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Ta so sanh phan hon:
\(\frac{2941}{2938}-1=\frac{3}{2938}\)
\(\frac{2940}{2937}-1=\frac{3}{2937}\)
vi\(2938>2937\)=>\(\frac{3}{2938}< \frac{3}{2937}\)hay\(\frac{2941}{2938}< \frac{2940}{2937}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 3 2 = 9 6 ; 2 3 = 4 6 . Do 9 6 > 4 6 > 1 6 ⇒ 3 2 > 2 3 > 1 6
b) 13 57 = 26 114 ; 29 38 = 87 114 . Do 26 114 < 87 114 ⇒ 13 57 < 29 38
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: BCNN (a,b) . ƯCLN (a,b) = a . b
Mà a . b = 2940 & BCNN (a,b) = 210
=> 210 . ƯCLN (a,b) = 2940
=> ƯCLN (a,b) = 2940 : 210
=> ƯCLN (a,b) = 14
Ta có: a = 14m ; b = 14n (m,n∈Z;m,n≠0)
=> a . b = 14m . 14n = 2940
=> 14m . 14n = 2940
=> 196 . mn = 2940
=> mn = 2940 : 196 = 15
Kết quả tìm kiếm | Học trực tuyến
https://h.vn/hoi-dap/tim-kiem?q=T%C3%ACm+2+s%E1%BB%91+t%E1%BB%B1+nhi%C3%AAn+a+v%C3%A0+b+bi%E1%BA%BFt+t%C3%A… 2/19
7 tháng 9 2016 lúc 8:51
=> Ta có các trường hợp:
m = 1; b = 15 =>\(\hept{\begin{cases}a=14.1=14\\b=14.15=210\end{cases}}\)
m = -1 ; b = -15 =>\(\hept{\begin{cases}a=14.\left(-1\right)==14\\b=14.\left(-15\right)=-210\end{cases}}\)
m = 15; b = 1 =>\(\hept{\begin{cases}a=14.15=210\\b=14.1=14\end{cases}}\)
m = -15 ; b = -1 =>\(\hept{\begin{cases}a=14.\left(-15\right)=-210\\b=14.\left(-1\right)=-14\end{cases}}\)
m = 3 ; b = 5 =>\(\hept{\begin{cases}a=14.3=42\\b=14.5=70\end{cases}}\)
m = -3 ; b = -5 =>\(\hept{\begin{cases}a=14.\left(-3\right)=-42\\b=14.\left(-5\right)=-70\end{cases}}\)
m = 5 ; b = 3 =>\(\hept{\begin{cases}a=14.5=70\\b=14.3=42\end{cases}}\)
m = -5 ; b = -3 => \(\hept{\begin{cases}a=14.\left(-5\right)=-70\\b=14.\left(-3\right)=-42\end{cases}}\)
Học ~ Giỏi
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2-\frac{2}{19}+\frac{2}{43}-\frac{2}{1995}}{3-\frac{3}{19}+\frac{3}{43}-\frac{3}{1995}}:\frac{4-\frac{4}{29}+\frac{4}{41}-\frac{4}{2941}}{5-\frac{5}{29}+\frac{5}{41}-\frac{5}{2941}}\)
\(\Rightarrow A=\frac{2\left(1-\frac{1}{19}+\frac{1}{43}-\frac{1}{1995}\right)}{3\left(1-\frac{1}{19}+\frac{1}{43}-\frac{1}{1995}\right)}:\frac{4\left(1-\frac{1}{29}+\frac{1}{41}-\frac{1}{2941}\right)}{5\left(1-\frac{1}{29}+\frac{1}{41}-\frac{1}{2941}\right)}\)
\(\Rightarrow A=\frac{2}{3}:\frac{4}{5}\)
\(\Rightarrow A=\frac{2}{3}.\frac{5}{4}\)
\(\Rightarrow A=\frac{10}{12}=\frac{5}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2941}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{1943}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2941}}{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{1943}}\)
\(=\dfrac{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2941}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{1943}}.\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{1943}}{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2941}}\)
\(=\dfrac{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2941}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{1943}\right)}.\dfrac{2\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{1943}\right)}{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2941}\right)}\)
\(=\dfrac{5}{3}.\dfrac{2}{4}=\dfrac{10}{12}=\dfrac{5}{6}\)
Vì đề bài không yêu cầu tính nên bn có thể không tính ra như mk cux đc!
![](https://rs.olm.vn/images/avt/0.png?1311)
UCLN(a;b) là:2940:210=14
Ta có:a=14m
b=14n
(m;n thuộc N và UCLN(m,n)=1)
Ta có:a.b=2940
hay 14m.14n=2940
196(m.n)=2940
m.n=2940:196
m.n=15
m 1 3 5 15
n 15 5 3 1
=>a 14 42 70 210
=> b 210 70 42 14
Vậy ta có các cặp số (a;b)={(14;210);(42;70);(70;42);(210;14)}
Tick minhf nhé bạn!
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(542 - 541)2938 = 12938 = 1
(542 + 541)0 = 10830 = 1
Vậy : (542 - 541)2938 = (542 + 541)0 (đpcm)
Ta có: 2941/2938-3/2938=1
2940/2937-3/2937=1
Vì 3/2938<3/2937
=>2941/2938<2940/2937