Ta có :
\(a.b=a:b\)
\(\Leftrightarrow b^2=1\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}b=1\\b=-1\end{array}\right.\)
(+) b = 1
\(\Rightarrow a+1=a\) ( vô lí )
(+) b = - 1
\(\Rightarrow a-1=-a\)
\(\Rightarrow a=\frac{1}{2}\) ( tm )
Vậy \(\left(a;b\right)=\left(\frac{1}{2};-1\right)\)
Giải:
Ta có:
\(a+b=ab\)
\(\Rightarrow a=b\left(a-1\right)\)
\(a+b=a:b\)
\(\Rightarrow a+b=b\left(a-1\right):b\)
\(\Rightarrow a+b=a-1\)
\(\Rightarrow b=-1\)
\(a+b=ab\)
\(\Rightarrow a-1=-a\)
\(\Rightarrow2a=1\)
\(\Rightarrow a=\frac{1}{2}\)
Vậy \(a=\frac{1}{2},b=-1\)
Ta có:
a + b = a.b => a = a.b - b = b.(a - 1)
=> a : b = a - 1 = a + b
=> b = -1
=> a = -1.(a - 1) = -a + 1
=> a + a = 1 = 2a
=> a = 1/2
Vậy a = 1/2; b = -1
Bài làm
Ta có:
a+b=ab+b=a:b
⇒a=b(a−1)⇒a=b(a−1)
a+b=a:ba+b=a:b
⇒a+b=b(a−1):b⇒a+b=b(a−1):b
⇒a+b=a−1⇒a+b=a−1
⇒b=−1⇒b=−1
a+b=aba+b=ab
⇒a−1=−a⇒a−1=−a
⇒2a=1⇒2a=1
⇒a=12⇒a=12
Vậy a=12,b=−1
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