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

Toán Lớp 8: Tìm x biết: a/ (2x-1)^2 – (3+x)^2 = (x-2)(x+2) + 2(x-1) b/(3-2x)^2 – (x-2)^2 = 2(x-3)(x+3) + (x-1)(x+2)

Toán Lớp 8: Tìm x biết:
a/ (2x-1)^2 – (3+x)^2 = (x-2)(x+2) + 2(x-1)
b/(3-2x)^2 – (x-2)^2 = 2(x-3)(x+3) + (x-1)(x+2)

Comments ( 2 )

  1. Giải đáp:
    a)x∈{$\sqrt[]{10}+3$;$3-\sqrt[]{10}$}
    b)x=25/9
    Lời giải và giải thích chi tiết:
    a)(2x-1)²-(3+x)²=(x-2)(x+2)+2(x-1)
    ⇔4x²-4x+1-(9+6x+x²)=x²-4+2x-2
    ⇔4x²-4x+1-9-6x-x²=x²-4+2x-2
    ⇔4x²-4x+1-9-6x-x²-x²+4-2x+2=0
    ⇔(4x²-x²-x²)-(4x+6x+2x)+(1-9+4+2)=0
    ⇔2x²-12x-2=0
    ⇔2(x²-6x-1)=0
    ⇔x²-6x-1=0
    ⇔x²-6x+9-10=0
    ⇔x²-6x+9=10
    ⇔(x-3)²=($\sqrt[]{10}$ )^2
    ⇔$\left[\begin{matrix} x-3=\sqrt[]{10}\\ x-3=-\sqrt[]{10}\end{matrix}\right.$
    ⇔$\left[\begin{matrix} x=\sqrt[]{10}+3\\ x=3-\sqrt[]{10}\end{matrix}\right.$
    Vậy x∈{$\sqrt[]{10}+3$;$3-\sqrt[]{10}$}
    b)(3-2x)²-(x-2)²=2(x-3)(x+3)+(x-1)(x+2)
    ⇔9-12x+4x²-(x²-4x+4)=2(x²-9)+x²+2x-x-2
    ⇔9-12x+4x²-x²+4x-4=2x²-18+x²+2x-x-2
    ⇔(4x²-x²)+(-12x+4x)+(9-4)=(2x²+x²)+(2x-x)-(18+2)
    ⇔3x²-8x+5=3x²+x-20
    ⇔3x²-8x-3x²-x=-20-5
    ⇔-9x=-25
    ⇔x=25/9
    Vậy x=25/9

  2. Giải đáp:
    a, x = √10 + 3 hoặc x = -√10 + 3
    b,x = $\frac{25}{9}$ 
     
    Lời giải và giải thích chi tiết:
    a, (2x – 1)² – (3 – x)² = (x – 2).(x + 2) + 2.(x – 1)
    ⇔ 4x² – 4x + 1 – 9 – 6x – x² = x² – 4 + 2x – 2
    ⇔ 2x² – 12x -2 = 0
    ⇔ x² – 6x – 1 = 0
    ⇔ (x – 3)² – 10 = 0
    ⇔ (x – 3)² = 10
    ⇒\(\left[ \begin{array}{l}x-3=√10\\x – 3 = -√10\end{array} \right.\) 
    ⇒\(\left[ \begin{array}{l}x=√10 +3\\x=-√10+3\end{array} \right.\) 
    b, (3-2x)² – (x-2)² = 2(x-3)(x+3) + (x-1)(x+2)
    ⇔ 9 – 12x + 4x² – x² + 4x – 4 = 2x² – 18 + x² + x – 2
    ⇔ -9x + 25 = 0
    ⇔ -9x = -25
    ⇔ x = $\frac{25}{9}$ 
    chúc bạn học tốt ~
    mirasuki
     

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