Ta có :
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=-4+4\)
\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)nên \(x+100=0\)
\(\Rightarrow x=0-100=-100\)
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yk nha
Đề không sai ?
\(x+\frac{1}{99}+x+\frac{2}{98}+x+\frac{3}{97}+x+\frac{4}{96}\)
\(\Rightarrow4x+\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+\frac{4}{96}=-4\)
\(\Rightarrow\left(-4\right)-\left(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+\frac{4}{96}\right)=4x\)
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Rightarrow\frac{x+100-99}{99}+\frac{x+100-98}{98}+\frac{x+100-97}{97}=\frac{x+100-96}{96}+\frac{x+100-95}{95}\)
\(+\frac{x+100-94}{94}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}-3\)
\(=\frac{x+100}{96}+\frac{x+100}{95}+\frac{x+100}{94}-3\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\right)=0\)
\(\Leftrightarrow x-100\)
\(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)
