\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)
\(=\left(\frac{1}{2}+\frac{2}{2}\right)\left(\frac{1}{3}+\frac{3}{3}\right)\left(\frac{1}{4}+\frac{4}{4}\right)...\left(\frac{1}{99}+\frac{99}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{100}{2}=50\)
Vậy \(A=50\).
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)
\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}=\frac{3.4.5.....100}{2.3.4.....99}\)
\(\Leftrightarrow A=\frac{100}{2}=50\)
mk ko viết lại đề bài đâu nhá:
\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}........\frac{99}{98}.\frac{100}{99}\)
\(A=\frac{100}{2}=50\)
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right).....\left(\frac{1}{99}+1\right)\)
\(\Leftrightarrow A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(\Leftrightarrow A=\frac{3.4.5.6....100}{2.3.4.5.6....99}=\frac{100}{2}=50\)
nếu đúng cho mk ha
Bài giải
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)
\(A=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(A=\frac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot99}=\frac{100}{2}=50\)
\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{3.4.5.5...100}{2.3.4...99}=50\)