\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
A=\(1-\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\)
A=\(\frac{2}{3}+\frac{1}{12}\)
A=\(\frac{3}{4}\)
B=\(\left(1+\frac{1}{2}\right)\)\(\left(1+\frac{1}{3}\right)\)\(\left(1+\frac{1}{4}\right)\)...\(\left(1+\frac{1}{99}\right)\)
B=\(\frac{3}{2}\).\(\frac{4}{3}\).\(\frac{5}{4}\)...\(\frac{100}{99}\)
B=\(\frac{3.4.5...100}{2.3.4...99}\)
B=\(\frac{100}{2}\)
B=50
\(\frac{4}{x}\)=\(\frac{-y}{6}\)=0.5
\(\frac{4}{x}\)=\(\frac{-y}{6}\)=\(\frac{1}{5}\)
=> \(\frac{4}{x}\)=\(\frac{1}{5}\)=>\(x\)=\(\frac{4.5}{1}\)=9
\(\frac{-y}{6}\)=\(\frac{1}{5}\)=>\(-y\)=\(\frac{6.1}{5}\)=\(\frac{6}{5}\)=> \(y\)=\(\frac{-6}{5}\)
Vậy \(x\)= 9
\(y\)=\(\frac{-6}{5}\)
Đề bài chỉ bảo tính \(x\)nhưng mình tính cả \(y\)nếu có bài tìm cả \(y\)thì áp dụng nha
