Tính giá trị biểu thức A biết:
\(A=16-\frac{-\frac{2}{9}-\frac{2}{10}-\frac{2}{11}-...-\frac{2}{2020}}{\frac{1}{27}+\frac{1}{30}+\frac{1}{33}+...+\frac{1}{6060}}\)
Tính giá trị biểu thức A biết:
\(A=16-\frac{-\frac{2}{9}-\frac{2}{10}-\frac{2}{11}-.....-\frac{2}{2020}}{\frac{1}{27}+\frac{1}{30}+\frac{1}{33}+.....+\frac{1}{6060}}\)
Giải cả bài ra cho mình.
Tính \(A=16-\frac{\frac{-2}{9}+\frac{-2}{10}+\frac{-2}{11}+...+\frac{-2}{2020}}{\frac{1}{27}+\frac{1}{30}+\frac{1}{33}+...+\frac{1}{6060}}\)
\(A=16-\frac{\left(-2\right)\cdot\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)}{\frac{1}{3}\cdot\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)}\)
\(A=16-\frac{-2}{\frac{1}{3}}=16-\left(-6\right)=22\)
Vậy A = 22
Tính giá trị biểu thức
\(\frac{1}{10}+\frac{2}{20}+\frac{9}{30}+\frac{16}{40}+\frac{25}{50}+\frac{36}{60}+\frac{49}{70}+\frac{69}{80}+\frac{81}{90}\)
=1/10+1/10+3/10+4/10+5/10+6/10+7/10+8/10+9/10
=1/10+45/10
=46/10=23/5
Tính giá trị biểu thức A x B
\(A=\frac{\frac{3}{10}+\frac{1}{2}-\frac{1}{6}}{\frac{1}{9}-\frac{1}{5}-\frac{1}{3}}\) \(B=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
tính giá trị biểu thức\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{2019}{1}+\frac{2019}{2}+\frac{2017}{3}+...+\frac{1}{2019}}\)
Sửa đề \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
Ta có: \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)
\(=\left(2019+1\right)+\left(\frac{2018}{2}+1\right)+...+\left(\frac{1}{2019}+1\right)-2019\)
\(=2020+\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}-2020\)
\(=\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}\)
\(=2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)\)Thay vào biểu thức A ta được:
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}}{2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)}=\frac{1}{2020}\)
Bài 1. Tính giá trị của biểu thức sau:
A=\(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
Bài 2. So sánh
A=\(\frac{20^{10}+1}{20^{10}-1}\)
B=\(\frac{20^{10}-1}{20^{10}-3}\)
Bài 3.Tính nhanh
P=\(\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}\)
Bài 1:
\(A=\frac{3333}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{3333}{101}\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{3333}{101}.\frac{4}{21}=\frac{1111.4}{101.7}=\frac{4444}{707}\)
Bài 2
\(A=\frac{2^{10}+1}{2^{10}-1}=\frac{2^{10}-1+2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(B=\frac{2^{10}-1}{2^{10}-3}=\frac{2^{10}-3+4}{2^{10}-3}=1+\frac{4}{2^{10}-3}\)
Ta thấy \(2^{10}-1>2^{10}-3\Rightarrow\frac{2}{2^{10}-1}< \frac{2}{2^{10}-3}< \frac{4}{2^{10}-3}\)
Từ đó \(\Rightarrow1+\frac{2}{2^{10}-1}< 1+\frac{4}{2^{10}-3}\Rightarrow A< B\)
Bài 3\(P=\frac{\left(\frac{2}{3}-\frac{1}{4}\right)+\frac{5}{11}}{\frac{5}{12}+\left(1-\frac{7}{11}\right)}=\frac{\frac{5}{12}+\frac{5}{11}}{\frac{5}{12}+\frac{4}{11}}=\frac{\frac{55+60}{11.12}}{\frac{55+48}{12.11}}=\frac{115}{103}\)
tính giá trị biểu thức
a, A=\(\frac{-1}{2}-\left[\frac{-3}{5}\right]+\left[\frac{-1}{9}\right]+\frac{1}{27}+\frac{7}{18}+\frac{4}{35}-\left[-\frac{2}{7}\right]\)
b, B=\(\frac{1}{3}-\frac{3}{4}-\left[\frac{-3}{5}-\frac{1}{57}+\frac{1}{36}+\frac{-1}{15}\right]-\frac{2}{9}\)
c, C=\(\left[-\frac{7}{15}\right]\times\frac{5}{8}\times\left[\frac{30}{-7}\right]\times\left[-16\right]\times\left[\frac{-1}{1000}\right]\)
d, D=\(\frac{1}{2}\times\frac{-11}{19}-50\%\times\left[-\frac{1}{19}\right]+\frac{10}{19}\times\frac{1111}{2222}\)
tính giá trị biểu thức chứ còn cái gì nữa
a, \(A=\frac{22}{27}\)
b,\(B=\frac{1}{57}\)
C,\(C=\frac{1}{50}\)
d, \(D=0\)
Tính giá trị biểu thức:
\(A=\frac{\frac{16}{10}:\left(1\frac{3}{5}\times \frac{5}{4}\right)}{\frac{64}{100}-\frac{1}{25}}+\frac{\left(\frac{108}{100}-\frac{2}{25}\right):\frac{4}{7}}{\left(5\frac{5}{9}-2\frac{1}{4}\right)\times 2\frac{2}{17}}+\frac{3}{5}\times \frac{1}{2}:\frac{2}{5}\)
Biến đổi biểu thức sau:
\(\frac{5}{1\cdot6}=\frac{1}{?}+\frac{1}{?}\)
a) Từ đó, tính giá trị biểu thức:
\(A=\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{2017\cdot2022}\)
b) Chứng minh \(B< A\)biết:
\(B=\frac{1}{6^2}+\frac{1}{11^2}+\frac{1}{16^2}+...+\frac{1}{2022^2}\)
a) \(A=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+......+\frac{1}{2017.2022}\)
\(5A=5.\left(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+.....+\frac{1}{2017.2022}\right)\)
\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+......+\frac{5}{2017.2022}\)
\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+........+\frac{1}{2017}-\frac{1}{2022}\)
\(5A=1-\frac{1}{2022}\)
\(5A=\frac{2022}{2022}-\frac{1}{2022}\)
\(5A=\frac{2021}{2022}\)
\(A=\frac{2021}{2022}\div5\)
\(A=\frac{20201}{10110}\)
TL:
\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
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HT