\(S=1+2+3-4-5+6+7+8-9-10+...+2011+2012+2013-2014-2015\)

\(=\left(1+2+3-4-5\right)+\left(6+7+8-9-10\right)+...+\left(2011+2012+2013-2014-2015\right)\)

\(=\left(-3\right)+2+...+2007\)

Từ 2 đến 2007 sẽ có: \(\dfrac{2007-2}{5}+1=402\left(số\right)\)

Tổng của dãy số 2;7;12;...;2007 sẽ là:

\(\dfrac{\left(2007+2\right)\cdot402}{2}=403809\)

=>S=403809-3=403806

a: \(A=21\cdot100-11\cdot100+90\cdot100+100\cdot125\cdot16\)

\(=100\left(21-11+90\right)+100\cdot2000\)

\(=100\left(10+90+2000\right)=2100\cdot100=210000\)

b: \(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{28}\cdot3^{18}\left(5\cdot3-7\cdot2\right)}=2\)

a. bằng77

b. bằng 13

lớp 2 đã học cái này rồi à

a. bằng77

b. bằng 13

BÀI 1

a,  \(5\times\frac{-7}{10}=\frac{-35}{10}=\frac{-7}{2}\)

b,  \(\frac{4}{5}\times\frac{-7}{10}=\frac{-28}{50}=\frac{-14}{25}\)

c,  \(\frac{4}{9}+\frac{4}{3}\times\frac{16}{4}=\frac{4}{9}+\frac{16}{3}=\frac{52}{9}\)

d,  \(\frac{11}{22}-\frac{3}{9}\times\frac{14}{21}=\frac{11}{22}-\frac{2}{9}=\frac{55}{198}=\frac{5}{18}\)  

BÀI 2

\(A=\frac{6}{13}\times\frac{5}{7}+\frac{6}{13}\times\frac{2}{7}+\frac{17}{13}\)

\(A=\frac{30}{91}+\frac{12}{91}+\frac{17}{13}\)

\(A=\frac{30}{91}+\frac{12}{91}+\frac{119}{91}\)

\(A=\frac{161}{91}=\frac{23}{13}\)

\(B=\frac{11}{15}\times\frac{4}{11}+\frac{11}{15}\times\frac{5}{11}+\frac{11}{15}\times\frac{2}{11}\)

\(B=\frac{4}{15}+\frac{1}{3}+\frac{2}{15}\)

\(B=\frac{11}{15}\)

\(C=\left(\frac{19}{64}-\frac{33}{22}+\frac{24}{51}\right)\times\left(\frac{1}{5}-\frac{1}{15}-\frac{2}{15}\right)\)

\(C=\frac{-797}{1088}\times0\)

\(C=0\)

\(D=\frac{8}{13}\times\frac{7}{12}+\frac{8}{13}\times\frac{5}{12}-\frac{1}{12}\)

\(D=\frac{14}{39}+\frac{10}{39}-\frac{1}{12}\)

\(D=\frac{83}{156}\)

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