Tính tích P=\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{99}\right)\)
Những câu hỏi liên quan
\(H=\frac{\left(1+97\right)\left(1+\frac{97}{2}\right)\left(1+\frac{97}{3}\right)\left(1+\frac{97}{4}\right)+...+\left(1+\frac{97}{99}\right)}{\left(1+99\right)\left(1+\frac{99}{2}\right)\left(1+\frac{99}{3}\right)\left(1+\frac{99}{4}\right)+...+\left(1+\frac{99}{97}\right)}\)
bài 1 : tính
1) A = \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)........\left(\frac{1}{99}+1\right)\)
2) B = \(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right).....\left(\frac{1}{99}-1\right)\)
1, A=\(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{100}{99}\)
A= \(\frac{100}{2}\)
A=50
2, B=\(\frac{-1}{2}.\frac{-2}{3}....\frac{-98}{99}\)
B= \(\frac{1}{99}\)
Đúng 0
Bình luận (0)
\(A=\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)......\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}......\frac{99}{98}\cdot\frac{100}{99}\)
\(=\frac{100}{2}\)
\(=50\)
\(B=\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)......\left(\frac{1}{99}-1\right)\)
\(=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot\left(-\frac{3}{4}\right).....\left(-\frac{97}{98}\right)\cdot\left(-\frac{98}{99}\right)\)
\(=-\frac{1}{99}\)
Đúng 0
Bình luận (0)
Tính \(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+......+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\) ta được B=
Tính tích P = \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(P=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(=\frac{100}{2}\)
=500
Đúng 0
Bình luận (0)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
Tính tích:
\(\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times\left(1+\frac{1}{4}\right)\times..............\times\left(1+\frac{1}{99}\right)\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\cdot\cdot\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\cdot\cdot\cdot\frac{100}{99}\)
\(=\frac{100}{2}=50\)
Đúng 0
Bình luận (0)
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}\)
\(=\frac{3.4.5...100}{2.3.4...99}\)
\(=\frac{100}{2}\)
\(=50\)
Đúng 0
Bình luận (0)
hình như thiếu 1/98
đề đó đúng ko ( 1+ 1/2) x (1+ 1/3) x (1+ 1/4) x...x (1+ 1/98) x (1+ 1/99)
Giải:
( 1+ 1/2) x (1+ 1/3) x (1+ 1/4) x...x (1+ 1/98) x (1+ 1/99)
= 3/2 x 4/3 x 5/4 x … x 99/98 x 100/99
= (3 x 4 x 5 x … x 99 x 100) / 2 x 3 x 4 x … 98 x 99
Giản ước ta được:
= 100/2
= 50
Đúng 0
Bình luận (1)
Tính nhanh
a, \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)\)
b, \(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)\)
a) \(=\frac{3}{2}.\frac{4}{3}....\frac{100}{99}=\frac{100}{2}=50\)
Đúng 0
Bình luận (0)
a) =3/2 . 4/3 . 5/4 ...100/99
=\(\frac{3.4.5...100}{2.3.4..99}\)
=\(\frac{100}{2}\)
b) =
Đúng 0
Bình luận (0)
b) = \(\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=-1\left(\frac{1.2.3...99}{2.3.4...100}\right)=-\frac{1}{100}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Tính B = \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+....\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\)
Còn thiếu mũ 99 ở cuối cùng nha
Tính \(\left(\frac{1}{2^2-1}\right)\left(\frac{1}{3^2-1}\right)\left(\frac{1}{4^2-1}\right)...\left(\frac{1}{98^2-1}\right)\left(\frac{1}{99^2-1}\right)\)