Question

Giải phương trình: \(\frac{263-x}{27}+\frac{268-x}{25}+\frac{305-x}{23}+\frac{320-x}{21}+\frac{331-x}{19}=15\)

P/S : Ai giải hộ t vs, t sắp ktra giữa kì rùi ....

Answer 1

\(đe\Leftrightarrow\frac{263-x}{27}+\frac{286-x}{25}+\frac{305-x}{23}+\frac{320-x}{21}+\frac{331-x}{19}-15=0\\ \Leftrightarrow\frac{263-x}{27}-1+\frac{286-x}{25}-2+\frac{305-x}{23}-3+\frac{320-x}{21}-4+\frac{331-x}{19}-5=0\\ \Leftrightarrow\frac{236-x}{27}+\frac{236-x}{25}\frac{236-x}{23}+\frac{236-x}{21}+\frac{236-x}{19}=0\\ \Leftrightarrow\left(236-x\right)\left(\frac{1}{27}+\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)

vì \(\left(\frac{1}{27}+\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)\ne0\)

=> \(236-x=0\Leftrightarrow x=236\)

Vậy \(S=\left\{236\right\}\)