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

Toán Lớp 8: Bài 1: Tìm x: 1. 3x .( x – 1 ) + x – 1 = 0 2. 2 ( x + 3 ) – $x^{2}$ – 3x = 0 3. 4$x^{2}$ – 25 – ( 2x – 5 ) ( 2x + 7 ) = 0 4. $x^{3}$

Tháng Sáu 27, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: Bài 1: Tìm x:
1. 3x .( x – 1 ) + x – 1 = 0
2. 2 ( x + 3 ) – $x^{2}$ – 3x = 0
3. 4$x^{2}$ – 25 – ( 2x – 5 ) ( 2x + 7 ) = 0
4. $x^{3}$ + 27 + ( x + 3 ) ( x – 9) = 0
5.( x + 2 )$^{2}$ – ( x – 2) ( x + 2) = 0
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Comments ( 1 )

  1. cobexinhdep
    cobexinhdep
    27 Tháng Sáu, 2022 at 12:49 chiều
    Reply
    1. 3x(x – 1) + x – 1 = 0
    ⇔3x (x – 1) +( x – 1) = 0
    ⇔(x – 1) (3x + 1) = 0
    ⇔\(\left[ \begin{array}{l}x – 1=0\\3x + 1=0\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=1\\x=-1/3\end{array} \right.\) 
    2. 2( x + 3) – x² – 3x = 0
    ⇔2 ( x + 3) -x (x + 3) = 0
    ⇔(x + 3) (2 – x ) = 0
    ⇔\(\left[ \begin{array}{l}x + 3=0\\2 – x= 0\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=-3\\x=2\end{array} \right.\) 
    3. 4x² – 25 – (2x + 5)(2x + 7) = 0
    ⇔(2x + 5) ( 2x – 5) – (2x – 5) ( 2x + 7) = 0
    ⇔ (2x – 5) (2x + 5 – 2x -7)=0
    ⇔(2x – 5) -2 = 0
    ⇔2x – 5 = 0
    ⇔2x = 5
    ⇔x = $\frac{5}{2}$ 
    4. x³ + 27 +(x + 3)(x – 9) =0
    ⇔(x + 3) ( x² -3x + 9) + (x + 3) (x – 9 )=0
    ⇔(x + 3) (x² -3x +9 +x -9 ) =0
    ⇔(x + 3)( x² – 2x )=0
    ⇔( x + 3) x(x – 2)=0
    ⇔ \(\left[ \begin{array}{l}x + 3=0\\x=0\\x – 2 = 0\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=-3\\x=0\\x = 2\end{array} \right.\) 
    5. (x + 2)² – (x – 2) (x + 2) = 0
    ⇔(x + 2) (x + 2 – x + 2)=0
    ⇔(x + 2)4 = 0
    ⇔x + 2 = 0
    ⇔ x = -2
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222-9+11+12:2*14+14 = ? ( )

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