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

Toán Lớp 9: Giải PT: ( theo giải PT bậc hai 1 ẩn) 1, ( x ² – x + 2)² – (2x+2) ² =0 2, (x ² + 2x -1)(x ² + 2x-2) =2 . Thank You!

Home/ Toán Lớp 9: Giải PT: ( theo giải PT bậc hai 1 ẩn) 1, ( x ² – x + 2)² – (2x+2) ² =0 2, (x ² + 2x -1)(x ² + 2x-2) =2 . Thank You!

Toán Lớp 9: Giải PT: ( theo giải PT bậc hai 1 ẩn) 1, ( x ² – x + 2)² – (2x+2) ² =0 2, (x ² + 2x -1)(x ² + 2x-2) =2 . Thank You!

Tháng Hai 16, 2023 Toán học 1 Comment 0 views

Toán Lớp 9: Giải PT: ( theo giải PT bậc hai 1 ẩn)
1, ( x ² – x + 2)² – (2x+2) ² =0
2, (x ² + 2x -1)(x ² + 2x-2) =2 . Thank You!

Comments ( 1 )

  1. tonhu12
    tonhu12
    16 Tháng Hai, 2023 at 5:18 sáng
    Reply
    1/ $(x^2-x+2)^2-(2x+2)^2=0\\↔(x^2-x+2-2x-2)(x^2-x+2+2x+2)=0\\↔(x^2-3x)(x^2+x+4)=0\\↔x(x-3)(x^2+x+4)=0$
    $x^2+x+4\\=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}\\=\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}$
    Nhận thấy: $\left(x+\dfrac{1}{2}\right)^2\ge 0$
    $↔\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}\ge \dfrac{15}{4}>0$
    $↔x^2+x+4>0$
    mà $x(x-3)(x^2+x+4)=0$
    $→\left[\begin{array}{1}x=0\\x-3=0\end{array}\right.\\↔\left[\begin{array}{1}x=0\\x=3\end{array}\right.$
    Vậy $S=\{0;3\}$
    2/ Đặt $x^2+2x-1=y$
    Pt $→y(y-1)=2$
    $↔y^2-y-2=0\\↔y^2-2y+y-2=0\\↔(y^2-2y)+(y-2)=0\\↔y(y-2)+(y-2)=0\\↔(y+1)(y-2)=0\\↔(x^2+2x-1+1)(x^2+2x-1-2)=0\\↔(x^2+2x)(x^2+3x-x-3)=0\\↔x(x+2)[(x^2+3x)-(x+3)]=0\\↔x(x+2)[x(x+3)-(x+3)]=0\\↔x(x+2)(x-1)(x+3)=0\\↔\left[\begin{array}{1}x=0\\x+2=0\\x-1=0\\x+3=0\end{array}\right.\\↔\left[\begin{array}{1}x=0\\x=-2\\x=1\\x=-3\end{array}\right.$
    Vậy $S=\{0;-2;1;-3\}$

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

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