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a, Ta có : \(7x+4y⋮37\)
\(\Rightarrow23\left(7x+4y\right)⋮37\)
\(\Rightarrow161x+92y⋮37\)
\(\Rightarrow\left(13x+18y\right)+148x+74y⋮37\)
Mà \(\hept{\begin{cases}148x⋮37\\74x⋮37\end{cases}\Rightarrow13x+18y⋮37}\)
Vậy \(13x+18y⋮37\)
b, Ta có : \(A=\frac{2014^{2012}+1}{2014^{2013}+1}\)
\(\Rightarrow2014A=\frac{2014^{2013}+2014}{2014^{2013}+1}=\frac{2014^{2013}+1+2013}{2014^{2013}+1}=1+\frac{2013}{2014^{2013}+1}\)
Ta có : \(B=\frac{2014^{2011}+1}{2014^{2012}+1}\)
\(\Rightarrow2014B=\frac{2014^{2012}+2014}{2014^{2012}+1}=\frac{2014^{2012}+1+2013}{2014^{2012}+1}=1+\frac{2013}{2014^{2012}+1}\)
Vì \(2014^{2013}+1>2014^{2012}+1\)
\(\Rightarrow\frac{1}{2014^{2013}+1}< \frac{1}{2014^{2012}+1}\Rightarrow1+\frac{1}{2014^{2013}+1}< 1+\frac{1}{2014^{2012}+1}\)
\(\Rightarrow2014A< 2014B\Rightarrow A< B\)
1: \(C=2010\cdot2012\)
\(C=\left(2011-1\right)\left(2011+1\right)\)
\(C=2011\left(2011+1\right)-\left(2011+1\right)\)
\(C=2011\cdot2011+2011-2011-1=2011\cdot2011-1\)
Mà \(D=2011\cdot2011\)
\(\Rightarrow C< D\)
2: Chia 1 số cho 60 thì dư 37.Vậy chia số đó cho 15 thì được số dư là 7
3: Chú thích: giá trị nhỏ nhất=GTNN
Để M có GTNN
thì \(2012-\frac{2011}{2012-x}\) có GTNN
Nên \(\frac{2011}{2012-x}\)có GTLN
nên 2012-x>0 và x thuộc N
Suy ra: 2012-x=1
Suy ra: x=2011
Vậy, M có GTNN là 2011 khi x=2011
Đặt \(A=\frac{37^{2013}+1}{37^{2012}+1}\) và \(B=\frac{37^{2014}+1}{37^{2013}+1}\) ta có :
\(\frac{1}{37}A=\frac{37^{2013}+1}{37^{2013}+37}=\frac{37^{2013}+37-36}{37^{2013}+37}=\frac{37^{2013}+37}{37^{2013}+37}-\frac{36}{37^{2013}+37}=1-\frac{36}{37^{2013}+37}\)
\(\frac{1}{37}B=\frac{37^{2014}+1}{37^{2014}+37}=\frac{37^{2014}+37-36}{37^{2014}+37}=\frac{37^{2014}+37}{37^{2014}+37}-\frac{36}{37^{2014}+37}=1-\frac{36}{37^{2014}+37}\)
Vì \(\frac{36}{37^{2013}+37}>\frac{36}{37^{2014}+37}\) nên \(1-\frac{36}{37^{2013}+37}< 1-\frac{36}{37^{2014}+37}\)
\(\Rightarrow\)\(\frac{1}{37}A< \frac{1}{38}B\)
\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
P=1+2012 +20122+20123+20124+...+201271+201272
2012P=2012(1+2012 +20122+20123+20124+...+201271+201272)
2012P=2012 +20122+20123+20124+...+201272+201273
2012P-P= (2012 +20122+20123+20124+...+201272+201273) - ( 1+2012 +20122+20123+20124+...+201271+201272 )
2011P=201273-1
P=(201273-1)/2011
Vì (201273-1)/2011< 201273-1 nên P<Q
NHỚ NHA!!!!!!
P=1+2012 +20122+20123+20124+...+201271+201272
2012P=2012(1+2012 +20122+20123+20124+...+201271+201272)
2012P=2012 +20122+20123+20124+...+201272+201273
2012P-P= (2012 +20122+20123+20124+...+201272+201273) - ( 1+2012 +20122+20123+20124+...+201271+201272 )
2011P=201273-1
P=(201273-1)/2011
Vì (201273-1)/2011< 201273-1 nên P<Q
NHỚ NHA!!!!!!
Ta có :M=\(\frac{2012^{37}+37^{2012}+1}{2012^{38}}\)=\(\frac{1}{2012}\)+\(\frac{37^{2012}}{2018^{38}}\)+\(\frac{1}{2012^{38}}\)
N=\(\frac{2012^{38}+37^{2012}+2}{2012^{39}}\)=\(\frac{1}{2012}\)+\(\frac{37^{2012}}{2012^{39}}\)+\(\frac{2}{2012^{39}}\)
Suy ra: M-N=\(\frac{37^{2012}}{2012^{38}}\left(1-\frac{1}{2012}\right)\)+\(\frac{1}{2012^{38}}\left(1-\frac{2}{2012}\right)\)
\(\Rightarrow\)M-N=\(\frac{37^{2012}}{2012^{38}}.\frac{2011}{2012}+\frac{1}{2012^{38}}.\frac{2010}{2012}\)
\(\Rightarrow\)M-N>0
\(\Rightarrow\)M>N
Vậy M>N