\(a,4.5^2-64:2^4\)

\(=100-4=96\)

\(b,1-2+3-4+5-...+2017\)

\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(2015-2016\right)+2017\)

Mỗi nhóm có kết quả là -1 và số nhóm sẽ là:

\(\left[\left(2016-1\right)+1\right]:2=1008\)

\(\Rightarrow=1008.\left(-1\right)+2017\)

\(=-1008+2017=1009\)

c, Gọi \(A=1-3+3^2-3^3+3^4-...-3^{2017}\)

\(\Rightarrow3A=3-3^2+3^3-3^4+..+3^{2018}\)

\(\Rightarrow3A+A=\left(3-3^2+3^3-3^4+...+3^{2018}\right)+\left(1-3+3^2-3^3+3^4-...-3^{2017}\right)\)

\(\Rightarrow4A=1+\left(3-3\right)+\left[\left(-3^2\right)+3^2\right]+...+\left(3^{2017}-3^{2017}\right)+3^{2018}\)

\(4A=1+3^{2018}\Rightarrow A=\dfrac{1+3^{2018}}{4}\)

\(-40.\left(-8\right).7.5=-40.\left[\left(-8\right).5\right].7\)

\(=-40.-40.7\)

\(=1600.7=11200\)

Vì \(B=\left\{x\in N;x=2k\left(k\in N\right);13< x\le56\right\}\)

\(\Rightarrow B=\left\{14;16;18;20;...;56\right\}\)

\(A=2016.20152015-2015.20162016\)

\(=2016.2015.10001-2015.2016.10001\)

\(=0\)

\(A=2016.20152015-2015.20162016\)

\(=2016.2015.10001-2015.2016.1001\)

\(=0\)

\(A=\dfrac{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2941}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{1943}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2941}}{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{1943}}\)

\(=\dfrac{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2941}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{1943}}.\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{1943}}{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2941}}\)

\(=\dfrac{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2941}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{1943}\right)}.\dfrac{2\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{1943}\right)}{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2941}\right)}\)

\(=\dfrac{5}{3}.\dfrac{2}{4}=\dfrac{10}{12}=\dfrac{5}{6}\)

Vì đề bài không yêu cầu tính nên bn có thể không tính ra như mk cux đc!

Cuối học kì rùi bn đăng lên làm j?