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

Toán Lớp 8: Bài 1 : Tìm x biết a) x^2 + 5x = 0 b) x^2 – 7x +6 = 0 c) (5- x)^2 – (x + 3)^3= 0

Toán Lớp 8: Bài 1 : Tìm x biết
a) x^2 + 5x = 0
b) x^2 – 7x +6 = 0
c) (5- x)^2 – (x + 3)^3= 0

Comments ( 2 )

  1. Giải đáp:
     
    Lời giải và giải thích chi tiết:
     a,x^2 +5x=0
    =>x(x+5)=0
    =>\(\left[ \begin{array}{l}x=0\\x+5=0\end{array} \right.\) $\\$=>\(\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.\) $\\$Vậy x in {-5;0}
    b,x^2 -7x+6=0
    =>x^2 -x-6x+6=0
    =>x(x-1)-6(x-1)=0
    =>(x-1)(x-6)=0
    =>\(\left[ \begin{array}{l}x-1=0\\x-6=0\end{array} \right.\) $\\$=>\(\left[ \begin{array}{l}x=1\\x=6\end{array} \right.\) $\\$Vậy x in {1;6}
    (5-x)^2 -(x+3)^2 =0
    =>[(5-x)-(x+3)][(5-x)+(x+3)]=0
    =>(5-x-x-3)(5-x+x+3)=0
    =>(2-2x).8=0
    =>2-2x=0
    =>2x=2
    =>x=1

  2. a) $x^{2}+5x=0$
    <=> x(x+5)=0
    <=> $\left[\begin{matrix} x=0\\ x+5=0\end{matrix}\right.$
    <=> $\left[\begin{matrix} x=0\\ x=-5\end{matrix}\right.$
    Vậy x=0; x=-5
    b) $x^{2}-7x+6=0$
    <=> $x^{2}-x-6x+6=0$
    <=> $x(x-1)-6(x-1)=0$
    <=> $(x-1)(x-6)=0$
    <=> $\left[\begin{matrix} x-1=0\\ x-6=0\end{matrix}\right.$
    <=> $\left[\begin{matrix} x=1\\ x=6\end{matrix}\right.$
    Vậy x=1; x=6
    c) $(5-x)^{2}-(x+3)^{2}=0$
    <=> $25-10x+x^{2}-x^{2}-6x-9=0$
    <=> $16-16x=0$
    <=> $-16x=-16$
    <=> x=1
    Vậy x=1
    – GIANG –

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