[sp]Write it as $\dfrac{1}{a+b} = \dfrac{1}{a+bc} + \dfrac{1}{b+ac} = \dfrac{(a+b)(1+c)}{(a+bc)(b+ac)}$. Then $$(a+b)^2(1+c) = (a+bc)(b+ac) = ab(1+c^2) + (a^2+b^2)c,$$ $$(a+b)^2 + 2abc = ab(1+c^2),$$ $$(a+b)^2 = ab(c-1)^2.$$ This shows that $ab = \Bigl(\dfrac{a+b}{c-1}\Bigr)^2$. In other words, $\sqrt{ab} = \Bigl|\dfrac{a+b}{c-1}\Bigr|,$ which is rational.anemone said:The rational numbers $a,\,b,\,c$ (for which $a+bc$, $b+ac$ and $a+b$ are all non-zero) satisfy the equality $\dfrac{1}{a+bc} =\dfrac{1}{a+b}-\dfrac{1}{b+ac}$.
Prove that $\sqrt{(c−3)(c+1)}$ is rational.
One way to prove an expression is rational is to show that it can be written as a ratio of two integers, where the denominator is not equal to zero.
Yes, the rational root theorem states that if a polynomial has rational roots, they must be in the form of p/q where p is a factor of the constant term and q is a factor of the leading coefficient.
Yes, all rational expressions can be written as a polynomial expression with a finite number of terms.
No, an expression cannot be both irrational and rational. It can only be one or the other.
Yes, the fundamental theorem of algebra states that any polynomial expression with complex coefficients can be factored into linear or quadratic factors, which can then be written as rational expressions.