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

Toán Lớp 8: 1)x^3-7x -6=0 2)x^3+4x^2-7x-10=0 3)x^3 -6x^2+11x-6=0 4)x^3-19x-30=0

Home/ Toán Lớp 8: 1)x^3-7x -6=0 2)x^3+4x^2-7x-10=0 3)x^3 -6x^2+11x-6=0 4)x^3-19x-30=0

Toán Lớp 8: 1)x^3-7x -6=0 2)x^3+4x^2-7x-10=0 3)x^3 -6x^2+11x-6=0 4)x^3-19x-30=0

Tháng Mười Hai 25, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: 1)x^3-7x -6=0
2)x^3+4x^2-7x-10=0
3)x^3 -6x^2+11x-6=0
4)x^3-19x-30=0

Comments ( 2 )

  1. maitrucloan
    maitrucloan
    25 Tháng Mười Hai, 2022 at 4:53 sáng
    Reply
    Giải đáp+Lời giải và giải thích chi tiết:
    a) x^3 – 7x – 6=0
    <=> x^3 +2x^2 – 2x^2 -4x-3x – 6=0
    <=> (x^3-2x^2-3x)+(2x^2 -4x -6)=0
    <=> x(x^2-2x-3)+2(x^2-2x-3)=0
    <=> (x+2)(x^2-2x-3)=0
    <=> (x+2)(x^2 +x-3x -3)=0
    <=> (x+2)[x(x+1)-3(x+1)]=0
    <=> (x+2)(x+1)(x-3)=0
    <=> \(\left[ \begin{array}{l}x+2=0\\x+1=0\\x-3=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=-2\\x=-1\\x=3\end{array} \right.\) 
         Vậy S={-2; -1; 3}
    2) x^3+4x^2-7x-10=0
    <=> x^3 +6x^2-2x^2  +5x-12x=0
    <=> (x^3+6x^2+5x)-(2x^2+12x+10)=0
    <=> x(x^2+6x+5)-2(x^2+6x+5)=0
    <=> (x-2)(x^2+6x+5)=0
    <=> (x-2)(x^2 +x + 5x +5)=0
    <=> (x-2)[x(x+1)+5(x+1)]=0
    <=> (x-2)(x+1)(x+5)=0
    <=>\(\left[ \begin{array}{l}x-2=0\\x+1=0\\x+5=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=2\\x=-1\\x=-5\end{array} \right.\) 
        Vậy S={2; -1; -5}
    3) x^3 – 6x^2 +11x-6=0
    <=> x^3 -2x^2 -4x^2+8x+3x-6=0
    <=> (x^3-4x^2+3x)-(2x^2-8x+6)=0
    <=> x(x^2 – 4x +3)-2(x^2-4x + 3)=0
    <=> (x-2)(x^2-4x+3)=0
    <=> (x-2)(x^2  – 3x -x +3 )=0
    <=> (x-2)[x(x-3)-(x-3)]=0
    <=> (x-2)(x-1)(x-3)=0
    <=>\(\left[ \begin{array}{l}x-2=0\\x-1=0\\x-3=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=2\\x=1\\x=3\end{array} \right.\) 
          Vậy S={1; 2; 3}
    4) x^3 – 19x – 30 =0
    <=> x^3 +5x^2-5x^2+6x-25x -30=0
    <=> (x^3+5x^2+6x)-(5x^2+25x+30)=0
    <=> x(x^2+5x +6)-5(x^2-5x+6)=0
    <=> (x-5)(x^2+5x+6)=0
    <=> (x-5)(x^2+3x+2x+6)=0
    <=> (x-5)[x(x+3)+2(x+3)]
    <=> (x-5)(x+2)(x+3)=0
    <=>\(\left[ \begin{array}{l}x-5=0\\x+2=0\\x+3=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=5\\x=-2\\x=-3\end{array} \right.\) 
                Vậy S={5; -2; -3}

  2. doanthulan
    doanthulan
    25 Tháng Mười Hai, 2022 at 4:53 sáng
    Reply
    Giải đáp:
    1)S={-1;-2;3}
    2)S={-1;2;-5}
    3)S={1;2;3}
    4)S={-2;-3;5}
    Lời giải và giải thích chi tiết:
    1)x³-7x-6=0
    ⇔x³+3x²-3x²+2x-9x-6=0
    ⇔(x³+3x²+2x)-(3x²+9x+6)=0
    ⇔x(x²+3x+2)-3(x²+3x+2)=0
    ⇔(x²+3x+2)(x-3)=0
    ⇔(x²+x+2x+2)(x-3)=0
    ⇔[x(x+1)+2(x+1)](x-3)=0
    ⇔(x+1)(x+2)(x-3)=0
    ⇔$\left[\begin{matrix} x+1=0\\ x+2=0\\x-3=0\end{matrix}\right.$
    ⇔$\left[\begin{matrix} x=-1\\ x=-2\\x=3\end{matrix}\right.$
    Vậy S={-1;-2;3}
    2)x³+4x²-7x-10=0
    ⇔x³-x²+5x²-2x-5x-10=0
    ⇔(x³-x²-2x)+(5x²-5x-10)=0
    ⇔x(x²-x-2)+5(x²-x-2)=0
    ⇔(x²-x-2)(x+5)=0
    ⇔(x²-2x+x-2)(x+5)=0
    ⇔[x(x-2)+(x-2)](x+5)=0
    ⇔(x+1)(x-2)(x+5)=0
    ⇔$\left[\begin{matrix} x+1=0\\ x-2=0\\x+5=0\end{matrix}\right.$
    ⇔$\left[\begin{matrix} x=-1\\ x=2\\x=-5\end{matrix}\right.$
    Vậy S={-1;2;-5}
    3)x³-6x²+11x-6=0
    ⇔x³-3x²-3x²+2x+9x-6=0
    ⇔(x³-3x²+2x)-(3x²-9x+6)=0
    ⇔x(x²-3x+2)-3(x²-3x+2)=0
    ⇔(x²-3x+2)(x-3)=0
    ⇔(x²-2x-x+2)(x-3)=0
    ⇔[x(x-2)-(x-2)](x-3)=0
    ⇔(x-1)(x-2)(x-3)=0
    ⇔$\left[\begin{matrix} x-1=0\\ x-2=0\\x-3=0\end{matrix}\right.$
    ⇔$\left[\begin{matrix} x=1\\ x=2\\x=3\end{matrix}\right.$
    Vậy S={1;2;3}
    4)x³-19x-30=0
    ⇔x³+5x²-5x²+6x-25x-30=0
    ⇔(x³+5x²+6x)-(5x²+25x+30)=0
    ⇔x(x²+5x+6)-5(x²+5x+6)=0
    ⇔(x²+5x+6)(x-5)=0
    ⇔(x²+2x+3x+6)(x-5)=0
    ⇔[x(x+2)+3(x+2)](x-5)=0
    ⇔(x+2)(x+3)(x-5)=0
    ⇔$\left[\begin{matrix} x+2=0\\ x+3=0\\x-5=0\end{matrix}\right.$
    ⇔$\left[\begin{matrix} x=-2\\ x=-3\\x=5\end{matrix}\right.$
    Vậy S={-2;-3;5}

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

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