(2x-1)^3+(x+2)^3=(3x+1)^3
Đặt $2x-1=a;x+2=b\Rightarrow a+b=2x-1+x+2=3x+1$
Khi đo pt trở thành:
$a^3+b^3=(a+b)^3
\\\Leftrightarrow a^3+b^3=a^3+b^3+3ab(a+b)
\\\Leftrightarrow 3ab(a+b)=0
\\\Leftrightarrow \left[\begin{matrix}a=0\\ b=0\\ a+b=0\end{matrix}\right.
\\\Leftrightarrow \left[\begin{matrix}2x-1=0\\ x+2=0\\ 3x+1=0\end{matrix}\right.
\\\Leftrightarrow ................$
phải là $\dfrac{x-241}{17} + \dfrac{x-220}{19} + \dfrac{x-195}{21} + \dfrac{x-166}{23} = 10$ chứ nhỉ? ^^
pt $\Leftrightarrow (\dfrac{x-241}{17}-1)+(\dfrac{x-220}{19}-2)+(\dfrac{x-195}{21}-3)+(\dfrac{x-166}{23}-4)=0$
$\Leftrightarrow \dfrac{x-258}{17}+\dfrac{x-258}{19}+\dfrac{x-258}{21}+\dfrac{x-258}{23}=0$
$\Leftrightarrow (x-258)(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23})=0$
$\Leftrightarrow x-258=0\Leftrightarrow x=258$ (vì $\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\neq 0$)
Vậy...
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