Tính hợp lí:
7) 6 - ( 5 - 5847 ) + 5 - 5847
8) 35 - 17 + 1993 - 35 - 1993
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\(a.35-17+1993-35-1993\\ =-17+\left(35-35\right)+\left(1993-1993\right)\\ =-17+0+0\\ =-17\\ b.-48725-\left(357-48725\right)+300\\ =-48725-357+48725+300\\ =\left(300-357\right)+\left(48725-48725\right)\\ =-57+0\\ =-57\\ c.3579-\left(49-5+3579\right)+49\\ =3579-49+5-3579+49\\ =5+\left(3579-3579\right)+\left(49-49\right)\\ =5+0+0\\ =5\)
a) \(35-17+1993-35-1993\)
\(=\left(35-35\right)+\left(1993-1993\right)-17\)
\(=-17\)
b) \(-48725-\left(357-48725\right)+300\)
\(=\left(-48725+48725\right)+\left(300-357\right)\)
\(=-57\)
c) \(3579-\left(49-5+3579\right)+49\)
\(=\left(3579-3579\right)+\left(49-49\right)+5\)
\(=5\)
a) \(\frac{3}{5}+\frac{6}{11}+\frac{7}{13}+\frac{2}{5}+\frac{16}{11}+\frac{19}{13}\)
\(=\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{6}{11}+\frac{16}{11}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)\)
\(=1+2+2=5\)
b) \(\frac{1995}{1997}x\frac{1990}{1993}x\frac{1997}{1994}x\frac{1993}{1995}x\frac{997}{995}=\frac{1995x1990x1997x1993x997}{1997x1993x1994x1995x995}=\frac{1990x997}{1994x995}=\frac{995x2x997}{997x2x995}=1\)
a) =(3/5+2/5)+(6/11+16/11)+(19/13+7/13)
=5/5+22/11+26/13
=1+2+3
=6
\(\frac{15}{25}=\frac{3\cdot5}{5\cdot5}=\frac{3}{5}=\frac{21}{35}< \frac{25}{35}=\frac{5\cdot5}{5\cdot7}=\frac{5}{7}\)
\(\frac{17}{49}< \frac{17}{48}< \frac{18}{48}=\frac{3\cdot6}{6\cdot8}=\frac{3}{8}\)
\(\frac{13}{60}< \frac{13}{50}=\frac{26}{100}< \frac{27}{100}\)
\(\frac{43}{47}>\frac{42}{47}>\frac{42}{49}=\frac{6\cdot7}{7\cdot7}=\frac{6}{7}=\frac{30}{35}>\frac{29}{35}\)
\(\frac{1993}{1995}< \frac{1994}{1996}=\frac{997}{998}\)
35.13+35.17+65.75-65.45
=35.(13+17)+65.(75-45)
=35.30+65.30
=30.(35+65)
=30.100
=3000
\(=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}+\dfrac{-4}{35}+\dfrac{5}{7}\right)+\dfrac{1}{41}=1+1+\dfrac{1}{41}=\dfrac{83}{41}\)
\(6-\left(5-5847\right)+5-5847\)
= \(6-\left[\left(5-5\right)+\left(5847-5847\right)\right]\)
= \(6-\left[0+0\right]\)
= \(6-0\)
= \(6\)
\(35-17+1993-35-1993\)
= \(\left[\left(35-35\right)-\left(1993-1993\right)\right]-17\)
= \(0-0-17\)
= \(0-17\)
= \(\left(-17\right)\)