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Giải:
Ta có:
\(\dfrac{a}{b}=\dfrac{-8}{11}\left(1\right)\Leftrightarrow1-\dfrac{a}{b}=1-\dfrac{-8}{11}\)
Hay \(\dfrac{b-a}{b}=\dfrac{11+8}{11}=\dfrac{19}{11}\left(2\right)\)
Thay \(b-a=190\) vào \(\left(2\right)\) ta được:
\(\dfrac{190}{b}=\dfrac{19}{11}\Leftrightarrow190.11=19b\Leftrightarrow b=110\)
Thay \(b=110\) vào \(\left(1\right)\) ta được:
\(\dfrac{a}{110}=\dfrac{-8}{11}\Leftrightarrow11a=-8.110\Leftrightarrow a=-80\)
Vậy phân số \(\dfrac{a}{b}\) cần tìm là \(\dfrac{-80}{110}\)
Thay b - a = 190 vào (1) ta được: 
Phân số a/b phải tìm là -80/110
a) \(\dfrac{11\cdot8-11\cdot3}{17-6}\)
\(=\dfrac{11\cdot\left(8-3\right)}{11}=5\)
b) \(\dfrac{24-12\cdot13}{12+4\cdot9}\)
\(=\dfrac{12\cdot\left(2-13\right)}{12\left(1+3\right)}=\dfrac{-11}{4}\)
a)\(\dfrac{-3}{29}+\dfrac{16}{58}\)\(=\dfrac{-3}{29}+\dfrac{8}{29}=\dfrac{5}{29}\)
b) \(\dfrac{8}{40}+\dfrac{-36}{45}=\dfrac{1}{5}+\dfrac{-4}{5}=\dfrac{-3}{5}\)
c) \(\dfrac{-8}{18}+\dfrac{-15}{27}=\dfrac{-4}{9}+\dfrac{-5}{9}=\dfrac{-9}{9}=-1\)
a) \(\dfrac{-3}{29}+\dfrac{16}{58}=\dfrac{-3}{29}+\dfrac{8}{29}=\dfrac{-3+8}{29}=\dfrac{5}{29}\)
b) \(\dfrac{8}{40}+\dfrac{-36}{45}=\dfrac{1}{5}+\dfrac{-4}{5}=\dfrac{1+\left(-4\right)}{5}=\dfrac{-3}{5}\)
c) \(\dfrac{-8}{18}+\dfrac{-15}{27}=\dfrac{-4}{9}+\dfrac{-5}{9}=\dfrac{-4+\left(-5\right)}{9}=\dfrac{-9}{9}=-1\)
Tính các tổng dưới đây sau khi đã rút gọn phân số :
a)\(\dfrac{7}{21}\) + \(\dfrac{9}{-36}\) = \(\dfrac{7}{21}\)+\(\dfrac{-9}{36}\)=\(\dfrac{1}{3}\)+\(\dfrac{-1}{4}\)=\(\dfrac{4}{12}\)+\(\dfrac{-3}{12}\)=\(\dfrac{1}{12}\)
b) \(\dfrac{-12}{18}\)+\(\dfrac{-21}{35}\)=\(\dfrac{-2}{3}\)+\(\dfrac{-3}{5}\)=\(\dfrac{-10}{15}\)+\(\dfrac{-9}{15}\)=\(\dfrac{-19}{15}\)
c) \(\dfrac{-3}{21}\)+\(\dfrac{6}{42}\)=\(\dfrac{-1}{7}\)+\(\dfrac{1}{7}\)=0
d) \(\dfrac{-18}{24}\)+\(\dfrac{15}{-21}\)=\(\dfrac{-18}{24}\)+\(\dfrac{-15}{21}\)=\(\dfrac{-3}{4}\)+\(\dfrac{-5}{7}\)=\(\dfrac{-21}{28}\)+\(\dfrac{-20}{28}\)=\(\dfrac{-41}{28}\)
\(\dfrac{3\cdot4+3\cdot7}{6\cdot5+9}=\dfrac{3\cdot11}{30+9}=\dfrac{33}{39}=\dfrac{11}{13}=\dfrac{77}{91}\)
\(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}=\dfrac{54-34}{189-119}=\dfrac{20}{70}=\dfrac{2}{7}=\dfrac{26}{91}\)
\(C=\dfrac{1010}{7710}=\dfrac{101}{771}=\dfrac{48783}{771\cdot483}\)
\(D=\dfrac{6+2\cdot4\cdot6+3\cdot6\cdot9+125\cdot6}{6\cdot3+6\cdot24+6\cdot81+6\cdot375}\)
\(=\dfrac{1+2\cdot4+3\cdot6+125}{3+24+81+375}=\dfrac{152}{483}=\dfrac{117192}{483\cdot771}\)
mà 48783<117192
nên C<D
\(\dfrac{21\cdot9}{7\cdot6-7\cdot3}\\ =\dfrac{21\cdot9}{7\cdot\left(6-3\right)}\\ =\dfrac{21\cdot9}{7\cdot3}\\ =\dfrac{21\cdot9}{21}=9\)
cảm ơn rất nhiều !