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

Toán Lớp 8: Giúp mình với đang cần gấp<(_ _)> Bài 5: Tìm x a) x^2-8x+16=0 ; b) 25x^2-9=0 ; c) x^3+1/27=0 ; d) x^3-3/2x^2+3/4x-1/8=0

Toán Lớp 8: Giúp mình với đang cần gấp<(_ _)>
Bài 5: Tìm x
a) x^2-8x+16=0 ; b) 25x^2-9=0 ;
c) x^3+1/27=0 ; d) x^3-3/2x^2+3/4x-1/8=0

Comments ( 2 )

  1. a) x^2-8x+16=0
    ⇔(x-4)²=0
    ⇔x-4=0
    ⇔x=4
    b)25x^2-9=0
    ⇔(5x-3)(5x+3)=0
    ⇔5x-3=0 hoặc 5x+3=0
    ⇔x=3/5 hoặc x=-3/5
    c)x³+1/27=0
    ⇔(x+1/3)(x²-1/3x+1/9)=0
    ⇔x=-1/3 hoặc x²-1/3x+1/9=0(vô nghĩa)
    d)x^3-3/(2x^2)+3/(4x)-1/8=0
    ⇔(x-1/2)³=0
    ⇔x-1/2=0
    ⇔x=1/2

  2. a)
    \(\begin{array}{l}{x^2} – 8x + 16 = 0\\ \Leftrightarrow {x^2} – 2.x.4 + {4^2} = 0\\ \Leftrightarrow {\left( {x – 4} \right)^2} = 0\\ \Leftrightarrow x – 4 = 0\\ \Leftrightarrow x = 4\end{array}\)
    Vậy \(x=4\) .
    b)
    \(25x^2-9=0\)
    \(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)
    \(\Leftrightarrow\left[{}\begin{matrix}5x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
    c)
    x^3 + 1/27 = 0
    <=> (x + 1/3)(x – 1/3 + 1/9) = 0
    <=> x + 1/3 = 0 hoặc x – 1/3 + 1/9 = 0
    <=> x = -1/3 hoặc x = 2/9
    d)
    x^3-3/2x^2+3/4x-1/8=0
    => (x – 1/2)^3 = 0
    => x – 1/2 = 0
    => x = 1/2

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