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

Toán Lớp 8: tìm x biết a.x^4-6x^2+3x+2=0 b.2x(x-5)-7(5-x)=0

Tháng Năm 25, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: tìm x biết a.x^4-6x^2+3x+2=0 b.2x(x-5)-7(5-x)=0

Comments ( 2 )

  1. minhtuyethue
    minhtuyethue
    25 Tháng Năm, 2022 at 11:55 chiều
    Reply
    Giải đáp+Lời giải và giải thích chi tiết:
     a) x^4-6x^2+3x+2=0
    <=>x^4-4x^2-2x^2+4x-x+2=0
    <=>x^2(x^2-4)-2x(x-2)-(x-2)=0
    <=>x^2(x-2)(x+2)-2x(x-2)-(x-2)=0
    <=>(x-2)[x^2(x+2)-2x-1]=0
    <=>(x-2)(x^3+2x^2-2x-1)=0
    <=>(x-2)(x^3-x^2+3x^2-3x+x-1)=0
    <=>(x-2)[x^2(x-1)+3x(x-1)+(x-1)]=0
    <=>(x-2)(x-1)(x^2+3x+1)=0
     TH1: x-2=0
     <=>x=2
     TH2: x-1=0
     <=>x=1
     TH3:x^2+3x+1=0
     <=>x^2+2.x. 3/2+(3/2)^2+5/4-5/4=0
    <=>(x+3/2)^2=5/4
    <=>x+3/2=+-(sqrt5)/2
    <=>$\left[\begin{matrix} x+\dfrac{3}{2}=\dfrac{\sqrt5}{2}\\ x+\dfrac{3}{2}=-\dfrac{\sqrt5}{2}\end{matrix}\right.$
    <=>$\left[\begin{matrix} x=\dfrac{-3+\sqrt5}{2}\\ x=\dfrac{-3-\sqrt5}{2}\end{matrix}\right.$
     Vậy x∈{1; 2; (-3+sqrt5)/2; (-3-sqrt5)/2}
     b) 2x(x-5)-7(5-x)=0
    <=>2x(x-5)+7(x-5)=0
    <=>(x-5)(2x+7)=0
    <=>$\left[\begin{matrix} x-5=0\\ 2x+7=0\end{matrix}\right.$
    <=>$\left[\begin{matrix} x=5\\ 2x=-7\end{matrix}\right.$
    <=>$\left[\begin{matrix} x=5\\ x=\dfrac{-7}{2}\end{matrix}\right.$
     Vậy x=5 hoặc x=(-7)/2

  2. hauthanhyen98
    hauthanhyen98
    25 Tháng Năm, 2022 at 11:55 chiều
    Reply
    #huy
    a)x^4-6x^2+3x+2=0
    <=>x^4-4x^2-2x^2+4x-x+2=0
    <=>x^2(x^2-4)-2x(x-2)-(x-2)=0
    <=>x^2(x-2)(x+2)-2x(x-2)-(x-2)=0
    <=>(x-2)[(x^2(x+2)-2x-1]=0
    <=>(x-2)(x^3+2x^2-2x-1)=0
    <=>(x-2)[x^2(x-1)+3x(x-1)+(x-1)]=0
    <=>(x-2)(x-1)(x^2+3x+1)=0
    <=>[(x-2=0),(x-1=0),(x^2+3x+1=0):}
    <=>[(x=2),(x=1),(x^2+2.x. 3/2+(3/2)^2+5/4-5/4=0):}
    <=>[(x=2),(x=1),((x+3/2)^2=5/4):}
    <=>[(x=2),(x=1),(x+3/2=±(\sqrt{5})/2):}
    <=>[(x=2),(x=1),([(x+3/2=(\sqrt{5})/2),(x+3/2=-(\sqrt{5})/2):}):}
    <=>[(x=2),(x=1),([(x=(-3+\sqrt{5})/2),(x=(-3-\sqrt{5})/2):}):}
    Vật x\in{1;2;(-3+\sqrt{5})/2;(-3-\sqrt{5})/2}
    b)2x(x-5)-7(5-x)=0
    <=>2x(x-5)-7(x-5)=0
    <=>(x-5)(2×7)=0
    <=>[(x-5=0),(2x+7=0):}
    <=>[(x=5),(2x=-7):}
    <=>[(x=5),(x=(-7)/2):}
    Vậy x=5 hoặc x=(-7)/2

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

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