1.tính giá trị biểu thức:
10/5.10+10/10.15+...+10/2015.2020
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Bài 1:
a) \(B=1-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-...-\frac{2}{61.63}-\frac{2}{63.65}\)
\(B=1-\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{61.63}+\frac{2}{63.65}\right)\)
\(B=1-\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{61}-\frac{1}{63}+\frac{1}{63}-\frac{1}{65}\right)\)
\(B=1-\left(\frac{1}{3}-\frac{1}{65}\right)\)
\(B=1-\frac{62}{195}\)
\(B=\frac{133}{195}\)
b) \(C=1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)
\(C=1-\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{95.100}\right)\)
\(C=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{95}-\frac{1}{100}\right)\)
\(C=1-\frac{1}{5}.\left(\frac{1}{5}-\frac{1}{100}\right)\)
\(C=1-\frac{1}{5}.\frac{19}{100}\)
\(C=1-\frac{19}{500}\)
\(C=\frac{481}{500}\)
bài 2 thì bn lm như bn Phùng Minh Quân nha!
Câu 1 : mình ko hiểu đề bài cho lắm ~.~
Câu 2 :
Ta có :
\(\left|\frac{1}{2}-x\right|\ge0\)
\(\Rightarrow\)\(A=10+\left|\frac{1}{2}-x\right|\ge10\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\left|\frac{1}{2}-x\right|=0\)
\(\Leftrightarrow\)\(\frac{1}{2}-x=0\)
\(\Leftrightarrow\)\(x=\frac{1}{2}\)
Vậy GTNN của \(A\) là \(10\) khi \(x=\frac{1}{2}\)
Chúc bạn học tốt ~
=(5/5-5/10+5/10-5/15+.........+5/2015-5/2020)
=(1/5-1/10+1/10-1/20+.......+1/2015-1/2020)
=1/5-1/2020
=403/2020
ai tích mk mk vs
\(\frac{5}{5.10}+\frac{5}{10.15}+.............+\frac{5}{2015.2020}\)
\(=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+..............+\frac{1}{2015}-\frac{1}{2020}\)
\(=\frac{1}{5}-\frac{1}{2020}\)
\(=\frac{403}{2020}\)
\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)
\(=2.\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2015.2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
\(=2.\frac{403}{2020}=\frac{403}{1010}\)
\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)
=\(\frac{2}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+...+\frac{5}{2015.2020}\right)\)
=\(\frac{2}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
=\(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
=\(\frac{2}{5}.\frac{403}{2020}\)
=\(\frac{403}{5005}\)
a: \(A=5\cdot2\cdot\left(-3\right)-10+3\cdot\left(-3\right)=-30-10-9=-49\)
b: \(B=8\cdot1\cdot\left(-1\right)^2-1\cdot\left(-1\right)-2\cdot1-10\)
=8+1-2-10
=-3
10.213
11. 2 giờ 55 phút
12. 4 ngày 21 giờ
13.
a: 6,008
b: 4500
c: 5,628
d: 4,009
=(45.5)^10/75^10
=(225/75)^10
=3^10
=59049
\(\frac{45^{10}\cdot5^{10}}{75^{10}}=\frac{3^{10}\cdot15^{10}\cdot5^{10}}{75^{10}}=\frac{3^{10}\cdot3^{10}\cdot5^{10}\cdot5^{10}}{3^{10}\cdot25^{10}}=\frac{3^{10}\cdot3^{10}\cdot5^{10}\cdot5^{10}}{3^{10}\cdot5^{10}\cdot5^{10}}=\frac{3^{20}\cdot5^{20}}{3^{10}\cdot5^{20}}=3^{10}\)
Đặt A = \(\frac{10}{5.10}+\frac{10}{10.15}+...+\frac{10}{2015.2020}\)
\(=10\left(\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{2015.2020}\right)\)
\(=\frac{10}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{2020}\right)\)\(=\frac{2}{5}-\frac{2}{2020}\)
\(=\frac{2}{5}-\frac{1}{1010}\)\(=\frac{404}{1010}-\frac{1}{1010}\)\(=\frac{403}{1010}\)
Vậy giá trị của biểu thức đã cho là 403/1010
\(\frac{10}{5.10}+\frac{10}{10.15}+...+\frac{10}{2015.2020}\)
\(=2.\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{2015.2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
\(\frac{2}{5}-\frac{1}{1010}\)
Tính nốt nha