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

Toán Lớp 8: tìm x để 16x^2-(2x-1)^2==0 9(x+2)^2-4.(x-3)^2=0

Tháng Sáu 19, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: tìm x để
16x^2-(2x-1)^2==0
9(x+2)^2-4.(x-3)^2=0

Comments ( 1 )

  1. dantam45
    dantam45
    19 Tháng Sáu, 2022 at 4:08 sáng
    Reply
    Giải đáp:
    16x^2-(2x-1)^2=0
    <=> (4x-2x+1)(4x+2x-1)=0
    <=> (2x+1)(6x-1)=0
    <=>\(\left[ \begin{array}{l}2x+1=0\\6x-1=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=\dfrac{-1}{2}\\x=\dfrac{1}{6}\end{array} \right.\) 
    Vậy S={-1/2;1/6}
    $\\$
    9(x+2)^2-4(x-3)^2=0
    <=> (3(x+2)-2(x-3))(3(x+2)+2(x-3))=0
    <=> (3x+6-2x+6)(3x+6+2x-6)=0
    <=> (x+12)5x=0
    <=> x(x+12)=0
    <=>\(\left[ \begin{array}{l}x=0\\x+12=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=0\\x=-12\end{array} \right.\) 
    Vậy S={-12;0}

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

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