Đáp án + Giải thích các bước giải:
Ta có :
`\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}`
`⇔(\frac{x^2-10x-29}{1971}-1)+(\frac{x^2-10x-27}{1973}-1)=(\frac{x^2-10x-1971}{29}-1)+(\frac{x^2-10x-1973}{27}-1)`
`⇔(\frac{x^2-10x-29}{1971}-\frac{1971}{1971})+(\frac{x^2-10x-27}{1973}-\frac{1973}{1973})=(\frac{x^2-10x-1971}{29}-\frac{29}{29})+(\frac{x^2-10x-1973}{27}-\frac{27}{27})`
`⇔\frac{x^2-10x-2000}{1971}+\frac{x^2-10x-2000}{1973}=\frac{x^2-10x-2000}{29}+\frac{x^2-10x-2000}{27}`
`⇔\frac{x^2-10x-2000}{1971}+\frac{x^2-10x-2000}{1973}-\frac{x^2-10x-2000}{29}-\frac{x^2-10x-2000}{27}=0`
`⇔(x^2-10x-2000)(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27})=0`
`⇔x^2-10x-2000=0` `[ Do (\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27})<0]`
`⇔(x^2+40x)-(50x+2000)=0`
`⇔x(x+40)-50(x+40)=0`
`⇔(x+40)(x-50)=0`
`⇔` \(\left[ \begin{array}{l}x+40=0\\x-50=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-40\\x=50\end{array} \right.\)
Vậy phương trình có tập nghiệm là : `S={-40;50}`
