​​a,3×3+6×2−4x=0 d,(2×2+3)2−10×3-15x=0 c,(x2+x+1)2=(4x−1)2 b,(x+1)3−x+1=(x−1)(x−2)

​​a,3×3+6×2−4x=0
d,(2×2+3)2−10×3-15x=0
c,(x2+x+1)2=(4x−1)2
b,(x+1)3−x+1=(x−1)(x−2)

0 bình luận về “​​a,3×3+6×2−4x=0 d,(2×2+3)2−10×3-15x=0 c,(x2+x+1)2=(4x−1)2 b,(x+1)3−x+1=(x−1)(x−2)”

  1. Đáp án:

    $a)  {\left[\begin{aligned}x=\frac{-6+\sqrt{84}}{6}\\x=\frac{-6-\sqrt{84}}{6}\end{aligned}\right.}\\
    d) {\left[\begin{aligned}x=1\\x=\frac{3}{2}\end{aligned}\right.}\\
    c)  {\left[\begin{aligned}x=0\\x=2\\ x=1 \\ x=-5\end{aligned}\right.}\\
    b) x=0$

    Giải thích các bước giải:

    $a) 3x^3+6x^2-4x=0\\
    \Leftrightarrow x(3x^2+6x-4)=0\\
    \Leftrightarrow {\left[\begin{aligned}x=0\\3x^2+6x-4=0\end{aligned}\right.}\\
    +) 3x^2+6x-4=0\\
    \Delta =6^2-4.3.(-4)=84>0\\
    \Rightarrow {\left[\begin{aligned}x=\frac{-6+\sqrt{84}}{6}\\x=\frac{-6-\sqrt{84}}{6}\end{aligned}\right.}\\
    d) (2x^2+3)^2-10x^3-15x=0\\
    \Leftrightarrow (2x^2+3)^2-5x(2x^2+3)=0\\
    \Leftrightarrow (2x^2+3)(2x^2+3-5x)=0\\
    \Leftrightarrow {\left[\begin{aligned}2x^2+3=0\\2x^2-5x+3=0\end{aligned}\right.} \\
    +) 2x^2+3=0\\
    \Delta =-4.2.3=-48<0$
    Vậy phương trình vô nghiệm
    $+) 2x^2-5x+3=0\\
    \Leftrightarrow 2x^2-2x-3x+3=0\\
    \Leftrightarrow 2x(x-1)-3(x-1)=0\\
    \Leftrightarrow (x-1)(2x-3)=0\\
    \Leftrightarrow {\left[\begin{aligned}x-1=0\\2x-3=0\end{aligned}\right.}\\
    \Leftrightarrow {\left[\begin{aligned}x=1\\x=\frac{3}{2}\end{aligned}\right.}\\
    c) (x^2+x+1)^2=(4x-1)^2\\
    \Leftrightarrow (x^2+x+1)^2-(4x-1)^2=0\\
    \Leftrightarrow (x^2+x+1-4x+1)(x^2+x+1+4x-1)=0\\
    \Leftrightarrow (x^2-3x+2)(x^2+5x)=0\\
    \Leftrightarrow x(x^2-3x+2)(x+5)=0\\
    \Leftrightarrow x(x^2-x-2x+2)(x+5)=0\\
    \Leftrightarrow x\left [x(x-1)-2(x-1)  \right ](x+5)=0\\
    \Leftrightarrow x(x-1)(x-2)(x-1)=0
    \Leftrightarrow {\left[\begin{aligned}x=0\\x-1=0\\ x-2=0\\ x+5=0\end{aligned}\right.}\\
    \Leftrightarrow {\left[\begin{aligned}x=0\\x=2\\ x=1 \\ x=-5\end{aligned}\right.}\\
    b) (x+1)^3-x+1=(x-1)(x-2)\\
    \Leftrightarrow x^3+3x^2+3x+1-x+1-(x-1)(x-2)=0\\
    \Leftrightarrow  x^3+3x^2+3x+1-x+1-(x^2-2x-x+2) =0\\
    \Leftrightarrow x^3+3x^2+3x+1-x+1-x^2+2x+x-2 =0\\
    \Leftrightarrow x^3+2x^2+5x =0\\
    \Leftrightarrow x(x^2+2x+5)=0\\
    \Leftrightarrow {\left[\begin{aligned}x=0\\x^2+2x+5=0\end{aligned}\right.}\\
    +) x^2+2x+5=0\\
    \Delta =2^2-4.1.5=-16<0$
    Do đó phương trình vô nghiệm

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