ta có :

\(\frac{2008}{2009}< 1\)

\(\frac{2009}{2010}< 1\)

\(\frac{2010}{2011}< 1\)

\(\frac{2011}{2012}< 1\)

\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}< 1+1+1+1\)

\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}< 4\)

vậy ....................

2008/2009+2009/2010+2010/2011+2011/2008         4

=2008/2008=1                                                            4

Vì 1<4 nên 2008/2009+2009/2010+2011/2008 <   4

40tuổi

con : 30 tuổi 

bố : 70 tuổi

\(A=10^{25}+\frac{1}{10^{26}}+1=1\cdot10^{25}\)

\(B=10^{26}+\frac{1}{10^{27}}+1=1\cdot10^{26}\)

\(1\cdot10^{25}< 1\cdot10^{26}\Rightarrow A< B\)

\(2x^2-72=0\Leftrightarrow2\left(x^2-36\right)=0\Leftrightarrow x^2=36\Leftrightarrow x=-6; x=6\)

\(\frac{1}{2}x+\frac{2}{3}\left(x-1\right)=\frac{1}{3}\)

\(\frac{1}{2}x+\frac{2}{3}x-\frac{2}{3}=\frac{1}{3}\)

\(\frac{7}{6}x=\frac{1}{3}+\frac{2}{3}\)

\(x=\frac{6}{7}\)

Ta có :

AOB = 50 độ 

BOM = 30 độ

mà AOB + BOM = AOM 

=> 50 + 30 = 80 độ

bé tự vẽ hình nha

ta co

góc aom = aob+bom

               =50+30

               =80 do

\(M=\frac{9}{49}-\frac{11}{60}+\frac{13}{84}-\frac{15}{112}\)

\(M=\frac{1}{2940}+\frac{1}{48}\)

\(M=\frac{83}{3920}\)

Bạn có thể giải thik rõ giúp mik được ko

Để \(A\)có giá trị nguyên thì \(6n+3⋮3n+1\)

Ta có :

\(6n+3=\left(3n+1\right).2+3-2=2\left(3n+1\right)+1\)

Ta thấy :

\(3n+1⋮3n+1\Rightarrow2\left(3n+1\right)⋮3n+1\)

Để \(6n+3⋮3n+1\)thì \(1⋮3n+1\)

\(\Rightarrow3n+1\in\left\{1;-1\right\}\)

\(\Rightarrow3n\in\left\{0;-2\right\}\)

\(\Rightarrow n\in\left\{0;\frac{-2}{3}\right\}\)

Vì \(n\inℤ\Rightarrow n=0\)

Vậy \(n=0\)

\(A=\frac{6n+3}{3n+1}=\frac{2\left(3n+1\right)+1}{3n+1}=2+\frac{1}{3n+1}\)

A có giá trị nguyên <=> \(\frac{1}{3n+1}\)có giá trị nguyên

                               <=> \(1⋮3n+1\)

                               <=> \(3n+1\inƯ\left(1\right)=\left\{\pm1\right\}\)

Đk n nguyên => n = 0