Question

[Cỏ dại]

Giải các phương trình:

\(a,\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(b,\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)

Answer 1

[Đạt TL]

a) \(\Leftrightarrow\frac{x+2}{98}+\frac{x+4}{96}+2=\frac{x+6}{94}+\frac{x+8}{92}+2\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}+\frac{1}{94}+\frac{1}{92}\right)=0\)

\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)

b)\(\Leftrightarrow\frac{x-12}{77}+\frac{x-11}{78}-2=\frac{x-74}{15}+\frac{x-73}{16}-2\)

\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)

\(\Leftrightarrow\left(x-89\right)\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)

\(\Leftrightarrow x-89=0\Leftrightarrow x=89\)