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

Toán Lớp 8: Đề bài: tìm x 1) (x-5)^2 – 81(2x-3)^2=0 2) (x+2)^3+(x-3)^3=0

Toán Lớp 8: Đề bài: tìm x
1) (x-5)^2 – 81(2x-3)^2=0
2) (x+2)^3+(x-3)^3=0

Comments ( 2 )

  1. Giải đáp + Lời giải và giải thích chi tiết:
    1, (x-5)^2 – 81(2x-3)^2 = 0
    => x^2 – 10x + 25 – 324x^2 + 972x – 729 = 0
    => -323x^2 + 962x – 704 = 0
    => -(323x^2-962x+704) = 0
    => -(323x^2-418x-544x+704) = 0
    => -[19x(17x-22)-32(17x-22)] = 0
    => -(17x-22)(19x-32) = 0
    => \(\left[ \begin{array}{l}17x-22=0\\19x-32=0\end{array} \right.\) 
    => \(\left[ \begin{array}{l}x=\dfrac{22}{17}\\x=\dfrac{32}{19}\end{array} \right.\) 
    Vậy S = {22/17,32/19}
    2, (x+2)^3-(x-3)^3=0
    => x^3 + 6x^2 + 12x + 8 + x^3 – 9x^2 + 27x – 27 = 0
    => 2x^3 – 3x^2 + 39x – 19 = 0
    => 2x^3 – x^2 – 2x^2 + x + 38x – 19 = 0
    => x^2(2x-1)-x(2x-1)+19(2x-1)=0
    => (2x-1)(x^2-x+19) = 0
    x^2-x+19=[x^2-2*x*1/2+(1/2)^2]+75/4 = (x-1/2)^2 + 75/4
    Vì (x-1/2)^2 \ge 0 => (x-1/2)^2 + 75/4 \ge 75/4 > 0
    => 2x – 1= 0
    => 2x = 1
    => x = 1/2
    Vậy S = {1/2}

  2. ***Lời giải***
     1)
    (x-5)^2 – 81(2x-3)^2=0
    <=>(x-5)^2 – [9(2x-3)]^2=0
    <=>[x-5-9(2x-3)][x-5+9(2x-3)]=0
    <=>(x-5-18x+27)(x-5+18x-27)=0
    <=>(-17x+22)(19x-32)=0
    <=>\(\left[ \begin{array}{l}-17x+22=0\\19x-32=0\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}-17x=-22\\19x=32\end{array} \right.\)
    <=>\(\left[ \begin{array}{l}x=\dfrac{22}{17}\\x=\dfrac{32}{19}\end{array} \right.\)
    Vậy S={22/17;32/19}
    2)
    (x+2)^3+(x-3)^3=0
    <=>[(x+2)+(x-3)][(x+2)^2-(x+2)(x-3)+(x-3)^2]=0
    <=>(x+2+x-3)[x^2+4x+4-(x^2-3x+2x-6)+x^2-6x+9]=0
    <=>(x+2+x-3)(x^2+4x+4-x^2+3x-2x+6+x^2-6x+9)=0
    <=>(2x-1)(x^2-x+19)=0
    Mà x^2-x+19=x^2-x+1/4+75/4=(x-1/2)^2+75/4
    Với ∀x Ta có: (x-1/2)^2≥0<=>(x-1/2)^2+75/4≥75/4>0
    =>2x-1=0
    <=>2x=1
    <=>x=1/2
    Vậy S={1/2}

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

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