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a: =35/17-18/17-9/5+4/5
=1-1=0
b: =-7/19(3/17+8/11-1)
=7/19*18/187=126/3553
c: =26/15-11/15-17/3-6/13
=1-6/13-17/3
=7/13-17/3=-200/39
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a: =-5/9-4/9+8/15+7/15-2/11=-2/11
b: =10/17+7/17-5/13-8/13+11/25
=11/25
c: =(9/12-2/12)*3/2=7/12*3/2=21/24=7/8
d: =(31/10-25/10)*3-2
=3/5*3-2
=9/5-2
=-1/5
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a: =-3/24-40/24
=-43/24
b: \(=\dfrac{6}{54}\cdot\dfrac{49}{35}=\dfrac{1}{9}\cdot\dfrac{7}{5}=\dfrac{7}{45}\)
c: \(=\dfrac{6}{5}+\dfrac{4}{3}=\dfrac{18+20}{15}=\dfrac{38}{15}\)
d: \(=\dfrac{31}{17}-\dfrac{14}{17}-\dfrac{5}{13}-\dfrac{8}{13}=1-1=0\)
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Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
\(\dfrac{2}{3}+\dfrac{5}{7}+\left(\dfrac{-2}{3}\right)\)
=\(\left[\dfrac{2}{3}+\left(\dfrac{-2}{3}\right)\right]+\dfrac{5}{7}\)
=\(0+\dfrac{5}{7}=\dfrac{5}{7}\)
\(\dfrac{31}{17}+\left(\dfrac{-5}{13}\right)+\left(\dfrac{-8}{13}\right)-\dfrac{14}{17}\)
=\(\left(\dfrac{31}{17}-\dfrac{14}{17}\right)+\left[\left(\dfrac{-5}{13}\right)+\left(\dfrac{-8}{13}\right)\right]\)
=\(1+\left(-1\right)\)
=\(0\)
\(\dfrac{2}{3}+\dfrac{5}{7}+\left(-\dfrac{2}{3}\right)=\dfrac{5}{7}\)
\(\dfrac{31}{17}+\left(-\dfrac{5}{13}\right)+\left(-\dfrac{8}{13}\right)-\dfrac{14}{17}=1-1=0\)