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

Toán Lớp 8: (2x+1)^2-(5x-2)^2 x^2+4x=5 (3x-2)^2 – (7x+9)^2=0

Toán Lớp 8: (2x+1)^2-(5x-2)^2
x^2+4x=5
(3x-2)^2 – (7x+9)^2=0

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    \text{@Tarminators}
    (2x+1)^2-(5x-2)^2
    = 4x^2+4x+1-25x^2+20x-4
    = (-25x^2+4x^2)+(20x+4x)+(1-4)
    = -21x^2 + 24x – 3
    = -(21x^2-24x+3)
    = -3[(7x^2-x)+(-7x+1)]
    = -3[x(7x-1)-1(7x-1)]
    = -3(7x-1)(x-1)
    x^2 + 4x = 5
    -> x^2 + 4x – 5 = 0
    -> (x^2-x)+(5x-5)=0
    -> x(x-1)+5(x-1)=0
    -> (x-1)(x+5) =0
    TH1 :
    x – 1 = 0
    -> x = 0 + 1
    -> x = 1
    TH2 :
    x + 5 = 0
    -> x = 0 – 5
    -> x = -5
    (3x-2)^2-(7x+9)^2=0
    -> 9x^2-12x+4-49x^2-126x-81=0
    -> -40x^2-138x-77=0
    -> (-40x^2-28x)+(-110x-77)=0
    -> -4x(10x+7)-11(10x+7) = 0
    -> (10x+7)(-4x-11)=0
    TH1 :
    10x + 7 = 0
    -> 10x = -7
    -> x = -7/10
    TH2 :
    -4x – 11 = 0
    -> -4x = 11
    -> x = -11/4

  2. Giải đáp + Lời giải và giải thích chi tiết:
    $(2x+1)^2-(5x-2)^2=(2x+1-5x+2)(2x+1+5x-2)=(3-3x)(7x-1)=3(1-x)(7x-1)$
    $x^2+4x=5$
    $⇔x^2+4x-5=0$
    $⇔(x^2-x)+(5x-5)=0$
    $⇔(x+5)(x-1)=0$
    $⇔\left[\begin{matrix} x=1\\ x=-5\end{matrix}\right.$
    $(3x-2)^2-(7x+9)^2=0$
    $⇔(3x-2-7x-9)(3x-2+7x+9)=0$
    $⇔-(4x+11)(10x+7)=0$
    $⇔\left[\begin{matrix} x=-\dfrac{11}{4}\\ x=-\dfrac{7}{10}\end{matrix}\right.$

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

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