Vậy phương trình có nghiệm duy nhất là x = -10.
\(\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\\ \Leftrightarrow\dfrac{x+1}{9}+1+\dfrac{x+2}{8}+1=\dfrac{x+3}{7}+1+\dfrac{x+4}{6}+1\\ \Leftrightarrow\dfrac{x+10}{9}+\dfrac{x+10}{8}=\dfrac{x+10}{7}+\dfrac{x+10}{6}\\ \Leftrightarrow\left(x+10\right).\dfrac{1}{9}+\left(x+10\right).\dfrac{1}{8}-\left(x+10\right).\dfrac{1}{7}-\left(x+10\right).\dfrac{1}{6}=0\\ \Leftrightarrow\left(x+10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\right)=0\)
vì \(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\ne0\)
nên \(x+10=0\Rightarrow x=-10\)
vậy phương trình có tập nghiệm là S={-10}
Ta có:
\(\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\)
\(\Leftrightarrow\dfrac{x+1}{9}+1+\dfrac{x+2}{8}+1=\dfrac{x+3}{7}+1+\dfrac{x+4}{6}+1\)
\(\Leftrightarrow\dfrac{x+1}{9}+\dfrac{9}{9}+\dfrac{x+2}{8}+\dfrac{8}{8}=\dfrac{x+3}{7}+\dfrac{7}{7}+\dfrac{x+4}{6}+\dfrac{6}{6}\)
\(\Leftrightarrow\dfrac{x+1+9}{9}+\dfrac{x+2+8}{8}=\dfrac{x+3+7}{7}+\dfrac{x+4+6}{6}\)
\(\Leftrightarrow\dfrac{x+10}{9}+\dfrac{x+10}{8}=\dfrac{x+10}{7}+\dfrac{x+10}{6}\)
\(\Leftrightarrow\dfrac{x+10}{9}+\dfrac{x+10}{8}-\dfrac{x+10}{7}-\dfrac{x+10}{6}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\right)=0\)
\(\Leftrightarrow\left(x+10\right)\cdot-\dfrac{37}{504}=0\)
\(\Rightarrow x+10=0\)
\(\Leftrightarrow x=-10\)
Vậy tập nghiệm của phương trình trên là S={-10}
Có:x+1/9+x+2/8=x+3/7+x+4/6
<=>x+1/9+1+x+2/8+1=x+3/7+1+x+4/6+1
<=>x+10/9+x+10/8=x+10/7+x+10/6
<=>x+10/9+x+10/8-x+10/7-x+10/6=0
<=>(x+10)(1/9+1/8-1/7-1/6)=0
mà:1/9+1/8-1/7-1/6>0
=>x+10=0
=>x=-10
