Cộng hai vào mỗi vễ của phương trình ta có
\(\Leftrightarrow\frac{x+2}{98}+\frac{x+4}{96}+2=\frac{x+6}{94}+\frac{x+8}{92}+2\)
\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{X+8}{92}+1\)
\(\Leftrightarrow\frac{x+2+98}{98}+\frac{x+4+96}{96}=\frac{x+6+94}{94}+\frac{x+8+92}{92}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
\(\Rightarrow x+100=0\Leftrightarrow x=-100\)
Vậy S={-100}
toan lop 7 ma khong biet con gia vo toan 8
Cộng 2 rồi xuống đổi 1, thế đổi 1 luôn từ đầu đi cho nhanh ạ.
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+2}{92}+1\)
\(\Leftrightarrow\frac{x+2}{98}+\frac{98}{98}+\frac{x+4}{96}+\frac{96}{96}=\frac{x+6}{94}+\frac{94}{94}+\frac{x+8}{92}+\frac{92}{92}\)
\(\Leftrightarrow\frac{x+2+98}{98}+\frac{x+4+96}{96}=\frac{x+6+94}{94}+\frac{x+8+92}{92}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)
Mà \(x+100=0\Leftrightarrow x=-100\)
Vậy x = -100
1 cách khá là KHỦNG :)))
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(< =>\frac{96\left(x+2\right)+98\left(x+4\right)}{98.96}=\frac{92\left(x+6\right)+94\left(x+8\right)}{94.92}\)
\(< =>\frac{96x+192+392+98x}{9408}=\frac{92x+94x+552+752}{8648}\)
\(< =>\frac{194x+584}{9408}=\frac{186x+1304}{8648}\)
\(< =>\frac{97x+292}{4704}=\frac{93x+652}{4324}\)
\(< =>\left(97x+292\right).4324=\left(93x+652\right).4704\)
\(< =>419428x+1262608=437472x+3067008\)
\(< =>1262608-3067008=437472x-419428x\)
\(< =>-1804400=18044x\)
\(< =>x=\frac{-1804400}{18044}=-100\)
Mù mắt chưa +.+
cách bình thường
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(< =>\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)(cộng mỗi vế cho 2)
\(< =>\frac{x+2+98}{98}+\frac{x+4+96}{96}=\frac{x+6+94}{94}+\frac{x+8+92}{92}\)
\(< =>\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(< =>\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Do \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)
Suy ra \(x+100=0\)\(< =>x=-100\)
