Tính hợp lý
a) 24.67 + 67.76
b) 2017.99 + 2017
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a: \(=100\cdot104=10400\)
d: \(=2\cdot9+4\cdot27=18+108=126\)
c: \(=520:\left\{\left[\left(80+20\right):2-2\right]+82\right\}+1\)
\(=520:\left\{48+82\right\}+1\)
=4+1
=5
a) 10-11+12-13+14-15+...-2017+2018
=(10-11)+(12-13)+(14-15)+...+(2016-2017)+2018
=(-1)+(-1)+(-1)+...+(-1)+2018
=(-1).1004+2018
=-1004+2018
=1004
b) -5+6-7+8-9+10-...-2017+2018
=-5+(6-7)+(8-9)+...+(2016-2017)+2018
=(-5)+(-1)+(-1)+...+(-1)+2018
=(-5)+(-1).1006+2018
=(-5)+(-1006)+2018
=1007
c) 3-4+5-6+7-8+...+101-102+103
= (3-4)+(5-6)+(7-8)+...+(101-102)+103
= (-1)+(-1)+(-1)+...+(-1)+103
= (-1).50+103 = (-50)+103
= 53
\(\left(-1-\frac{1}{12}\right).\left(-1-\frac{1}{13}\right).\left(-1-\frac{1}{14}\right)...\left(-1-\frac{1}{2017}\right)\)
\(=\frac{-13}{12}.\frac{-14}{13}.\frac{-15}{14}...\frac{-2018}{2017}\)
\(=\frac{-13}{12}.\frac{14}{-13}.\frac{-15}{14}...\frac{2018}{-2017}\)
\(=\frac{\left(-13\right).14.\left(-15\right)...2018}{12.\left(-13\right).14...2017}=\frac{2018}{12}=\frac{1009}{6}\)
a) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+....+\frac{1}{32}-\frac{1}{64}=\frac{1}{1}-\frac{1}{64}=\frac{63}{64}\)
b) \(\frac{63}{64}=0,98...;\frac{2016}{2017}=0,99...\)
\(\Rightarrow\frac{63}{64}< \frac{2016}{2017}\)
\(a)\)\(24.67+67.76\)
\(=\)\(67.\left(24+76\right)\)
\(=\)\(67.100\)
\(=\)\(6700\)
\(b)\)\(2017.99+2017\)
\(=\)\(2017.\left(99+1\right)\)
\(=\)\(2017.100\)
\(=\)\(201700\)
a) 24.67 + 67.76
=67(24+76)
=67.100
=6700
b) 2017.99 + 2017
=2017(99+1)
=2017.100
=201700