4 + (x – 3)^4 = 82

Tháng Sáu 20, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: 13.
a) 2x^2 – 7x^2 – 4 = 0
b) 4x^4 + 3x^2 – 1 = 0
14.
a) (x + 1)^4 + (x – 3)^4 = 82

Comments ( 1 )

  1. tonhu12
    tonhu12
    20 Tháng Sáu, 2022 at 10:47 sáng
    Reply
    Giải đáp:
    13.
    a) 2x^4 – 7x^2 – 4 = 0
    Đặt t = x^2 >= 0. Thì x^4 = t^2
    Ta có: 2x^4 – 7x^2 – 4 = 0
    <=> 2t^2 – 7t – 4 = 0
    <=> 2t^2 – 8t + t – 4 = 0
    <=> 2t.(t – 4) + (t – 4) = 0
    <=> (2t + 1).(t -4) = 0
    <=> $\left[\begin{matrix} 2t+1=0\\ t-4=0\end{matrix}\right.$ <=>$\left[\begin{matrix} t = \dfrac{-1}{2}\\ t = 4\end{matrix}\right.$ <=>$\left[\begin{matrix} x^2 = \dfrac{-1}{2} (l)\\ x^2 = 4\end{matrix}\right.$ <=> $\left[\begin{matrix} x = 2\\ x = -2\end{matrix}\right.$
    Vậy x ∈ {4,-4}
    b) 4x^4 + 3x^2 – 1=  0
    Đặt t = x^2 >= 0. Thì x^4 = t^2
    Ta có: 4x^4 + 3x^2 – 1=  0
     <=> 4t^2 + 3t – 1 = 0
    <=> 4t^2 + 4t – t – 1 = 0
    <=> 4t.(t + 1) – (t + 1) = 0
    <=> (4t – 1).(t + 1) = 0
    <=> $\left[\begin{matrix} 4t-1=0\\ t+1=0\end{matrix}\right.$ <=>$\left[\begin{matrix} t = \dfrac{1}{4}\\ t = -1\end{matrix}\right.$ <=>$\left[\begin{matrix} x^2 = \dfrac{1}{4} \\ x^2 = -1(l)\end{matrix}\right.$ <=> $\left[\begin{matrix} x = \dfrac{1}{2}\\ x = \dfrac{-1}{2}\end{matrix}\right.$
    Vậy x ∈ {1/2; -1/2}
    ————————
    14.
    a) (x +1)^4 + (x – 3)^4 = 82
    Đặt t = x – 1 => x= t + 1
    => $\begin{cases} x + 1= t + 1 + 1 = t + 2\\x – 3 = t + 1 – 3 = t -2 \end{cases}$
    Ta có: (x +1)^4 + (x – 3)^4 = 82
    <=> (t + 2)^4 + (t – 2)^4 = 82
    <=> (t^4+4t^3 .2+6t^2 .2^2+4t . 2^3+2^4)+(t^4-4t^3 . 2+6t^2 . 2^2-4t . 2^3+2^4) = 82
    <=> 2t^4 + 48t – 50 = 0
    <=> 2.(t^4 + 24t – 25) = 0
    <=> t^4 – t + 25t – 25 = 0
    <=> t^2.(t^2 – 1) + 25.(t – 1) = 0
    <=> (t^2 -1)(t^2 + 25) = 0
    <=> (t – 1)(t + 1) = 0
    <=> $\left[\begin{matrix} t-1=0\\ t+ 1=0\end{matrix}\right.$ <=>$\left[\begin{matrix} t = 1\\ t = -1\end{matrix}\right.$ <=>$\left[\begin{matrix}x – 1 = 1 \\ x – 1 = -1\end{matrix}\right.$ <=> $\left[\begin{matrix} x = 2\\ x = 0 \end{matrix}\right.$
    Vậy x ∈ {2; 0}

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

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