\(VT\ge0\Rightarrow\)\(-5\le y\le5\)

\(VT=8k^2\Rightarrow25-y^2=8k^2\Rightarrow k^2\le3\)
\(k^2=\left\{0,1\right\}\)

\(k=0\Rightarrow\hept{\begin{cases}x=2009\\y=+-5\end{cases}}\)

\(k^2=1\Rightarrow y^2=17\left(loai\right)\)

KL

\(\left(x,y\right)=\left(2009,-5\right);\left(2009,5\right)\)

a, x ⋮ 25 và x < 100

Vì x ⋮ 25 

nên x ∈ B(25) = { 0;25;50;75;100;... }

Mà x < 100

=> x = { 0 ; 25 ; 50 ; 75 }

 b,5x + 3x = 3^6 : 3^3 .4 + 12

   x.( 5 +3 )= 3^3 . 4 + 12

    x . 8       = 27 . 4 + 12

    x . 8       = 108 + 12

    x . 8       = 120

    x            = 120 : 8

    x            = 15

                                                               ~HT~

\(25-y^2-8.\left(x-2009\right)^2\)

ta thấy vế phải \(8.\left(x-2009\right)^2\ge0\) \(\forall x\)

\(\Rightarrow VT:25-y^2\ge0\)

\(\Rightarrow0\le y^2\le25\)

\(\Rightarrow y^2\in\left\{0;1;4;9;16;25\right\}\)

mà \(8.\left(x-2009\right)^2\) chẵn\(\Rightarrow25-y^2\)chẵn \(\Rightarrow y^2lẻ\)

\(\Rightarrow y^2\in\left\{1;9;25\right\}\)

\(\Rightarrow y\in\left\{1;3;5\right\}\) (do \(y\in N\))

\(TH1:y=1\)

\(\Rightarrow8.\left(x-2009\right)^2=24\)

\(\Leftrightarrow\left(x-2009\right)^2=3\left(koTM\right)\)(do \(x\in N\))

\(TH2:y=3\)

\(\Rightarrow8.\left(x-2009\right)^2=16\)

\(\left(x-2009\right)^2=2\left(koTM\right)\)(do \(x\in N\))

\(TH3:y=25\)

\(\Rightarrow8.\left(x-2009\right)^2=0\)

\(\Rightarrow\left(x-2009\right)^2=0\Rightarrow x=2009\left(TM\right)\)

vậy cặp số \(\left(x,y\right)\) thỏa mãn \(25-y^2-8.\left(x-2009\right)^2\)  là  \(\left(2009;25\right)\)

Ta có :

\(25-y^2=8\left(x-2009\right)^2\)

\(\Rightarrow8\left(x-2009\right)^2\le25\)

\(\Leftrightarrow\left(x-2009\right)^2\le\frac{25}{8}\)

\(\Rightarrow0\le\left(x-2009\right)^2\le3\)

\(\Rightarrow\left(x-2009\right)^2\in\left\{0;1\right\}\)

+) Trường hợp 1 :

\(\Rightarrow\left(x-2009\right)^2=0\)

\(\Rightarrow x=2009\)

\(\Rightarrow y=5\)

\(\Leftrightarrow\hept{\begin{cases}x=2009\\y=5\end{cases}}\)

+) Trường hợp 2 :

\(\left(x-2009\right)^2=1\)

\(\Rightarrow x-2009=1\)

\(\Rightarrow x=2010\)

\(\Rightarrow25-y^2=8\)

\(\Rightarrow y^2=17\) (loại)

+) Trường hợp 3 :

\(\left(x-2009\right)^2=1\)

\(\Rightarrow x=2008\)

\(\Rightarrow25-y^2=8\)(loại)

Vậy ......

\(\)

Do \(8\left(x-2009\right)^2\ge0\Rightarrow25-y^2\ge0\)

\(\Leftrightarrow y^2\le25\).Mà \(y\inℕ\) nên \(0\le y^2\le25\Leftrightarrow0\le y\le5\)

Mà \(8\left(x-2009\right)^2⋮8\Rightarrow25-y^2⋮8\)

\(\Rightarrow y\in\left\{1;3;5\right\}\)

Thay vào tìm x. :) Nhớ đk: \(x,y\inℕ\)

Ta có: \(25-y^2=8.\left(x-2009\right)^2\)

\(\Rightarrow8.\left(x-2009\right)^2+y^2=25\left(1\right)\)

Vì \(y^2\ge0\)nên \(\left(x-2009\right)^2\le\frac{25}{8}\)

\(\Rightarrow\left(x-2009\right)^2=0\)hoặc \(\left(x-2009\right)^2=1\)

Với \(\left(x-2009\right)^2=1\)thay vào \(\left(1\right)\), ta có:

\(8.1+y^2=25\)

\(\Rightarrow8+y^2=25\)

\(\Rightarrow y^2=17\)( loại )

Với \(\left(x-2009\right)^2=0\)thay vào \(\left(1\right)\), ta có:

\(8.0+y^2=25\)

\(\Rightarrow0+y^2=25\)

\(\Rightarrow y^2=25\)

\(\Rightarrow\orbr{\begin{cases}y=5\\y=-5\end{cases}}\)

Mà \(y\in N\)

\(\Rightarrow y=5,x=2009\)

Vậy \(x=2009,y=5\)

Ta có: \(\left(x-2009\right)^2\ge0\)nên \(8\left(x-2009\right)^2\ge0\)

VP \(\ge0\)nên \(25-y^2\ge0\Leftrightarrow y^2\le25\)(1)

Mặt khác, do \(\left[8\left(x-2009\right)^2\right]⋮2\)nên \(\left(25-y^2\right)⋮2\)

\(\Leftrightarrow y^2\)lẻ \(\Leftrightarrow y\)lẻ (2)

Kết hợp (1), (2) và \(y\inℕ\),ta được: \(y\in\left\{1;3;5\right\}\)(suy ra từ \(y^2\in\left\{1;9;25\right\}\))

*Với y = 1 thì \(25-1^2=8\left(x-2009\right)^2\Leftrightarrow8\left(x-2009\right)^2=24\Leftrightarrow\left(x-2009\right)^2=3\)(loại)

*Với y = 3 thì \(25-3^2=8\left(x-2009\right)^2\Leftrightarrow8\left(x-2009\right)^2=16\Leftrightarrow\left(x-2009\right)^2=2\)(loại)

*Với y = 5 thì \(25-5^2=8\left(x-2009\right)^2\Leftrightarrow8\left(x-2009\right)^2=0\Leftrightarrow\left(x-2009\right)^2=0\)\(\Leftrightarrow x=2009\)

Vậy x = 5 và y = 2009.

giải rõ ra giùm cái

y=5:x=2009

nhe