giúp em với : Akai Haruma

\(a,15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\\ =15\frac{3}{13}-3\frac{4}{7}-8\frac{3}{13}\\ =7\frac{3}{13}-3\frac{4}{7}\\ =\frac{94}{13}-\frac{25}{7}\\ =\frac{94\cdot7-25\cdot13}{13\cdot7}\\ =\frac{333}{91}\)

\(b,\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\\ =7\frac{4}{9}-3\frac{4}{9}+4\frac{7}{11}\\ =4\frac{4}{9}+4\frac{7}{11}\\ =\frac{40}{9}+\frac{51}{11}\\ =\frac{40\cdot11+51\cdot9}{11\cdot9}\\ =\frac{899}{99}\)

\(c,=-\frac{7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}\right)+\frac{52}{9}\\ =-\frac{7}{9}\cdot1+\frac{52}{9}\\ =\frac{-7+52}{9}\\ =\frac{45}{9}=5\)

\(d,=\frac{50}{100}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{75}{100}\\ =\frac{50\cdot2\cdot2\cdot2\cdot5\cdot7\cdot25\cdot3}{50\cdot2\cdot3\cdot5\cdot7\cdot25\cdot2\cdot2}=1\)

\(e,=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\\ =1-\frac{1}{43}\\ =\frac{42}{43}\)

a) Ta có: \(15\frac{3}{13}-\left(3\frac{4}{7}+8\frac{3}{13}\right)\)

\(=15+\frac{3}{13}-3-\frac{4}{7}-8-\frac{3}{13}\)

\(=4-\frac{4}{7}=\frac{24}{7}\)

b) Ta có: \(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)

\(=7+\frac{4}{9}+4+\frac{7}{11}-3-\frac{4}{9}\)

\(=8+\frac{7}{11}=\frac{95}{11}\)

c) Ta có: \(\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+5\frac{7}{9}\)

\(=\frac{-7}{9}\cdot\frac{4}{11}+\frac{-7}{9}\cdot\frac{7}{11}+\frac{-7}{9}\cdot\frac{-52}{7}\)

\(=\frac{-7}{9}\cdot\left(\frac{4}{11}+\frac{7}{11}-\frac{52}{7}\right)\)

\(=\frac{-7}{9}\cdot\frac{45}{-7}=5\)

d) Ta có: \(50\%\cdot1\frac{1}{3}\cdot10\cdot\frac{7}{35}\cdot0.75\)

\(=\frac{1}{2}\cdot\frac{4}{3}\cdot10\cdot\frac{7}{35}\cdot\frac{3}{4}\)

\(=5\cdot\frac{7}{35}=1\)

e) Ta có: \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=1-\frac{1}{43}=\frac{43}{43}-\frac{1}{43}\)

\(=\frac{42}{43}\)

a)

\(\begin{array}{l}\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}}\\ = \frac{7}{{11}}.\left( {\frac{{13}}{{23}} + \frac{{10}}{{23}}} \right)\\ = \frac{7}{{11}}.\frac{23}{23}\\ = \frac{7}{{11}}.1\\ = \frac{7}{{11}}\end{array}\)                                   

b)

\(\begin{array}{l}\frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\\ = \frac{5}{9}.\left( {\frac{{23}}{{11}} - \frac{1}{{11}} + 1} \right)\\ = \frac{5}{9}.\left( {2 + 1} \right)\\ = \frac{5}{9}.3 = \frac{5}{3}\end{array}\)

c)

\(\begin{array}{l}\left[ {\left( { - \frac{4}{9} + \frac{3}{5}} \right):\frac{{13}}{{17}}} \right] + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}}\\ = \left( { - \frac{4}{9} + \frac{3}{5}} \right).\frac{{17}}{{13}} + \left( {\frac{2}{5} - \frac{5}{9}} \right).\frac{{17}}{{13}}\\ = \frac{{17}}{{13}}.\left( { - \frac{4}{9} + \frac{3}{5} + \frac{2}{5} - \frac{5}{9}} \right)\\ = \frac{{17}}{{13}}.\left[ {\left( { - \frac{4}{9} - \frac{5}{9}} \right) + \left( {\frac{3}{5} + \frac{2}{5}} \right)} \right]\\ =\frac{{17}}{{13}}. (\frac{-9}{9}+\frac{5}{5})\\= \frac{{17}}{{13}}.\left( { - 1 + 1} \right)\\ = \frac{{17}}{{13}}.0 = 0\end{array}\)          

d)

\(\begin{array}{l}\frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)\\ = \frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{6}{{22}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{4}{{10}}} \right)\\ = \frac{3}{{16}}:\frac{{ - 3}}{{22}} + \frac{3}{{16}}:\frac{{ - 3}}{{10}}\\ = \frac{3}{{16}}.\frac{{ - 22}}{3} + \frac{3}{{16}}.\frac{{ - 10}}{3}\\ = \frac{3}{{16}}.\left( {\frac{{ - 22}}{3} + \frac{{ - 10}}{3}} \right)\\ = \frac{3}{{16}}.\frac{{ - 32}}{3}\\ =  - 2\end{array}\)

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

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

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

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

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

\(C=0\)

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

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

\(\Rightarrow D=1\)

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

\(A=21\frac{4}{11}-\left(1\frac{3}{5}+7\frac{4}{11}\right)\)

\(A=\frac{235}{11}-\left(\frac{8}{5}+\frac{81}{11}\right)\)

\(A=\left(\frac{235}{11}-\frac{81}{11}\right)+\frac{8}{5}\)

\(A=\frac{154}{11}+\frac{8}{5}\)

\(\Rightarrow A=\frac{78}{5}\)

\(B=\left(7\frac{8}{9}+2\frac{3}{13}\right)-\left(4\frac{8}{9}-7\frac{10}{13}\right)\)

\(B=\left(\frac{71}{9}+\frac{29}{13}\right)-\left(\frac{44}{9}-\frac{101}{13}\right)\)

\(B=\left(\frac{71}{9}-\frac{44}{9}\right)+\left(\frac{29}{13}-\frac{101}{13}\right)\)

\(B=\frac{27}{9}+\frac{-72}{13}\)

\(B=3+\frac{-72}{13}\)

\(\Rightarrow B=\frac{-33}{13}\)

P/s: Hoq chắc :v

\(C=\frac{-3}{5}+\frac{2}{7}+\frac{-3}{7}.\frac{3}{5}+\frac{-3}{7}\)

\(C=\left(\frac{-3}{5}+\frac{3}{5}\right).\left(\frac{2}{7}+\frac{-3}{7}+\frac{-3}{7}\right)\)

\(C=0.\frac{-4}{7}\)

\(C=0\)

= 1/3 - 1/3 + 5/7 - 5/7 - 7/9 + 7/9 +9/11 - 9/11 -11/13 + 11/13 +13/15

= 0 + 0 - 0 + 0 -0 + 13/15

= 0 + 13/15

= 13/15

\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{13}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)

\(=\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{3}{5}-\frac{3}{5}\right)+\left(\frac{5}{7}-\frac{5}{7}\right)+\left(\frac{7}{9}-\frac{7}{9}\right)+\left(\frac{9}{11}-\frac{9}{11}\right)+\left(\frac{11}{13}-\frac{11}{13}\right)+\frac{13}{15}\)

\(=0+0+0+0+0+0+\frac{13}{15}\)

\(=\frac{13}{15}\)