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a) Rút gọn:
\(\dfrac{-1}{6};\dfrac{1}{5};\dfrac{-1}{2}\)
Quy đồng:
\(\dfrac{-5}{30};\dfrac{6}{30};\dfrac{-15}{30}\)
b) Rút gọn:
\(\dfrac{-3}{5};\dfrac{-5}{8};\dfrac{-4}{9}\)
Quy đồng:
\(\dfrac{-216}{360};\dfrac{-225}{360};\dfrac{-160}{360}\)
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) \(\dfrac{-18}{24}\)+\(\dfrac{15}{-21}\)=\(\dfrac{-18}{24}\)+\(\dfrac{-15}{21}\)=\(\dfrac{-3}{4}\)+\(\dfrac{-5}{7}\)=\(\dfrac{-21}{28}\)+\(\dfrac{-20}{28}\)=\(\dfrac{-41}{28}\)
Vì cộng cả tử và mẫu của phân số \(\dfrac{23}{40}\) với cùng một số tự nhiên n rồi rút gọn ta được \(\dfrac{3}{4}\) nên ta được:
\(\dfrac{23+n}{40+n}=\dfrac{3}{4}\)
\(\Rightarrow\)\(4.\left(23+n\right)=3.\left(40+n\right)\)
hay 92 + 4.n = 120 + 3.n
4.n - 3.n = 120 - 92
\(\Rightarrow\) n = 28
Vậy số tự nhiên n cần tìm là 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
a) \(\dfrac{4.7}{9.32}=\dfrac{4.7}{9.4.8}=\dfrac{7}{72}\) b) \(\dfrac{3.21}{14.15}=\dfrac{3.7.3}{7.2.3.5}=\dfrac{3}{10}\)
c) \(\dfrac{2.5.13}{26.35}=\dfrac{2.5.13}{2.13.5.7}=\dfrac{1}{7}\) d)\(\dfrac{9.6-9.3}{18}\) \(=\dfrac{9.\left(6-3\right)}{9.2}=\dfrac{9.3}{9.2}=\dfrac{3}{2}\)
e) \(\dfrac{17.5-17}{3-20}=\dfrac{17.\left(5-1\right)}{-17}=-\dfrac{17.4}{17}=-4\) f) \(\dfrac{49+7.49}{49}=\dfrac{49.\left(1+7\right)}{49}=8\)
\(=-\dfrac{1}{14}-\dfrac{5}{14}=\dfrac{-1-5}{14}=-\dfrac{6}{14}\)
\(=\dfrac{-1}{14}-\dfrac{5}{14}=\dfrac{-6}{14}=\dfrac{-3}{7}\)