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

Toán Lớp 8: Tìm `x` biết: `(x-2)(x-5)-x(2x-1)=x^2+4x` `(2x-1)^2-2x(x-1)=-9`

Toán Lớp 8: Tìm x biết:
(x-2)(x-5)-x(2x-1)=x^2+4x
(2x-1)^2-2x(x-1)=-9

Comments ( 2 )

  1. a/ (x-2)(x-5)-x(2x-1)=x^2+4x
    ⇔ x^2-7x+10-2x^2+x=x^2+4x
    ⇔ -x^2-6x+10=x^2+4x
    ⇔ -2x^2-10x+10=0
    ⇔ x^2+5x-5=0
    ⇔ x=\frac{-5\pm 3\sqrt 5}{2}
    Vậy S={\frac{-5\pm 3\sqrt 5}{2}}
    b/ (2x-1)^2-2x(x-1)=-9
    ⇔ 4x^2-4x+1-2x^2+2x=-9
    ⇔ 2x^2-2x+1=-9
    ⇔ 2x^2-2x+10=0 (vô nghiệm)
    Vậy $S= \{\varnothing \}$

  2. Giải đáp +  Lời giải và giải thích chi tiết:
    1. (x-2)(x-5)-x(2x-1) = x^2 + 4x
    <=> x^2 – 7x + 10 – 2x^2 + x -x^2 – 4x = 0
    <=> -2x^2 -10x + 10 = 0
    <=> -2(x^2 + 5x – 5 ) = 0
    <=> x^2 +5x-5=0
    C1: Dùng Delta nếu như em đã học =)))))
    Delta = 5^2 – 4.1.(-5) = 45
    => text{Pt có hai nghiệm} \(\left[ \begin{array}{l}x=\dfrac{-5+ \sqrt{45}}{2}\\x=\dfrac{-5- \sqrt{45}}{2}\end{array} \right.\) 
    C2:
    <=> [x^2 + 2.x. 5/2 + (5/2)^2] – 45/4 =0
    <=> (x+5/2)^2 = 45/4
    <=> (x+5/2)^2 = (\sqrt45/2)^2
    <=> \(\left[ \begin{array}{l}x+\dfrac52= \dfrac{\sqrt{45}}{2}\\x+\dfrac52= \dfrac{-\sqrt{45}}{2}\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x= \dfrac{-5 + \sqrt{45}}{2}\\x= \dfrac{-5-\sqrt{45}}{2}\end{array} \right.\) 
    KL: Vậy  x = (-5 + \sqrt{45})/2  hoặc x = (-5 – \sqrt{45})/2
    2. (2x-1)^2 – 2x(x-1) = -9
    => 4x^2 – 4x + 1 – 2x^2 + 2x = -9
    => 2x^2 – 2x + 10=0
    => 2( x^2 – x + 5 ) = 0
    => x^2 – x + 5 = 0
    => (x^2 – 2.x. 1/2 + 1/4) + 19/4 = 0
    => (x-1/2)^2 + 19/4 = 0
    Do (x-1/2)^2 $\geq$ 0 AA xinRR
    => (x-1/2)^2  +19/4 $\geq$ 19/4 
    => text{Phương trình vô nghiệm}

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

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