nguowch đề :))

\(ab^2+b+7⋮a^2b+a+b\Leftrightarrow a\left(ab^2+b+7\right)-b\left(a^2b+a+b\right)⋮a^2b+a+b\Leftrightarrow7a-b^2⋮a^2b+a+b\left(1\right)\)

\(+,7a=b^2\Rightarrow\left(a;b\right)=\left(7k^2;7k\right)\left(k\text{ nguyên dương}\right)\)

\(+,7a>b^2\text{ từ 1}\Rightarrow7a-b^2\ge a^2b+a+b\Leftrightarrow6a\ge a^2b+b+b^2\text{ mà: b là số nguyên dương}\Rightarrow b< 3\Leftrightarrow b\in\left\{1;2\right\}\)

làm tiếp

\(+,7a< b^2\text{ từ (1)}\Rightarrow b^2-7a\ge a^2b+a+b\Leftrightarrow voli\text{ :)}.Tự\text{ kết luận}\)

em ko bik

wow

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Tham khảo qua link đây nhé cậu :

https://books.google.com.vn/books?id=PhymDAAAQBAJ&pg=PA59&lpg=PA59&dq=T%C3%ACm+c%C3%A1c+s%E1%BB%91+nguy%C3%AAn+a,b,c+th%E1%BB%8Fa+m%C3%A3n+a%5E2%2Bab%3D5a%2B2b%2B9&source=bl&ots=8bzSP0h3kN&sig=ACfU3U2A_d9ME7r47hAsdMemtJWUaW1w_A&hl=vi&sa=X&ved=2ahUKEwjg89-r0_HoAhUkK6YKHWIXCnkQ6AEwAnoECAwQKw#v=onepage&q=T%C3%ACm%20c%C3%A1c%20s%E1%BB%91%20nguy%C3%AAn%20a%2Cb%2Cc%20th%E1%BB%8Fa%20m%C3%A3n%20a%5E2%2Bab%3D5a%2B2b%2B9&f=false

Học tốt

\(a^2 + ab = 5a + 2b + 9 \)

\(\Leftrightarrow a^2+ab-5a-2b+6=15\)

\(\Leftrightarrow a\left(a+b\right)-2\left(a+b\right)-3\left(a-2\right)=15\)

\(\Leftrightarrow\left(a-2\right)\left(a+b\right)-3\left(a-2\right)=15\)

\(\Leftrightarrow\left(a-2\right)\left(a+b-3\right)=15\)

Do \(a,b\in Z\Rightarrow\hept{\begin{cases}a-2\in Z\\a+b-3\in Z\end{cases}}\)

\(\Rightarrow\left(a-2\right);\left(a+b-3\right)\inƯ\left(15\right)=\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

\(\left(+\right)\hept{\begin{cases}a-2=1\\a+b-3=15\end{cases}\Rightarrow}a=3;b=15\)

\(\left(+\right)\hept{\begin{cases}a-2=-1\\a+b-3=-15\end{cases}\Rightarrow}a=1;b=-13\)

\(\left(+\right)\hept{\begin{cases}a-2=3\\a+b-3=5\end{cases}\Rightarrow}a=5;b=3\)

\(\left(+\right)\hept{\begin{cases}a-2=-3\\a+b-3=-5\end{cases}}\Rightarrow a=-1;b=-1\)

\(\left(+\right)\hept{\begin{cases}a-2=15\\a+b-3=1\end{cases}\Rightarrow}a=17;b=-13\)

\(\left(+\right)\hept{\begin{cases}a-2=-15\\a+b-3=-1\end{cases}\Rightarrow}a=-13;b=15\)

\(\left(+\right)\hept{\begin{cases}a-2=5\\a+b-3=3\end{cases}\Rightarrow a=7;b=-1}\)

\(\left(+\right)\hept{\begin{cases}a-2=-5\\a+b-3=-3\end{cases}\Rightarrow}a=-3;b=3\)

Vậy  \(\left(a,b\right)=\left(3,15\right);\left(1,-13\right);\left(5,3\right);\left(-1,-1\right);\left(17,-13\right);\left(-13,15\right);\left(7,-1\right);\left(-3,3\right)\)

đơn giản

ab+2a-(b+2)=1

<=>a(b+2)-(b+2)=1

<=>(a-1)(b+2)=1=1.1=(-1).(-1)

+)(a-1)(b+2)=1.1

=>a-1=1 và b+2=1

=>a=2 và b=-1

+)(a-1)(b+2)=(-1).(-1)

=>a-1=-1 và b+2=-1

=>a=0 và b=-3

Vậy \(\left(a;b\right)\in\left\{\left(0;-3\right);\left(2;-1\right)\right\}\)
 

Pt <=>(a-3+b)(a-2)=15

a²b+a+b             ab²+b+1

a²b+a+\(\dfrac{a}{b}\)            \(\dfrac{a}{b}\)

_________

          \(b-\dfrac{a}{b}\)

-Để a²b+a+b chia hết cho ab²+b+1 thì:

\(b-\dfrac{a}{b}=0\Leftrightarrow\dfrac{b^2-a}{b}=0\Rightarrow b^2=a\)

-Vậy với \(b^2=a\) và \(b\ne0\) thì ta có đpcm.