Ta có:\(\frac{x+1}{11}+\frac{x+2}{10}=\frac{x+3}{9}+\frac{x+4}{8}\)
\(\Rightarrow1+\frac{x+1}{11}+1+\frac{x+2}{10}=1+\frac{x+3}{9}+1+\frac{x+4}{8}\)
\(\Rightarrow\frac{x+12}{11}+\frac{x+12}{10}=\frac{x+12}{9}+\frac{x+12}{8}\)
\(\Rightarrow\frac{x+12}{11}+\frac{x+12}{10}-\frac{x+12}{9}-\frac{x+12}{8}=0\)
\(\Rightarrow\left(x+12\right)\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)
Mà \(\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)>0\)
\(\Rightarrow x+12=0\Rightarrow x=-12\)
\(\frac{x+1}{11}+\frac{x+2}{10}=\frac{x+3}{9}+\frac{x+4}{8}\)
<=> \(\frac{x+1}{11}+\frac{x+2}{10}-\frac{x+3}{9}-\frac{x+4}{8}=0\)
<=> \(\left(\frac{x+1}{11}+1\right)+\left(\frac{x+2}{10}+1\right)-\left(\frac{x+3}{9}+1\right)-\left(\frac{x+4}{8}+1\right)=0\)<=> \(\frac{x+12}{11}+\frac{x+12}{10}-\frac{x+12}{9}-\frac{x+12}{8}=0\)
<=> \(\left(x+12\right)\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)
<=> x + 12 = 0.Vì \(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\ne0\)
<=> x = -12
\(\frac{x+1}{11}+\frac{x+2}{10}=\frac{x+3}{9}+\frac{x+4}{8}\)
\(\Rightarrow\left(\frac{x+1}{11}+1\right)+\left(\frac{x+2}{10}+1\right)=\left(\frac{x+3}{9}+1\right)+\left(\frac{x+4}{8}+1\right)\)
\(\Rightarrow\frac{x+12}{10}+\frac{x+12}{11}=\frac{x+12}{9}=\frac{x+12}{8}\)
\(\Rightarrow\frac{x+12}{11}+\frac{x+12}{10}-\frac{x+12}{9}-\frac{x+12}{8}=0\)
\(\Rightarrow\left(x+12\right)\left(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)
Mà \(\frac{1}{11}+\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\ne0\)
\(\Rightarrow x+12=0\)
\(\Rightarrow x=-12\)
Vậy \(x=-12\)
lộn thay dấu > thành dấu \(\ne\) nhé
