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\(=\dfrac{3}{4}.\dfrac{-8}{5}+\dfrac{1}{4}=-\dfrac{6}{5}+\dfrac{1}{4}=-\dfrac{19}{20}\)
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\(\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right):\dfrac{2}{9}-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right):\dfrac{2}{3}=\)
\(\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right)\cdot\dfrac{3}{2}-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right)\cdot\dfrac{3}{2}=\)
\(\left[\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right)-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right)\right]\cdot\dfrac{3}{2}=\)
\(\left[\dfrac{1}{2020}-\dfrac{2018}{2019}-\dfrac{2019}{2020}-\dfrac{1}{2019}\right]\cdot\dfrac{3}{2}=\)
\(0\)
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\(\left(\dfrac{4}{13}-\dfrac{13}{5}\right):\dfrac{4}{5}+\left(\dfrac{9}{13}+\dfrac{3}{5}\right):\dfrac{4}{5}\)
\(=\left(\dfrac{4}{13}-\dfrac{13}{5}\right)\cdot\dfrac{5}{4}+\left(\dfrac{9}{13}+\dfrac{3}{5}\right)\cdot\dfrac{5}{4}\)
\(=\left[\dfrac{4}{13}-\dfrac{13}{5}+\dfrac{9}{13}+\dfrac{3}{5}\right]\cdot\dfrac{5}{4}\)
\(=\dfrac{-5}{4}\)
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Câu 6:
a) a vuông góc với IJ
b vuông góc với IJ
=> a//b
b) KLJ + IKL = 180° ( 2 góc trong cùng phía)
75° + IKL = 180°
IKL = 180° - 75°
IKL = 105°
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d)\(\left(\dfrac{1}{3}\right)^{50}.\left(-9\right)^{25}-\dfrac{2}{3}:4=\dfrac{1}{3^{50}}.\left(-3^{50}\right)-\dfrac{1}{6}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)
e) \(=\left(4.\dfrac{3}{4}-\dfrac{1}{2}\right).\dfrac{6}{5}-17=\left(3-\dfrac{1}{2}\right).\dfrac{6}{5}-17=\dfrac{5}{2}.\dfrac{6}{5}-17=3-17=-14\)
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=-4/15 - 18/19-20/19-11/15
= (-4/15-11/15) - (18/19+20/19)
= -1 - 2
= - 3
\(\left(\frac{-4}{15}-\frac{18}{19}\right)-\left(\frac{20}{19}+\frac{11}{15}\right)=\frac{-4}{15}-\frac{18}{19}-\frac{20}{19}-\frac{11}{15}\)
\(=\left(\frac{-4}{15}-\frac{11}{15}\right)-\left(\frac{18}{19}+\frac{20}{19}\right)=\frac{-15}{15}-\frac{38}{19}=-1-2=-3\)
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Đặt A=\(\frac{6}{-7}\)+ \(\frac{9}{4}\). \(\frac{2}{15}\)- \(\frac{2}{14}\)
A = \(\frac{6}{-7}\)+ \(\frac{3}{10}\)- \(\frac{2}{14}\)
A= \(\frac{-60}{70}\)+ \(\frac{21}{70}\)- \(\frac{10}{70}\)
A= \(\frac{-7}{10}\)
\(\dfrac{1}{2}-\dfrac{1}{9}:2+1=\dfrac{13}{9}\)