Có ab-2a-b=3=>ab-2b-b=3

=>a(b-2)-b=5-2

=>a(b-2)-b+2=5

=>a(b-2)-(b-2)=5       (Đặt dấu trừ đằng trước)

=>(b-2)(a-1)=5

=>b-2 và a-1 thuộc ước của 5 là cộng trừ 1 cộng trừ 5

+Nếu b-2=5 thì a-1=1 rồi giải như tìm x

+BẠN LÀM CÁC TRƯỜNG HỢP TIẾP THEO TƯƠNG TỰ NHÉ!!!!!!!!!

tìm a,b thuoc z để ab-2a-b=3

\(\dfrac{5}{2}=\dfrac{1}{6}+\dfrac{b}{3}\)

\(\Leftrightarrow\dfrac{5}{2a}-\dfrac{1}{6}-\dfrac{b}{3}=0\)

msc : 18a

\(\Leftrightarrow\dfrac{45}{18a}-\dfrac{3a}{18a}-\dfrac{6ab}{18a}=0\)

\(\Leftrightarrow45-3a-6ab=0\)

\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{15}{1+2b}\\b=\dfrac{15}{2a}-\dfrac{1}{2}\end{matrix}\right.\)

\(\dfrac{5}{2a}\) hay \(\dfrac{5}{2}a\) vậy bạn?

\(\dfrac{5}{2a}=\dfrac{1}{6}+\dfrac{b}{3}\)

\(\Rightarrow\dfrac{15}{6a}=\dfrac{a}{6a}+\dfrac{2ab}{6a}\)

\(\Rightarrow a+2ab=15\)

\(\Rightarrow a\left(1+2b\right)=15\)

\(\Rightarrow a=\dfrac{15}{2b+1}\).

-Vì \(a,b\) nguyên nên để \(\dfrac{15}{2b+1}\) là số nguyên thì: \(15⋮\left(2b+1\right)\)

\(\Rightarrow2b+1\inƯ\left(15\right)\)

\(\Rightarrow2b+1\in\left\{1;3;5;15;-1;-3;-5;-15\right\}\)

\(\Rightarrow b\in\left\{0;1;2;7;-1;-2;-3;-8\right\}\)

*Với \(b=0\) thì \(a=\dfrac{15}{2.0+1}=15\)

*Với \(b=1\) thì \(a=\dfrac{15}{2.1+1}=5\)

*Với \(b=2\) thì \(a=\dfrac{15}{2.2+1}=3\)

*Với \(b=7\) thì \(a=\dfrac{15}{2.7+1}=1\)

*Với \(b=-1\) thì \(a=\dfrac{15}{2.\left(-1\right)+1}=-15\)

*Với \(b=-2\) thì \(a=\dfrac{15}{2.\left(-2\right)+1}=-5\)

*Với \(b=-3\) thì \(a=\dfrac{15}{2.\left(-3\right)+1}=-3\)

*Với \(b=-8\) thì \(a=\dfrac{15}{2.\left(-8\right)+1}=-1\)

a) \(\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{1}{a-b}\left(đk:a,b\ne0,a\ne b\right)\Leftrightarrow\dfrac{b-a}{ab}=\dfrac{1}{a-b}\)

\(\Leftrightarrow-\left(a-b\right)^2=ab\Leftrightarrow a^2-ab+b^2=0\)

\(\Leftrightarrow\left(a^2-ab+\dfrac{1}{4}b^2\right)+\dfrac{3}{4}b^2=0\Leftrightarrow\left(a-\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2=0\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a-\dfrac{1}{2}b=0\\\dfrac{3}{4}b^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}b\\b=0\end{matrix}\right.\) \(\Leftrightarrow a=b=0\left(ktm\right)\)

Vậy k có a,b thõa mãn 

b) \(\dfrac{5}{2a}=\dfrac{1}{6}+\dfrac{b}{3}\left(a\ne0\right)\Leftrightarrow\dfrac{2b+1}{6}-\dfrac{5}{2a}=0\Leftrightarrow\dfrac{a\left(2b+1\right)-15}{6a}=0\)

\(\Leftrightarrow a\left(2b+1\right)-15=0\Leftrightarrow a\left(2b+1\right)=15\)

Do \(a,b\in Z,a\ne0\) nên ta có bảng sau:

Vậy...

a: Sửa đề: \(B=\left(\dfrac{2a}{a+3}+\dfrac{2}{3-a}+\dfrac{3}{a^2-9}\right):\dfrac{a+1}{a-3}\)

\(=\dfrac{2a^2-6a-2a-6+3}{\left(a-3\right)\left(a+3\right)}\cdot\dfrac{a-3}{a+1}=\dfrac{2a^2-8a-3}{\left(a+3\right)\left(a+1\right)}\)

b: |a|=2

=>a=2 hoặc a=-2

Khi a=2 thì \(B=\dfrac{2\cdot2^2-8\cdot2-3}{\left(2+3\right)\left(2+1\right)}=\dfrac{-11}{15}\)

Khi a=-2 thì \(B=\dfrac{2\cdot\left(-2\right)^2-8\cdot\left(-2\right)-3}{\left(-2+3\right)\left(-2+1\right)}=-21\)