NT
6
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HN
3
6 tháng 3 2017
A=\(\dfrac{-2}{9}\)+\(\dfrac{-3}{4}\)+\(\dfrac{3}{5}\)+\(\dfrac{1}{15}\)+\(\dfrac{1}{57}\)+\(\dfrac{1}{3}\)+\(\dfrac{-1}{36}\)
A=(\(\dfrac{-2}{9}\)+\(\dfrac{-3}{4}\)+\(\dfrac{-1}{36}\))+(\(\dfrac{3}{5}\)+\(\dfrac{1}{15}\)+\(\dfrac{1}{3}\))
A=-1+1=0
B=\(\dfrac{1}{2}\)+\(\dfrac{-1}{5}\)+\(\dfrac{-5}{7}\)+\(\dfrac{1}{6}\)+\(\dfrac{-3}{35}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{41}\)
B=(\(\dfrac{-1}{5}\)+\(\dfrac{-5}{7}\)+\(\dfrac{-3}{35}\))+(\(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{3}\))+\(\dfrac{1}{41}\)
B=-1+1+\(\dfrac{1}{41}\)=\(\dfrac{1}{41}\)
IL
0
BT
1
NL
Nguyễn Lê Phước Thịnh
CTVHS
28 tháng 1 2024
\(\dfrac{-2}{9}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{15}+\dfrac{1}{57}+\dfrac{1}{3}-\dfrac{1}{36}\)
\(=\left(-\dfrac{2}{9}-\dfrac{3}{4}-\dfrac{1}{36}\right)+\left(\dfrac{3}{5}+\dfrac{1}{15}+\dfrac{1}{3}\right)+\dfrac{1}{57}\)
\(=\dfrac{-8-27-1}{36}+\dfrac{9+1+5}{15}+\dfrac{1}{57}\)
\(=-\dfrac{36}{36}+\dfrac{15}{15}+\dfrac{1}{57}\)
\(=\dfrac{1}{57}\)
A=−29+−34+35+115+157+13−136A=-29+-34+35+115+157+13-136
→A=(−29+−34+13−136)+(35+115)+
Đúng(1)
Gọi ƯCLN(12n+1,30n+2)=d(d>0,d∈Z)ƯCLN(12n+1,30n+2)=d(d>0,d∈Z)
⇒12n+1⋮d,30n+2⋮d⇒12n+1⋮d,30n+2⋮d
⇒60n+5⋮d,60n+4⋮d⇒60n+5⋮d,60n+4⋮d
⇒60n+5−60n−4⋮d⇒60n+5-60n-4⋮d
⇒1⋮d⇒1⋮d
⇒d∈Ư(1
Gọi ƯCLN(12n+1,30n+2)=d(d>0,d∈Z)ƯCLN(12n+1,30n+2)=d(d>0,d∈Z)
⇒12n+1⋮d,30n+2⋮d⇒12n+1⋮d,30n+2⋮d
⇒60n+5⋮d,60n+4⋮d⇒60n+5⋮d,60n+4⋮d
⇒60n+5−60n−4⋮d⇒60n+5-60n-4⋮d
⇒1⋮d⇒1⋮d
⇒d∈Ư(1
Chứng minh các phân số sau là phân số tối giản với mọi số nguyên n: A=\frac{12n+1}{30n+2}
lúc nãy cháu mình nó trả lời lung tung sory nha bù lè
\(A=\frac{2}{9}+\frac{\left(-3\right)}{4}+\frac{3}{5}+\frac{1}{15}+\frac{1}{57}+\)\(\frac{1}{3}-\frac{1}{36}\)
\(A=\left(\frac{1}{3}-\frac{1}{36}+\frac{2}{9}+-\frac{3}{4}\right)\)\(+\left(\frac{3}{5}+\frac{1}{15}\right)+\frac{1}{57}\)
\(A=\frac{1}{57}\)
Khó quá , em mới lớp 2 thôi.