thực hiện phép tính sau một cách nhanh nhất:
\(1\frac{5}{7}\cdot0.75-\frac{6}{7}\cdot1\frac{1}{3}+\frac{6}{7}\)
\(2017\cdot2018\left(\frac{2016}{2017}-\frac{2016}{2018}\right)\)
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{100}-1\right)\)
\(\frac{1023}{2+2^2+2^3+....+2^{10}}\)GIÚP MÌNH VỚI!HELP ME!
\(=\frac{12}{7}\cdot\frac{3}{4}-\frac{6}{7}\cdot\frac{4}{3}+\frac{6}{7}\)
\(=\frac{6}{7}\left(\frac{3}{2}-\frac{4}{3}+1\right)\)
\(=\frac{6}{7}\left(\frac{1}{6}+1\right)=\frac{6}{7}\cdot\frac{7}{6}=1\)
2.
\(=2017\cdot2018\cdot\left[\left(2016\cdot2018\right)-\left(2016\cdot2017\right)\right]\)
\(=2017\cdot2018\cdot2016\left(2018-2017\right)=2016\cdot2017\cdot2018\)
3.
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{100}-1\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{99}{100}\)
\(=\frac{1}{100}\)
4.
\(=\frac{1+2+2^2+2^4+...+2^9}{2\left(1+2+2^2+2^3+2^4+...+2^9\right)}\)
\(=\frac{1}{2}\)
mình chỉ làm được câu 3 thôi
có \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)....\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}\times\frac{-2}{3}\times....\times\frac{-99}{100}\)
\(=\frac{\left(-1\right)\left(-2\right)....\left(-99\right)}{2\times3\times....\times100}\)
\(=\frac{-\left(1\times2\times....\times99\right)}{2\times3\times....\times100}\)
\(=\frac{-1}{100}\)
câu 1
\(=\frac{12}{7}.0,75+\frac{6}{7}\left(\frac{-4}{3}+1\right)\)
\(=\frac{9}{7}+\frac{6}{7}.\frac{-1}{3}\)
\(=\frac{9}{7}+\frac{-2}{7}\)
\(=1\)
Câu 2
có \(2017.2018\left(\frac{2016}{2017}-\frac{2016}{2018}\right)\)
\(=2017.2018.\frac{2016}{2017}-2017.2018.\frac{2016}{2018}\)
\(=2018.2016-2017.2016\)
\(=2016\left(2018-2017\right)\)
\(=2016\)
Câu 4
Đặt A=2+22+23+....+210
\(\Rightarrow2A=2^2+2^3+....+2^{11}\)
\(\Rightarrow2A-A=2^{11}-2\)
\(\Rightarrow A=2^{11}-2\)\(=2046\)
nên \(\frac{1023}{2+2^2+....+2^{10}}\)\(=\frac{1023}{2046}\)\(=\frac{1}{2}\)
hình như câu 2 bạn sai rồi Xuân Anh
ra 2016 mà
