1) Thay x=25 vào A ta có:
\(A=1-\dfrac{2\sqrt{25}}{25+1}=\dfrac{8}{13}\)
2) \(B=\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1}\)
\(B=\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{\sqrt{a}\left(a+1\right)+\left(a+1\right)}\)
\(B=\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}\)
\(B=\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\)
\(B=\dfrac{a-2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(a+1\right)}\)
\(B=\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(a+1\right)}\)
Nên:
\(P=A:B\)
\(P=\left(1-\dfrac{2\sqrt{a}}{a+1}\right):\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(a+1\right)}\)
\(P=\left(\dfrac{a+1-2\sqrt{a}}{a+1}\right):\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(a+1\right)}\)
\(P=\dfrac{\left(\sqrt{a}-1\right)^2}{a+1}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(a+1\right)}{\left(\sqrt{a}-1\right)^2}\)
\(P=\sqrt{a}+1\)
3) \(P=\dfrac{2\sqrt{a}+5}{4}\) khi:
\(\Leftrightarrow\sqrt{a}+1=\dfrac{2\sqrt{a}+5}{4}\)
\(\Leftrightarrow4\sqrt{a}+4=2\sqrt{a}+5\)
\(\Leftrightarrow2\sqrt{a}=1\)
\(\Leftrightarrow\sqrt{a}=\dfrac{1}{2}\)
\(\Leftrightarrow a=\dfrac{1}{4}\left(tm\right)\)
