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\(B=\frac{1}{3}-\frac{3}{4}+0,6+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow B=\frac{3}{15}-\frac{48}{64}+\frac{9}{15}+\frac{1}{64}-\frac{8}{36}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow B=\frac{3}{15}+\frac{9}{15}+\frac{1}{15}+\left(-\frac{48}{64}+\frac{1}{64}\right)+\left(-\frac{8}{36}-\frac{1}{36}\right)\)
\(\Rightarrow B=\frac{13}{15}-\frac{47}{64}-\frac{1}{4}\)
\(\Rightarrow B=-\frac{113}{960}\)
\(C=0\)
\(D=\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow D=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+...-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)
\(\Rightarrow D=1\)
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
\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{3}{8}+\dfrac{5}{8}=1\)
=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
\(P=\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)
\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(P=\frac{3\cdot\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\cdot\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}=\frac{3}{11}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}+\frac{1}{3^{2016}}\)
\(\frac{1}{B}=3+3^2+3^3+...+3^{2015}+3^{2016}\)
\(\frac{3}{B}=3^2+3^3+3^4+...+3^{2016}+3^{2017}\)
\(\frac{3}{B}-\frac{1}{B}=\left(3^2+3^3+3^4+...+3^{2016}+3^{2017}\right)-\left(3+3^2+3^3+...+3^{2015}+3^{2016}\right)\)
\(\frac{2}{B}=3^{2017}-3\)
\(B=\frac{2}{3^{2017}-3}\)
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=\(\left(\frac{1}{3}\right)^1-\left(\frac{1}{3}\right)^{2017}\)
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:
\(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{3}}=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{3}}=\frac{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}=\frac{3}{11}\)
Câu 1;
13 −17 −113 23 −27 −213 ·34 −316 −364 −3256 1−14 −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: