Mình không ghi lại đề:
\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}=\frac{1}{5}\)
\(\frac{2}{\left(x+1\right)\left(x+3\right)}+...+\frac{2}{\left(x+7\right)\left(x+9\right)}=\frac{2}{5}\)
\(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+...+\frac{1}{x+7}-\frac{1}{x+9}=\frac{2}{5}\)
\(\frac{1}{x+1}-\frac{1}{x+9}=\frac{2}{5}\)
\(\frac{8}{\left(x+1\right)\left(x+9\right)}=\frac{2}{5}\)
<=>40=2(x+1)(x+9)
<=>\(x^2+10x-11=0\)
<=>\(\left(x-1\right)\left(x+11\right)=0\)
<=>x=1 hoặc x=-11
Ta có:
\(1^2+\left(-11\right)^2=122\)
Ai thấy mình làm đúng thì tích nha.Ai tích mình mình tích lại
