Question

\(\frac{x-2}{71}+\frac{x-4}{69}=\frac{x-6}{67}+\frac{x-8}{65}\)

giải hộ tớ vơi. mai tớ ktra rồi

Answer 1

\(\frac{x-2}{71}+\frac{x-4}{69}=\frac{x-6}{67}+\frac{x-8}{65}\)

\(\Leftrightarrow\frac{x-2}{71}-1+\frac{x-4}{69}-1=\frac{x-6}{67}-1+\frac{x-8}{65}-1\)

\(\Leftrightarrow\frac{x-73}{71}+\frac{x-73}{69}=\frac{x-73}{67}+\frac{x-73}{65}\)

\(\Leftrightarrow\frac{x-73}{71}+\frac{x-73}{69}-\frac{x-73}{67}-\frac{x-73}{65}=0\)

\(\Leftrightarrow\left(x-73\right)\left(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\right)=0\)

Mà \(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\ne0\)

\(x-73=0\Leftrightarrow x=73\)

Answer 2

\(\frac{x-2}{71}-1+\frac{x-4}{69}-1=\frac{x-6}{67}-1+\frac{x-8}{65}-1\)

\(\Leftrightarrow\frac{x-73}{71}+\frac{x-73}{69}=\frac{x-73}{67}+\frac{x-73}{65}\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\right)=0\)

\(\Rightarrow...\)

Answer 3

Ta có: \(\frac{x-2}{71}+\frac{x-4}{69}=\frac{x-6}{67}+\frac{x-8}{65}\)

\(\frac{x-2}{71}+\frac{x-4}{69}-2=\frac{x-6}{67}+\frac{x-8}{65}\)

\(\frac{x-2}{71}-1+\frac{x-4}{69}-1=\frac{x-6}{67}-1+\frac{x-8}{65}-1=0\)

\(\frac{x-73}{71}+\frac{x-73}{69}-\frac{x-6}{67}-\frac{x-8}{65}=0\)

\(\left(x-73\right)\left(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\right)=0\)

\(\left(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\right)\ne0\)

=> x - 73 = 0 <=> x = 73

Vậy x = 73.