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

Toán Lớp 8: Bài 3: Tìm x, biết: a) 3x^33 – 27x = 0 b) x^3 – 3x^2 – 4x + 12 = 0 c) 3x(x –

Tháng Bảy 30, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: Bài 3: Tìm x, biết:
a) 3x^33 – 27x = 0
b) x^3 – 3x^2 – 4x + 12 = 0
c) 3x(x – 1) – 12x + 12 = 0
e) x^2 – 2x – 15 = 0
d) x^33 + x^2 – 4x – 4 = 0

Comments ( 2 )

  1. bichquyenlien
    bichquyenlien
    30 Tháng Bảy, 2022 at 2:41 chiều
    Reply
    Giải đáp:
     
    Lời giải và giải thích chi tiết:
     a,3x^3 -27x=0
    =>3x(x^2 -9)=0
    =>x(x-3)(x+3)=0
    Th1:
    x=0
    Th2:
    x-3=0
    =>x=3
    Th3:
    x+3=0
    =>x=-3
    Vậy x in{0;-3;3}
    d,x^3 +x^2 -4x-4=0
    =>x^2 (x+1)-4(x+1)=0
    =>(x+1)(x-2)(x+2)=0
    Th1:
    x+1=0
    =>x=-1
    Th2:
    x-2=0
    =>x=2
    TH3:
    x+2=0
    =>x=-2
    Vậy x in{-1;2;-2}
    b,x^3 -3x^2 -4x+12=0
    =>x^2 (x-3)-4(x-3)=0
    =>(x-3)(x-2)(x+2)=0
    Th1:
    x-3=0
    =>x=3
    TH2:
    x-2=0
    =>x=2
    Th3:
    x+2=0
    =>x=-2
    Vậy x in{3;2;-2}
    e,x^2 -2x-15=0
    =>x^2 -5x+3x-15=0
    =>x(x-5)+3(x-5)=0
    =>(x-5)(x+3)=0
    Th1:
    x-5=0
    =>x=5
    Th2:
    x+3=0
    =>x=-3
    Vậy x in{5;-3}
    c,3x(x-1)-12x+12=0
    =>3x(x-1)-12(x-1)=0
    =>3(x-1)(x-4)=0
    Th1:
    x-1=0
    =>x=1
    Th2:
    x-4=0
    =>x=4
    Vậy x in{1;4}

  2. cobexinhdep
    cobexinhdep
    30 Tháng Bảy, 2022 at 2:40 chiều
    Reply
    Giải đáp:
     $a) 3x³-27x=0$
    $⇔ 3x.(x²-9)=0$
    $⇔ 3x.(x-3).(x+3)=0$
    $⇔$ \(\left[ \begin{array}{l}3x=0\\x-3=0\\x+3=0\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}x=0\\x=3\\x=-3\end{array} \right.\) 
    $Vậy$ $\text{S={0;±3}}$
    $d) x³+x²-4x-4=0$
    $⇔ x².(x+1)-4.(x+1)=0$
    $⇔ (x-2).(x+2).(x+1)=0$
    $⇔$ \(\left[ \begin{array}{l}x-2=0\\x+2=0\\x+1=0\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}x=2\\x=-2\\x=-1\end{array} \right.\) 
    $\text{Vậy S={±2;-1}}$
    $b) x³-3x²-4x+12=0$
    $⇔ x².(x-3)-4.(x-3)=0$
    $⇔ (x-2).(x+2).(x-3)=0$
    $⇔$ \(\left[ \begin{array}{l}x-2=0\\x+2=0\\x-3=0\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}x=2\\x=-2\\x=3\end{array} \right.\) 
    $\text{Vậy S={±2;3}}$
    $e) x²-2x-15=0$
    $⇔ x²+3x-5x-15=0$
    $⇔ x.(x+3)-5.(x+3)=0$
    $⇔ (x-5).(x+3)=0$
    $⇔$ \(\left[ \begin{array}{l}x-5=0\\x+3=0\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}x=5\\x=-3\end{array} \right.\) 
    $\text{Vậy S={5;-3}}$
    $c) 3x.(x-1)-12x+12=0$
    $⇔ 3x.(x-1)-(12x-12)=0$
    $⇔ 3x.(x-1)-12.(x-1)=0$
    $⇔ (3x-12).(x-1)=0$
    $⇔$ \(\left[ \begin{array}{l}3x-12=0\\x-1=0\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}3x=12\\x=1\end{array} \right.\) $⇔$  \(\left[ \begin{array}{l}x=4\\x=1\end{array} \right.\) 
    $\text{Vậy S={4;1}}$
     

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

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