tim x biet
a)\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{2499}{2500}\)
b) \(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{50^2}{49.51}\)
Tính nhanh:
\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{49.51}{50.50}\)
Rồi phần sau ntn các bạn làm hộ mình với
Tính
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.......\frac{50^2}{49.51}\)
\(\text{= 2/1 . 2/3 . 3/2 . 3/4 . 4/3 . 4/5 ....... 50/49.50/51 }\)
Dùng phương pháp khử liên tiếp ta có
\(=\frac{2}{1}-\frac{50}{51}=\frac{52}{51}\)
B=\(\frac{2^2}{1.3}\)+\(\frac{3^2}{2.4}+\frac{4^2}{3.5}+\frac{5^2}{4.6}+........+\frac{99^2}{98.100}\).TÌM PHẦN NGUYÊN CỦA B
\(C=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+....+\frac{2499}{2500}\). CM:C>48
\(N=\frac{1.4}{2.3}+\frac{2.5}{3.4}+\frac{3.6}{4.5}+....+\frac{98.101}{99.100}\). CM : 97<N<98
Tính giá trị biểu thức sau: N = \(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{50^2}{49.51}\)
\(\Leftrightarrow N=\frac{\left(2.3.4....50\right)\left(2.3.4...........50\right)}{\left(1.2.3.........49\right)\left(3.4.5...........51\right)}=\frac{50.2}{51}=\frac{100}{51}\)
Tính giá trị biểu thức sau: N = \(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}....+\frac{50^2}{49.51}\)
\(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+....+\frac{50^2}{49.51}\)
\(=\frac{2^2-1}{1.3}+\frac{3^2-1}{2.4}+....+\frac{50^2-1}{49.51}+\frac{1}{1.3}+\frac{1}{2.4}+....+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(1+1+...+1\right)+\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{49}-\frac{1}{51}\)
Tự làm tiếp :))
tớ nhầm đoạn này tí :((
\(=\left(1+1+....+1\right)+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)(49 chữ số 1)
\(=49+\frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\right)\right]\)
\(=49+\left(\frac{3}{2}-\frac{1}{50}-\frac{1}{51}\right):2\)Tự tính
sao bạn Khuyển Dạ Xoa làm 2 bài vậy?
Tính
E = \(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+.............+\frac{50^2}{49.51}\)
D = \(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}......\frac{2499}{2500}\) DẤU "." Ở DÂN LÀ DẤU NHÂN
C = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{999}\right).\left(1-\frac{1}{1000}\right)\)
G = \(\left(1-\frac{1}{11}\right).\left(1-\frac{2}{11}\right)......\left(1-\frac{20}{11}\right)\)
ĐANG CẦN GẤP
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{50^2}{49.51}\) tính
giải giúp mình nha
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.,,\frac{50^2}{49.51}\)
=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.,,\frac{50.50}{49.51}\)
=\(\frac{\left(2.3.4...50\right).\left(2.3.4...50\right)}{\left(1.2.3....49\right).\left(3.4.5....51\right)}\)
=\(\frac{50.2}{1.51}\)
=\(\frac{100}{51}\)
\(=\frac{2.3.4...50}{1.2.3...49}.\frac{2.3.4...50}{3.4.5...51}=50.\frac{2}{51}=\frac{100}{51}\)
Tính giá trị biểu thức sau: N = \(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{50^2}{49.51}\)
Cho \(H=2+\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}.\) Chứng minh rằng H > 50 .