101 + (–60) + 29 + (–40)
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a) 37.24+37.76+63.79+21.63
= 37(24+76)+63(79+21)
= 37.100+63.100
= (37+63)100
= 100.100
= 10000
b) 101+(-60)+29+(-40)
= (101+29)+(-60-40)
= 130-100
= 30
c) 40+(139-172+99)-(139+199-127)
= 40+139-172+99-139-199+127
= (139-139)+(-127+127)+(99-199)+40
= 0+0-100+40
= -60
d) (158.129-158.39):180
= [158(129-39)]:180
= (150.90):180
= 75
=
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\(g,101+\left(-60\right)+29+\left(-40\right)=41+29+\left(-40\right)=70+\left(-40\right)=30\)
\(h,167+\left(-252\right)+52+\left(-67\right)=\left(-85\right)+52+\left(-67\right)=\left(-33\right)+\left(-67\right)=-100\)
\(i,38-138+250-350=\left(-100\right)+250-350=150-350=-200\)
\(k,118+107-\left(118-93\right)-50=225-25-50=100-50=50\)
\(m,40+\left(139-172+99\right)-\left(139+199-172\right)=40+\left[\left(-33\right)+99\right]-\left(338-172\right)\)
\(=40+66-166=106-166=-60\)
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\(\frac{29}{60}< \frac{4}{9}\)
\(\frac{6}{28}< \frac{101}{400}\)
\(\frac{131313}{393939}=\frac{1}{3}\)
a) ta có: \(\frac{29}{60}=0,483;\frac{4}{9}=0,44\)
\(\Rightarrow0,483>0,44\Rightarrow\frac{29}{60}>\frac{4}{9}\)
b) ta có: \(\frac{6}{28}=\frac{3}{14}=0,21;\frac{101}{400}=0,25\)
\(\Rightarrow0,21< 0,25\Rightarrow\frac{6}{28}< \frac{101}{400}\)
c) ta có: \(\frac{131313}{393939}=\frac{13x10101}{39x10101}=\frac{13}{39}=\frac{1}{3}\)
\(\Rightarrow\frac{131313}{393939}=\frac{1}{3}\)
= ( 101 + 29 ) + ((-60) + (-40))
= 130 + (-100)
= 30
HT