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a) \(2\frac{3}{4}\cdot\left(-0,4\right)-1\frac{3}{5}\cdot2,75+1,2:\frac{4}{11}\)
\(=2\frac{3}{4}\cdot\left(-\frac{2}{5}\right)-1\frac{3}{5}\cdot\frac{11}{4}+\frac{6}{5}:\frac{4}{11}\)
\(=\frac{11}{4}\cdot\left(-\frac{2}{5}\right)-1\frac{3}{5}\cdot\frac{11}{4}+\frac{6}{5}\cdot\frac{11}{4}\)
\(=\frac{11}{4}\left(-\frac{2}{5}-1\frac{3}{5}+\frac{6}{5}\right)\)
\(=\frac{11}{4}\left(-\frac{2}{5}-\frac{8}{5}+\frac{6}{5}\right)\)
\(=\frac{11}{4}\cdot\left(-\frac{4}{5}\right)=\frac{11}{1}\cdot\left(-\frac{1}{5}\right)=-\frac{11}{5}\)
b) \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)....\left(\frac{1}{31}+1\right)\)
\(=\left(\frac{1}{2}+\frac{2}{2}\right)\left(\frac{1}{3}+\frac{3}{3}\right)\left(\frac{1}{4}+\frac{4}{4}\right)...\left(\frac{1}{31}+\frac{31}{31}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{32}{31}\)
\(=\frac{3\cdot4\cdot5\cdot...\cdot32}{2\cdot3\cdot4\cdot...\cdot31}=\frac{32}{2}=16\)
c) Đặt \(C=1+2+3+...+30\)
Số số hạng là : \(\left(30-1\right):1+1=30\)(số)
Tổng của dãy số là : \(\frac{\left(1+30\right)\cdot30}{2}=465\)
Do đó : \(\frac{930}{C}=\frac{930}{465}=2\)
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
a) \(\frac{1}{9}+3,25+5\frac{3}{16}+4\frac{1}{3}+2,8+0,5=\frac{1}{9}+\frac{13}{4}+\frac{83}{16}+\frac{13}{3}+\frac{14}{5}+\frac{1}{2}\)
\(=\frac{11651}{720}\)
B) \(2\frac{1}{3}+0,45+4,25+\frac{1}{81}+6\frac{8}{27}=\frac{7}{3}+\frac{9}{20}+\frac{17}{4}+\frac{1}{81}+\frac{170}{27}\)
\(=\frac{10807}{810}\)
C) \(1,25+2\frac{1}{4}+4\frac{2}{5}+0,3+2,14+4\frac{1}{8}=\frac{5}{4}+\frac{9}{4}+\frac{22}{5}+\frac{3}{10}+\frac{107}{50}+\frac{33}{8}\)
\(=\frac{2893}{200}\)
CHÚC BN HỌC TỐT!!!!!
\(C=1+4^1+4^2+3^3+...+4^{31}\)
\(4C=4\left(1+4^1+4^2+4^3+...+4^{31}\right)\)
\(4C=4+4^2+4^3+4^4+...+4^{32}\)
\(4C-C=\left(4+4^2+4^3+4^4+...+4^{32}\right)-\left(1+4^1+4^2+4^3+...+4^{31}\right)\)
\(3C=4^{32}-1\)
\(C=\dfrac{4^{32}-1}{3}\)
4C = \(4+4^2+4^3+...+4^{31}\)
3C = \(4^{31}-1\)
\(C=\dfrac{4^{31}-1}{3}\)