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a) \(\frac{31}{23}-\left(\frac{7}{23}+\frac{8}{23}\right)\)
\(=\frac{31}{23}-\frac{15}{23}\)
\(=\frac{16}{23}\)
b) \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{-67}{67}+\frac{41}{41}\)
\(=\frac{1}{3}-1+1\)
\(=\frac{1}{3}\)
c) \(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{52}-\frac{3}{11}\right)\)
\(=\frac{38}{45}-\frac{8}{45}+\frac{17}{52}+\frac{3}{11}\)
\(=\frac{30}{45}+\frac{17}{52}+\frac{3}{11}\)
\(=\frac{2}{3}+\frac{17}{52}+\frac{3}{11}\)
\(=\frac{104+51}{156}+\frac{3}{11}\)
\(=\frac{155}{156}+\frac{3}{11}\)
\(=\frac{156}{156}-\frac{1}{156}+\frac{3}{11}\)
\(=1-\frac{1}{156}+\frac{3}{11}\)
\(=1-\left(\frac{11-468}{1716}\right)\)
\(=1-\frac{-457}{1716}\)
\(=1+\frac{457}{1716}\)
\(=\frac{2173}{1716}\)
a)31/23-(7/32+8/23)=31/23-7/32-8/23=(31/23-8/23)-7/32=1-7/32=25/32
a)Ta có:
\(\frac{419}{-723}< 0< \frac{-697}{-313}\)
\(\Rightarrow\frac{419}{-723}< \frac{-697}{-313}\)
c)\(\frac{17}{215}>\frac{17}{314}\)
d)Ta có:
\(\frac{11}{54}< \frac{22}{54}< \frac{22}{37}\)
\(\Rightarrow\frac{11}{54}< \frac{22}{37}\)
e)Ta có:
\(\frac{-385}{-126}>0>\frac{-57}{3461}\)
\(\Rightarrow\frac{-385}{-126}>\frac{-57}{3461}\)
f)Ta có:
\(\frac{123}{109}>1>\frac{556}{789}\)
\(\Rightarrow\frac{123}{109}>\frac{556}{789}\)
g)Ta có:
\(\frac{-56}{57}>-1>\frac{-49}{47}\)
\(\Rightarrow\frac{-56}{57}>\frac{-49}{47}\)
a) Ta có: \(\frac{-697}{-313}=\frac{697}{313}>0\)
\(\frac{419}{-723}< 0\)
\(\Rightarrow\frac{419}{-713}< \frac{-697}{-313}\)
b) Ta có: \(\frac{190}{191}< 1\)
\(\frac{2019}{2018}>1\)
\(\Rightarrow\frac{190}{191}< \frac{2019}{2018}\)
c) Ta có: \(\frac{19}{27}< 1\)
\(\Rightarrow\frac{19}{27}< \frac{19.10+3}{27.10+3}\)
\(\Rightarrow\frac{19}{27}< \frac{193}{273}\)
d) Ta có: \(\frac{53}{47}< \frac{57}{47}\)
\(\frac{57}{43}>\frac{57}{47}\)
\(\Rightarrow\frac{53}{47}< \frac{57}{43}\)
a) 419/-723 < 0 < -697/-313
=> 419/-723 < -697/-313
b) 190/191 < 1 < 2019/2018
=> 190/191 < 2019/2018
d) 53/47 < 57/47 < 57/43
=> 53/47 < 57/43
a/=(74-(-1937)1)
=74-(-1937)
=2011
b/=4/7+5/6:5-3/8*(-4)
=4/7+1/6-(-3/2)
=31/42-(-3/2)
=47/21
minh chi biet bay nhieu
a) 13/57=13+16/57+16=29/73 ( Ghi nhớ SKG Toán 6)
-=> 13/57 < 29/73
b) 17/42 = 17-4/42-4 = 13/38
=> 17/42 > 13/38
c)7/41 = 7+6/41+6= 13/47
=> 7/41<13/47
1)
A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\)
A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{99}-\frac{1}{101}\)
A = \(\frac{1}{1}-\frac{1}{101}\)
A = \(\frac{100}{101}\)
Vậy A = \(\frac{100}{101}\)
B = \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
B = \(\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}.\frac{100}{101}\)
B = \(\frac{250}{101}\)
Vậy B = \(\frac{250}{101}\)
2)
Gọi ƯCLN ( 2n + 1 ; 3n + 2 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\Rightarrow1⋮d}\)
\(\Rightarrow d=1\)
Vậy \(\frac{2n+1}{3n+2}\)là p/s tối giản
Gọi ƯCLN ( 2n+3 ; 4n+4 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+4⋮d\end{cases}\Rightarrow\hept{\begin{cases}2n+3⋮d\\\left(4n+4\right):2⋮d\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\2n+2⋮d\end{cases}\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d}\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Vậy ...
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