Toán Lớp 8: 125x^3-(2x+1)^3-(3x-1)^3=0
(x-3)^3+(x+1)^3=8(x-1)^3
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Toán Lớp 8: 125x^3-(2x+1)^3-(3x-1)^3=0
(x-3)^3+(x+1)^3=8(x-1)^3
Comments ( 1 )
$125x^3-\left(2x+1\right)^3-\left(3x-1\right)^3=0$
$⇔15x\left(3x-1\right)\left(2x+1\right)=0$
$⇔\begin{cases} x=0\\3x-1=0\\2x+1=0 \end{cases}$ $⇔\begin{cases} x=0\\x=\dfrac{1}{3}\\\:x=-\dfrac{1}{2} \end{cases}$
$\text{Vậy pt có nghiệm {0 ;$\dfrac{1}{3}$;$-\dfrac{1}{2}$}}$
$—————-$
$\left(x-3\right)^3+\left(x+1\right)^3=8\left(x-1\right)^3$
$⇔2x^3-6x^2+30x-26=8x^3-24x^2+24x-8$
$⇔2x^3-6x^2+30x-26-\left(8x^3-24x^2+24x-8\right)=8x^3-24x^2+24x-8-\left(8x^3-24x^2+24x-8\right)$
$⇔-6x^3+18x^2+6x-18=0$
$⇔-6\left(x-3\right)\left(x+1\right)\left(x-1\right)=0$
$⇔\begin{cases} x-3=0\\x+1=0\\\:x-1=0 \end{cases}$ $⇔\begin{cases} x=3\\x=-1\\x=1\end{cases}$
$\text{Vậy pt có nghiệm {3 ; ± 1}}$