Giá trị biểu thức
\(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{20}\)là
\(\frac{1}{2}S=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{21}\)
\(\Rightarrow\left(\frac{1}{2}S\right)-S=\left(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{21}\right)-\left(\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{20}\right)\)
\(\Rightarrow-\frac{1}{2}S=\left(\frac{1}{2}\right)^{21}-\left(\frac{1}{2}\right)\)
\(\Rightarrow S=\frac{\left(\left(\frac{1}{2}\right)^{21}-\frac{1}{2}\right)}{-\frac{1}{2}}\)
Giá trị biểu thức:
\(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+......+\left(\frac{1}{2}\right)^{20}\)
Đặt \(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{20}\)(1)
\(\Rightarrow2A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{19}\)(2)
Lấy (2) trừ đi (1) ta có :
\(\Rightarrow2A-A=1-\left(\frac{1}{2}\right)^{20}\)
\(\Rightarrow A=1-\left(\frac{1}{2}\right)^{20}\)
tính giá trị của biểu thức
\(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)....\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)....\left(20^4+\frac{1}{4}\right)}\)
tính giá trị biểu thức C=\(\sqrt{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2}+\frac{1}{3}\right)...\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)}\)
tính giá trị của biểu thức A = \(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}\)
bài này hình như có trong sách Nâng cao phát triển toán 8?
tính giá trị của biểu thức A = \(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right).......\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)........\left(20^4+\frac{1}{4}\right)}\)
Nhân từng thừa số ở cả tử và mẫu với 16 là ra ý mà
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{n+1}\right)\left(n\in N\right)\)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+.......+\frac{1}{20}\left(1+2+3+4....+20\right)\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n}{n+1}\)
\(=\frac{1}{n+1}\)
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)...+\frac{1}{20}.\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+\frac{1}{4}.4.5:2+...+\frac{1}{20}.20.21:2\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+5+...+21}{2}=115\)
Tính giá trị của biểu thức \(A=\left(\frac{1}{125}-\frac{1}{1^3}\right).\left(\frac{1}{125}-\frac{1}{2^3}\right).\left(\frac{1}{125}-\frac{1}{3^3}\right)...\left(\frac{1}{125}-\frac{1}{19^3}\right).\left(\frac{1}{125}-\frac{1}{20^3}\right)\)
1/ Tính tổng:
M =\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
2/Tìm X:
\(\frac{1}{21}+\frac{1}{28} +\frac{1}{36} +.....+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
3/Tính tích sau rồi tìm số nghịch đảo của kết quả:
\(T=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{8}\right)\left(1-\frac{1}{10}\right)\)
4/ TÍnh giá trị của biểu thức:
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}\frac{4^2}{4.5}\)
\(1.\)\(M=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{6}-\frac{1}{7}\)
\(M=1-\frac{1}{7}=\frac{6}{7}\)
Mình làm câu 1 thoi nha!
1.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
