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

Toán Lớp 8: giải phương trình sau 1/ x^3-16x=0 2/ x^2-25x=0 3/ (3x-5)^2 -( x-4)^2 =0 4/ 25x^2 =(4x-3)^2

Tháng Mười Hai 3, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: giải phương trình sau
1/ x^3-16x=0
2/ x^2-25x=0
3/ (3x-5)^2 -( x-4)^2 =0
4/ 25x^2 =(4x-3)^2

Comments ( 2 )

  1. truongthanhhung
    truongthanhhung
    3 Tháng Mười Hai, 2022 at 8:33 chiều
    Reply
    1, x^3-16x=0
    <=> x(x^2-16)=0
    <=> x(x-4)(x+4)=0
    <=> [(x=0),(x-4=0),(x+4=0):}
    <=> [(x=0),(x=4),(x=-4):}
    Vậy S={0;4;-4}
    2, x^2-25x=0
    <=> x(x-25)=0
    <=> [(x=0),(x-25=0):}
    <=> [(x=0),(x=25):}
    Vậy S={0;25}
    3, (3x-5)^2-(x-4)^2=0
    <=> (3x-5-x+4)(3x-5+x-4)=0
    <=> (2x-1)(4x-9)=0
    <=> [(2x-1=0),(4x-9=0):}
    <=>\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=\dfrac{9}{4}\end{array} \right.\) 
    Vậy S={1/2;9/4}
    4, 25x^2=(4x-3)^2
    <=> (5x)^2-(4x-3)^2=0
    <=> (5x-4x+3)(5x+4x-3)=0
    <=> (x+3)(9x-3)=0
    <=> 3(x+3)(3x-1)=0
    <=> [(x+3=0),(3x-1=0):}
    <=>\(\left[ \begin{array}{l}x=-3\\x=\dfrac{1}{3}\end{array} \right.\) 
    Vậy S={-3;1/3}
     

  2. cattien
    cattien
    3 Tháng Mười Hai, 2022 at 8:33 chiều
    Reply
    ~ Bạn tham khảo ~
    1, x^3 – 16x = 0
    <=> x(x^2 – 16) = 0
    <=> x(x-4)(x+4)=0
    \(⇔\left[ \begin{array}{l}x=0\\x-4=0\\x+4=0\end{array} \right.\)
    \(⇔\left[ \begin{array}{l}x=0\\x=4\\x=-4\end{array} \right.\)
    Vậy S={0;4;-4}
    2, x^2-25x=0
    <=> x(x-25)=0
    \(⇔\left[ \begin{array}{l}x=0\\x-25=0\end{array} \right.\)
    \(⇔\left[ \begin{array}{l}x=0\\x=25\end{array} \right.\)
    Vậy S={0;25}
    3,(3x-5)^2 -( x-4)^2 =0
    <=> (3x-5-x+4)(3x-5+x-4)=0
    <=> (2x-1)(4x-9)=0
    \(⇔\left[ \begin{array}{l}2x-1=0\\4x-9=0\end{array} \right.\)
    \(⇔\left[ \begin{array}{l}2x=1\\4x=9\end{array} \right.\)
    \(⇔\left[ \begin{array}{l}x=\dfrac12\\x=\dfrac94\end{array} \right.\)
    Vậy S={1/2;9/4}
    4, 25x^2 =(4x-3)^2
    <=> (5x)^2 – (4x-3)^2=0
    <=> (5x-4x+3)(5x+4x-3)=0
    <=> (x+3)(9x-3)=0
    <=> 3(x+3)(3x-1)=0
    \(⇔\left[ \begin{array}{l}x+3=0\\3x-1=0\end{array} \right.\)
    \(⇔\left[ \begin{array}{l}x=-3\\3x=1\end{array} \right.\)
    \(⇔\left[ \begin{array}{l}x=-3\\x=\dfrac13\end{array} \right.\)
    Vậy S={-3;1/3}

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

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