giải các phươngtrifnhsau
a, \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
b,(\(\frac{x+2}{98}+1\))+(\(\frac{x+3}{97}+1\))=(\(\frac{x+4}{96}+1\))+(\(\frac{x+5}{95}+1\))
c, \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
d, \(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)
\(a.\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\\\Leftrightarrow \left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\\\Leftrightarrow x-23=0\left(vi\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\ne0\right)\\ \Leftrightarrow x=23\)
Này tớ làm tắt có gì cậu không hiểu nói tớ nhé
\(b.\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\\ \Leftrightarrow\frac{x+2}{98}+1+\frac{x+3}{97}+1-\left(\frac{x+4}{96}+1+\frac{x+5}{95}+1\right)=0\\\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\\\Leftrightarrow \left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\\ \Leftrightarrow x+100=0\left(Vi\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\right)\\\Leftrightarrow x=-100\)
\(c.\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\\ \Leftrightarrow\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{+2001}+1\\\Leftrightarrow \frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\\ \Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\\\Leftrightarrow x+2005=0\left(Vi\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\right)\\ \Leftrightarrow x=2005\)
\(d.\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\\\Leftrightarrow \frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{95}+1=0\\\Leftrightarrow \frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\\\Leftrightarrow \left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\\\Leftrightarrow 300-x=0\left(Vi\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\ne0\right)\\\Leftrightarrow x=300\)
