Question

[Trần Kim Anh]

Tính giá trị biểu thức:

A=\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right).....\left(\frac{1}{99}+1\right)\)

Answer 1

[.]

\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)

\(=\left(\frac{1}{2}+\frac{2}{2}\right)\left(\frac{1}{3}+\frac{3}{3}\right)\left(\frac{1}{4}+\frac{4}{4}\right)...\left(\frac{1}{99}+\frac{99}{99}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{100}{2}=50\)

Vậy \(A=50\).

Answer 2

[Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )]

\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)

\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}=\frac{3.4.5.....100}{2.3.4.....99}\)

\(\Leftrightarrow A=\frac{100}{2}=50\)