Question

(x+2/98+1)+(x+3/97+1)=(x+4/96+1)+(x+5/95+1)

Answer 1

\(\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+1\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

\(\Leftrightarrow x+100=0\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\right)\)

<=> x=-100

Answer 2

ko chép đề nhé 

\(\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95} \)

=> \((x+100)(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95})=0\)

vì \((\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}) khác 0\)

=>\(x+100=0\)

=>x=-100

Answer 3

\(\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+1\)

\(< =>\frac{x+2+98}{98}+\frac{x+3+97}{97}=\frac{x+4+96}{96}+\frac{x+5+95}{95}\)

\(< =>\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)

\(< =>\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)

\(< =>\left(x+100\right).\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

\(Do:\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)

\(< =>x+100=0\)

\(< =>x=-100\)

Vậy nghiệm của phương trình trên là {-100}

Answer 4

\(\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+1\)

\(\frac{x+2}{98}+\frac{98}{98}+\frac{x+3}{97}+\frac{97}{97}=\frac{x+4}{96}+\frac{96}{96}+\frac{x+5}{95}+\frac{95}{95}\)

\(\frac{x+2+98}{98}+\frac{x+3+97}{97}=\frac{x+4+96}{96}+\frac{x+5+95}{95}\)

\(\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}=0\)

\(\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)

\(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

\(Th1:x+100=0\Leftrightarrow x=-100\)

\(Th2:\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)

Vậy pt cs nghiệm là -100