a) ta có: \(a.b=a:b=\frac{a}{b}\Rightarrow a.b:\frac{a}{b}=1\) hay \(a.b:\frac{a}{b}=\frac{a.b.b}{a}=b^2=1\) => b = 1 hoặc b = -1
ta có: \(a+b=a.b\Rightarrow\frac{a+b}{ab}=1\) hay \(\frac{a+b}{ab}=\frac{a}{ab}+\frac{b}{ab}=\frac{1}{b}+\frac{1}{a}=1\)
Nếu b = 1
=> 1/1 + 1/a = 1 => 1/a = 0 => không tìm được a
Nếu b = -1
=> 1/-1 + 1/a = 1 => 1/a = 2 => a = 1/2
KL: b = -1; a = 1/2
b) ta có: a-b = 2.(a+b)
=> a- b = 2a + 2b
=> a-2a = 2b +b
-a = 3b
=> a = -3b
=> a : b = -3b : b = -3
=> a - b = 2.(a+b) = -3
=> a - b = -3 ; 2.(a+b) = -3 => a + b = -3/2
=> a = ( -3/2 + (-3) ) : 2 = -9/4
=> a + b = -3/2
b = -3/2 - (-9/4)
b = 3/4
KL:...
b, a - b = 2(a+b)
=>a-b=2a+2b
=>a-2a=2b+b
=>-a=3b
=>a=-3b
=>\(\frac{a}{b}=\frac{-3b}{b}=-3\)
Mà a-b=2(a+b)=a/b
=>\(\hept{\begin{cases}a-b=-3\\2\left(a+b\right)=-3\end{cases}\Rightarrow\hept{\begin{cases}a-b=-3\\a+b=-\frac{3}{2}\end{cases}}}\)
=>\(a-b+a+b=-3-\frac{3}{2}\)
=> \(2a=-\frac{9}{2}\Rightarrow a=\frac{-9}{4}\Rightarrow b=\frac{3}{4}\)
Vậy a=-9/4,b=3/4
