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29 tháng 6 2017

Câu 1;

\(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\cdot\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}\cdot\frac{3\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{4\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)

\(=\frac{1}{2}\cdot\frac{3}{4}+\frac{5}{8}=\frac{3}{8}+\frac{5}{8}=1\)

Câu 2:

\(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{3}}=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{3}}=\frac{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}=\frac{3}{11}\)

29 tháng 6 2017

Câu 1;

13 17 113 23 27 213  ·34 316 364 3256 114 116 164  +58 

=13 17 113 2(13 17 113 ) ·3(14 116 164 1256 )4(14 116 164 1256 ) +58 

=12 ·34 +58 =38 +58 =1

Câu 2:

\(B=\frac{1}{3}-\frac{3}{4}+0,6+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)

\(\Rightarrow B=\frac{3}{15}-\frac{48}{64}+\frac{9}{15}+\frac{1}{64}-\frac{8}{36}-\frac{1}{36}+\frac{1}{15}\)

\(\Rightarrow B=\frac{3}{15}+\frac{9}{15}+\frac{1}{15}+\left(-\frac{48}{64}+\frac{1}{64}\right)+\left(-\frac{8}{36}-\frac{1}{36}\right)\)

\(\Rightarrow B=\frac{13}{15}-\frac{47}{64}-\frac{1}{4}\)

\(\Rightarrow B=-\frac{113}{960}\)

\(C=0\)

\(D=\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Rightarrow D=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+...-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)

\(\Rightarrow D=1\)

11 tháng 8 2019

D= \(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}......-\frac{1}{3.2}-\frac{1}{2.1}\)

=\(\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{97.98}+\frac{1}{98.99}\right)\)

=\(\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{98}-\frac{1}{99}\right)\)

=\(\frac{1}{99}-\left[1-(\frac{1}{2}-\frac{1}{2}+......+\frac{1}{98}-\frac{1}{99})\right]\)

=\(\frac{1}{99}-\left(1-0-0-.....-0-\frac{1}{99}\right)\)

=\(\frac{1}{99}-1-\frac{1}{99}\)

=1

26 tháng 7 2018

Câu hỏi của Quỳnh Như - Toán lớp 7 | Học trực tuyến

\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{3}{8}+\dfrac{5}{8}=1\)

23 tháng 6 2015

=3/4-3/5+3/7+3/13 / 11/4-11/5+11/7+11/13   +   3/4-3/5+3/7+3/13 / 11/4-11/5+11/7+11/13

=3.1/4-3.1/5+3.1/7+3.1/13 / 11.1/4-11.1/5+11.1/7+11.1/13    +     3.1/4-3.1/5+3.1/7+3.1/13  /  11. 1/4-11.1/5+11.1/7+11.1/13

=3.(1/4-1/5+1/7+1/13) / 11.(1/4-1/5+1/7+1/13)     +   3.(1/4-1/5+1/7+1/13)   /  11.(1/4-1/5+1/7+1/13)

=3/11+3/11

=6/11

\(P=\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)

\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)

\(P=\frac{3\cdot\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\cdot\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}=\frac{3}{11}\)

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}+\frac{1}{3^{2016}}\)

\(\frac{1}{B}=3+3^2+3^3+...+3^{2015}+3^{2016}\)

\(\frac{3}{B}=3^2+3^3+3^4+...+3^{2016}+3^{2017}\)

\(\frac{3}{B}-\frac{1}{B}=\left(3^2+3^3+3^4+...+3^{2016}+3^{2017}\right)-\left(3+3^2+3^3+...+3^{2015}+3^{2016}\right)\)

\(\frac{2}{B}=3^{2017}-3\)

\(B=\frac{2}{3^{2017}-3}\)

3 tháng 8 2015

P=\(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)

P=\(\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{3}+\frac{11}{7}+\frac{11}{3}}\)

P=\(\frac{\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{\frac{1}{11}.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)

P=\(\frac{\frac{1}{3}}{\frac{1}{11}}=\frac{1}{3}:\frac{1}{11}=\frac{11}{3}\)

B=\(\frac{1}{3}+\frac{1}{^{3^2}}+\frac{1}{3^3}+................+\frac{1}{3^{2015}}+\frac{1}{3^{2016}}\)

B=\(\left(\frac{1}{3}\right)^1+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{2015}+\left(\frac{1}{3}\right)^{2016}\)

2B=\(\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+\left(\frac{1}{3}\right)^4+...+\left(\frac{1}{3}\right)^{2016}+\left(\frac{1}{3}\right)^{2017}\)

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B=\(\left(\frac{1}{3}\right)^1+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{2015}+\left(\frac{1}{3}\right)^{2016}\)


B=\(\left(\frac{1}{3}\right)^1-\left(\frac{1}{3}\right)^{2017}\)

2 tháng 11 2016

toán bồi à mình làm 1 câu như vậy rùi

2 tháng 11 2016

giup minh vs ban