giup minh voi

\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)

\(=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)

\(=\frac{2}{7}:\frac{1}{\frac{7}{2}}=\frac{2}{7}:\frac{2}{7}=1\)

\(A=\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)

\(A=\frac{-4}{5}+\frac{4}{3}+\frac{-5}{4}+\frac{14}{5}-\frac{7}{3}\)

\(A=\left(\frac{-4}{5}+\frac{14}{5}\right)+\left(\frac{4}{3}-\frac{7}{3}\right)+\frac{-5}{4}\)

\(A=2+\left(-1\right)+\frac{-5}{4}\)

\(A=\frac{-1}{4}\)

A=\(\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)\(\frac{7}{3}\)

  =\(\frac{-4}{5}+\frac{4}{3}+\frac{-5}{4}+\frac{14}{5}+\frac{-7}{3}\)=\(\left(\frac{-4}{5}+\frac{14}{5}\right)+\left(\frac{4}{3}+\frac{-7}{3}\right)+\frac{-5}{4}\)

  =\(\frac{10}{5}+\frac{-3}{3}+\frac{-5}{4}\)=\(2-1+\frac{-5}{4}\)=\(1+\frac{-5}{4}\)=\(\frac{4}{4}+\frac{-5}{4}\)=\(\frac{4-5}{4}\)=\(\frac{-1}{4}\)

\(A=\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)

\(=\frac{-4}{5}+\frac{4}{3}+\frac{-5}{4}+\frac{14}{5}-\frac{7}{3}\)

\(=\frac{-4}{5}+\frac{14}{5}+\frac{4}{3}-\frac{7}{3}+\frac{14}{5}\)

\(=-2-1+\frac{14}{5}\)

\(=\frac{14}{5}-\frac{15}{5}\)

\(=\frac{-1}{5}\)

\(a)\frac{-3}{4}+\frac{3}{7}+-\frac{1}{4}+\frac{4}{9}+\frac{4}{7}\)

\(=\left(\frac{-3}{4}+-\frac{1}{4}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{9}\)

\(=-1+1+\frac{4}{9}\)

\(=\frac{4}{9}\)

\(\frac{-3}{4}+\frac{3}{7}+\frac{-1}{4}+\frac{4}{9}+\frac{4}{7}\)

\(=\left(\frac{-3}{4}+\frac{-1}{4}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{9}\)

\(=\left(-1\right)+1+\frac{4}{9}\)

\(=0+\frac{4}{9}\)

\(=\frac{4}{9}\)

GIÚP MIK IK XIN CÁC BN ĐÓ

dài thế ai mà tính đc

\(\frac{1}{3}\)

\(A=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(A=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4.(\frac{1}{9}-\frac{1}{7}-\frac{1}{11})}+\frac{3.(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625})}{4.(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625})}\)

\(A=\frac{1}{4}+\frac{3}{4}\)(Vì\(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\ne0\)và\(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\ne0\))

\(A=1\)

Vậy A = 1

\(B=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-...-\frac{1}{6}-\frac{1}{2}\)

\(B=\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(-B=-\frac{1}{10.9}+\frac{1}{9.8}+\frac{1}{8.7}+...+\frac{1}{2.1}\)

\(-B=-\frac{1}{9}-\frac{1}{10}+\frac{1}{8}-\frac{1}{9}+\frac{1}{7}-\frac{1}{8}+...+1-\frac{1}{2}\)

\(-B=-\frac{1}{9}.2-\frac{1}{10}+1\)

\(-B=-\frac{2}{9}-\frac{1}{10}+1\)

\(-B=\frac{-20}{90}-\frac{9}{90}+\frac{90}{90}\)

\(-B=\frac{61}{90}\)

\(B=\frac{-61}{90}\)

Vậy\(B=\frac{-61}{90}\)

Linz

c)\(=\frac{-3.5}{7.3^2}+\frac{2^2\cdot-3}{3^2\cdot7}+\frac{17}{7}=\frac{-5}{21}-\frac{4}{21}+2\frac{3}{7}=\frac{-9}{21}+2\frac{3}{7}=\frac{-3}{7}+2\frac{3}{7}=2\)

d)\(=8\frac{2}{7}-3\frac{4}{9}-4\frac{2}{7}=4-3\frac{4}{9}=\frac{36-31}{9}=\frac{5}{9}\)

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

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

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

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

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

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

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

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

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

giúp em với : Akai Haruma

a) \(-\frac{1}{4}.13\frac{9}{11}-0,25.6\frac{2}{11}\)

\(=-\frac{1}{4}.13\frac{9}{11}-\frac{1}{4}.6\frac{2}{11}\)

\(=-\frac{1}{4}\left(13\frac{9}{11}+6\frac{2}{11}\right)\)

\(=-\frac{1}{4}.20\)

\(=-5\)

b) \(B=\frac{-5}{6}.\frac{4}{19}+\frac{-7}{12}.\frac{4}{19}-\frac{40}{57}\)

\(=\frac{4}{19}\left(\frac{-5}{6}+\frac{-7}{12}\right)-\frac{40}{57}\)

\(=\frac{4}{19}.\frac{-17}{12}-\frac{40}{57}\)

\(=\frac{-17}{57}-\frac{40}{57}\)

\(=-1\)

c)  \(\frac{3}{7}.\frac{9}{26}-\frac{1}{14}.\frac{1}{13}-\frac{1}{7}\)

\(=\frac{3}{7}.\frac{9}{26}-\frac{1}{2}.\frac{1}{7}.\frac{1}{13}-\frac{1}{7}\)

\(=\frac{1}{7}\left(3.\frac{9}{26}-\frac{1}{2}.\frac{1}{13}-1\right)\)

\(=\frac{1}{7}.0\)

\(=0\)

d) \(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)

\(=\left(\frac{4}{9}+6\frac{5}{9}\right):\left(-\frac{1}{7}\right)\)

\(=7:\left(-\frac{1}{7}\right)\)

\(=-49\)

\(A=-\frac{1}{4}x3\frac{9}{11}-0,25x6\frac{2}{11}\)

\(A=-0,25x3\frac{9}{11}-0,25x6\frac{2}{11}\)

\(A=-0,25x\left(\frac{9}{11}+\frac{2}{11}\right)\)

\(A=-0,25x1\)

\(A=-0,25\)

\(B=-\frac{5}{6}x\frac{4}{19}+-\frac{7}{12}x\frac{4}{19}-\frac{40}{57}\)

\(B=-\frac{5}{6}x\frac{4}{19}+-\frac{7}{12}x\frac{4}{19}-\frac{4}{19}x190\)

\(B=\frac{4}{19}x\left(-\frac{5}{6}-\frac{7}{12}-190\right)\)

\(B=\frac{4}{19}x\left(-\frac{2297}{12}\right)\)

\(B=\frac{-2297}{57}\)

\(C=\frac{3}{7}x\frac{9}{26}-\frac{1}{14}x\frac{1}{13}-\frac{1}{7}\)

\(C=\frac{1}{7}x3x\frac{9}{26}-\frac{1}{7}x\frac{1}{2}x\frac{1}{13}-\frac{1}{7}\)

\(C=\frac{1}{7}x\left(3x\frac{9}{26}-\frac{1}{2}x\frac{1}{13}-1\right)\)

\(C=\frac{1}{7}x0\)

\(C=\frac{1}{7}\)

\(D=\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)

\(D=\left(\frac{4}{9}+6\frac{5}{9}\right):\left(-\frac{1}{7}\right)\)

\(D=7:\left(-\frac{1}{7}\right)\)

\(D=-49\)