Để A lớn nhất thì:

    3-y = 0 Và x-y = 0

=> y =x = 3

=> GTLN của A là 2013 - I3 - 3I - (3 - 3)²

                           = 2013 - 0

                           = 2013

Vậy GTLN của A là 2013

\(A=2013-\left|3-y\right|-\left(x-y\right)^2\)

Vì \(\left|3-y\right|\ge0;\left(x-y\right)^2\ge0\)

\(\Rightarrow A\le2013\Rightarrow A_{max}=2013\)

\(\Leftrightarrow\hept{\begin{cases}3-y=0\\x-y=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=3\\x=3\end{cases}}}\)

Tôi ko bt=)

(139139 . 133 - 133133 . 139 ) : ( 2 + 4 + 6 + ... + 2002 )

=(139 . 1001 . 133 - 133 . 1001 . 139 ) :{ ( 2 + 2002) .[ ( 2002 - 2 ) : 2 + 1 ] : 2}

= [139 . (1001 . 133 - 133 . 1001)] : {2004 . 1001 : 2 }

= 139 . 0 - 1003002

= 0 - 1003002

= -1003002

P/S : Sai thì thôi ~ 

\(a)\frac{8}{9}x-\frac{2}{3}=\frac{1}{3}x+1\frac{1}{3}\)

\(\Rightarrow\frac{8}{9}x-\frac{1}{3}x=\frac{2}{3}+1\frac{1}{3}\)

\(\Rightarrow\frac{5}{9}x=\frac{2}{3}+\frac{4}{3}\)

\(\Rightarrow\frac{5}{9}x=2\Rightarrow x=2\div\frac{5}{9}=\frac{18}{5}\)

\(b)(\frac{-2}{5}+\frac{3}{7})-(\frac{4}{9}+\frac{12}{20}-\frac{13}{25})+\frac{7}{35}\)

\(=\frac{1}{35}-(\frac{4}{9}+\frac{3}{5}-\frac{13}{25})+\frac{1}{5}\)

\(=\frac{1}{35}-(\frac{4}{9}+\frac{15}{25}-\frac{13}{25})+\frac{1}{5}\)

\(=\frac{1}{35}-(\frac{4}{9}+\frac{2}{25})+\frac{1}{5}\)

\(=\frac{1}{35}-\frac{118}{25}+\frac{1}{5}\)

Làm nốt

\(B=x^2+\left|2y-6\right|-9\ge-9\)

Do \(x^2\ge0;\left|2y-6\right|\ge0\forall x,y\)

Dấu "=" xảy ra khi x = 0 và \(2y-6=0\Leftrightarrow y=3\)

Vậy \(B_{min}=9\Leftrightarrow\hept{\begin{cases}x=0\\y=3\end{cases}}\)

Trả lời zùm đi

éo biết

\(A=2013-\left|3-y\right|\)

Vì \(\left|3-y\right|\ge0\Rightarrow A\le2013\)

\(\Rightarrow A_{max}=2013\Leftrightarrow3-y=0\)

\(\Leftrightarrow y=3\)

P/s ko chắc :)

\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)

                                      \(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)

                                      \(=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)

\(3600:\left[\left(5x+335\right):x\right]=50\)

\(\left[\left(5x+335\right):x\right]=3600:50\)

\(\left[\left(5x+335\right):x\right]=72\)

đến đây thì mik chịu ! Xin lỗi nhé !!!

3600 : [ ( 5x + 335) : x ] = 50

( 5x +335 ) : x = 3600 : 50

( 5x + 335 ) : x = 72

( 5x + 335 ) = 72x

=> 72x-5x = 335

67x = 335

=> x = 5