Đáp số là 1 vì cả hai phân số đều bằng 2/7 và 2/7:2/7=14/14=1.Đúng 100000000000000000000% luôn bạn ơi!

đưa tử về 1 thử coi bn 

trước đây cô giáo cũng bày mk thế 

Ta có A=2/7.7.1/2=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=\(\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)\(\frac{7}{3}\)

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

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

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

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

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

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

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

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

\(=-\frac{5}{7}\cdot\left(\frac{2}{11}+\frac{9}{14}\right)+\frac{12}{7}\)

\(=-\frac{5}{7}\cdot\frac{127}{154}+\frac{12}{7}\)

\(=-\frac{635}{1078}+\frac{12}{7}=\frac{1213}{1078}\)

kt lại đề bài xem hình như sai ở đâu mà số to thế

=-5/7.(2/11+9/14)+1 5/7

=-5/7.127/154+12/7

=635/1078+12/7

=2483/1078

MK ko chắc nữa nha

Sai đề bạn ơi , phải là :

\(C=-\frac{5}{7}.\frac{2}{7}+-\frac{5}{7}.\frac{9}{11}+1\frac{5}{7}\)

\(C=-\frac{5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)

\(C=-\frac{5}{7}.1+\frac{12}{7}\)

\(C=-\frac{5}{7}+\frac{12}{7}\)

\(\Rightarrow C=1\)

Vậy C=1

Hok_Tốt

#Thiên_Hy

dài quá! mình bó tay luôn!

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

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

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

\(=-49\)

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

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

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

\(A=-0,25x1\)

\(A=-0,25\)

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

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

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

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

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

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

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

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

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

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

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

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

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

\(D=-49\)

\(a,\frac{7}{12}\cdot\frac{6}{11}+\frac{7}{12}\cdot\frac{5}{11}+2\frac{7}{12}\)

\(=\frac{7}{12}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+2\frac{7}{12}\)

\(=\frac{7}{12}+\frac{31}{12}\)

\(=\frac{38}{12}=\frac{19}{6}\)

\(b,\frac{-5}{9}\cdot\frac{-6}{13}+\frac{5}{-9}\cdot\frac{-5}{13}-\frac{5}{9}\)

\(=\frac{-5}{9}\cdot\frac{-6}{13}+\frac{-5}{9}\cdot\frac{-5}{13}+\frac{-5}{9}\cdot1\)

\(=\frac{-5}{9}\cdot\left(\frac{-6}{13}+\frac{-5}{13}+1\right)\)

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

\(=\frac{-5}{9}\cdot\frac{2}{13}\)

\(=\frac{-10}{117}\)

\(c,\)\(0,8\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-1\frac{2}{5}\)

\(=\frac{4}{5}\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-\frac{7}{5}\)

\(=\frac{4}{5}\cdot\left(\frac{-15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)

\(=\frac{4}{5}\cdot\left(-2\right)-\frac{7}{5}\)

\(=\frac{-8}{5}-\frac{7}{5}\)

\(=-3\)

\(d,\)\(75\%\cdot\frac{6}{7}+5\%\cdot\frac{6}{7}+\frac{7}{10}\cdot1\frac{1}{7}\)

\(=\frac{3}{4}\cdot\frac{6}{7}+\frac{1}{20}\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)

\(=\left(\frac{3}{4}+\frac{1}{20}\right)\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)

\(=\frac{4}{5}\cdot\frac{6}{7}+\frac{4}{5}\cdot1\)

\(=\frac{4}{5}\cdot\left(\frac{6}{7}+1\right)\)

\(=\frac{4}{5}\cdot\frac{13}{7}\)

\(=\frac{52}{35}\)

a)7/12.6/11+7/12.5/11-2.7/12

=7/12(6/11+5/11-2)

=7/12(1-2)

=7/12.(-1)

=-7/12

     \(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)

\(=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{2}{6}-\frac{2}{8}+\frac{2}{10}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)

\(=\frac{2.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)

\(=\frac{2}{7}-\frac{2}{7}\)

\(=0\)

\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)

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

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

\(=\frac{2}{7}-\frac{2}{7}=0\)

Study well ! >_<

O

a) \(\frac{{ - 3}}{7}.\frac{2}{5} + \frac{2}{5}.\left( { - \frac{5}{{14}}} \right) - \frac{{18}}{{35}}\)

\(\begin{array}{l} = \frac{2}{5}.\left( {\frac{{ - 3}}{7} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\left( {\frac{{ - 6}}{{14}} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\frac{{ - 11}}{{14}} - \frac{{18}}{{35}} = \frac{{ - 11}}{{35}} - \frac{{18}}{{35}} =  \frac{{ -29}}{{35}}\end{array}\)

b) \(\left( {\frac{2}{3} - \frac{5}{{11}} + \frac{1}{4}} \right):\left( {1 + \frac{5}{{12}} - \frac{7}{{11}}} \right)\)

\(\begin{array}{l} = \left( {\frac{{2.11.4}}{{3.11.4}} - \frac{{5.3.4}}{{11.3.4}} + \frac{{1.3.11}}{{4.3.11}}} \right):\left( {\frac{11.12}{11.12} + \frac{{5.11}}{{12.11}} - \frac{{7.12}}{{11.12}}} \right)\\ = \left( {\frac{{88 - 60 + 33}}{{121}}} \right):\left( { \frac{{121+55 - 84}}{{121}}} \right)\\ = \frac{{61}}{{121}}:\frac{{92}}{{121}} = \frac{{61}}{{121}}.\frac{{121}}{{92}}= \frac{{61}}{{92}}\end{array}\)

c) \(\left( {13,6 - 37,8} \right).\left( { - 3,2} \right)\)

\( = \left( { - 24,2} \right).\left( { - 3,2} \right) = 77,44\)

d) \(\left( { - 25,4} \right).\left( {18,5 + 43,6 - 16,8} \right):12,7\)

\(\begin{array}{l} = \left( { - 25,4} \right).\left( {62,1 - 16,8} \right):12,7\\ = \left( { - 25,4} \right).45,3:12,7\\ = \left( { - 25,4} \right):12,7.45,3\\ =  (- 2).45,3 =  - 90,6\end{array}\)

a: \(=\dfrac{2}{5}\cdot\left(-\dfrac{3}{7}-\dfrac{5}{14}\right)-\dfrac{18}{35}\)

\(=\dfrac{2}{5}\cdot\dfrac{-6-5}{14}-\dfrac{18}{35}\)

\(=\dfrac{2}{5}\cdot\dfrac{-11}{14}-\dfrac{18}{35}=-\dfrac{22}{70}-\dfrac{18}{35}=\dfrac{-58}{70}=-\dfrac{29}{35}\)

b: \(=\dfrac{88-60+33}{132}:\dfrac{132+55-84}{132}\)

\(=\dfrac{61}{132}\cdot\dfrac{132}{103}=\dfrac{61}{103}\)

c: \(=-24.2\cdot\left(-3.2\right)=24.2\cdot3.2=77.44\)

d: \(=\dfrac{-25.4}{12.7}\cdot45.3=-2\cdot45.3=-90.6\)