2
\(\text{a) }\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+.....+\frac{1}{98.99.100}\right)x=-3\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{98.99}-\frac{1}{99.100}\right)x=-3\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)x=-3\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)x=-3\)
\(\Rightarrow\frac{1}{2}.\left(\frac{4950}{9900}-\frac{1}{9900}\right)x=-3\)
\(\Rightarrow\left(\frac{1}{2}.\frac{4949}{9900}\right).x=-3\)
\(\Rightarrow\frac{4949}{19800}x=-3\)
\(\Rightarrow x=\left(-3\right).\frac{19800}{4949}\)
\(\Rightarrow x=\frac{-59400}{4949}\)
P/s : ko chắc nha
Câu 2.b)
ta có\(\frac{x^2-1}{x+1}=\frac{\left(x-1\right)\left(x+1\right)}{x+1}=1+\frac{x-1}{x+1}\)
vậy \(x+1\varepsilonƯ_{x-1}\)
DẾn đây bạn tự làm tiếp ik