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\(\dfrac{5}{7}x-x=1\dfrac{1}{7}\\ < =>\dfrac{5}{7}x-x=\dfrac{8}{7}\\ < =>\left(\dfrac{5}{7}-\dfrac{7}{7}\right)x=\dfrac{8}{7}\\ < =>-\dfrac{2}{7}x=\dfrac{8}{7}\\ =>x=\dfrac{\dfrac{8}{7}}{-\dfrac{2}{7}}=-4\)
A = - ( 1+2+3 +....+ 202) = - 203. 101 = -20503
B= ( 1+2-3-4) + ( 5+6-7-8) +..........+( 97+98 -99-100) + ( 101+102)
= -4 + (-4) .........+ (-4) + 203
= -4 .25 + 203 = 103
=\(\dfrac{6}{7}+\dfrac{5}{8}.\dfrac{1}{5}-\dfrac{3}{16}.\left(-4\right)\)
= \(\dfrac{6}{7}+\dfrac{5.1}{8.5}-\dfrac{3.\left(-4\right)}{16}\)= \(\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{-3}{4}\)= \(\dfrac{48}{56}+\dfrac{7}{56}-\dfrac{-42}{56}\)
= \(\dfrac{48+7-\left(-42\right)}{56}\)=\(\dfrac{97}{56}\)
\(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{3}+\dfrac{3}{15}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)
\(1/4 + 3/4 : -6/-7 = 1/4 + 3/4 x -7/-6 = 1/4 + 7/8 = 2/8 + 7/8 = 9/8\)
2,
\(M=\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\) =\(\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)
\(=\dfrac{3}{4}\)
=1/4+7/8=9/8
\(\dfrac{1}{4}+\dfrac{3}{4}:\dfrac{-6}{-7}=\dfrac{1}{4}+\dfrac{3}{4}:\dfrac{6}{7}=\dfrac{1}{4}+\dfrac{3}{4}\times\dfrac{7}{6}=\dfrac{1}{4}+\dfrac{7}{8}=\dfrac{9}{8}\)