Bài 3:
a) Ta có: \(x=9+4\sqrt{5}=\left(\sqrt{5}\right)^2+2\cdot2\cdot\sqrt{5}+2^2=\left(\sqrt{5}+2\right)^2\)
Thay x vào A ta có:
\(A=\dfrac{\sqrt{\left(\sqrt{5}-2\right)^2}+7}{\sqrt{\left(\sqrt{5}-2\right)^2}-1}=\dfrac{\sqrt{5}-2+7}{\sqrt{5}-2-1}=\dfrac{\sqrt{5}+5}{\sqrt{5}-3}\)
b) \(B=\dfrac{1}{\sqrt{x}+2}+\dfrac{3}{1-\sqrt{x}}+\dfrac{x+8}{x+\sqrt{x}-2}\)
\(B=\dfrac{1}{\sqrt{x}+2}-\dfrac{3}{\sqrt{x}-1}+\dfrac{x+8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{x+8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{\sqrt{x}-1-3\sqrt{x}-6+x+8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
c) Ta có: \(P=A\cdot B\)
\(P=\dfrac{\sqrt{x}+7}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(P=\dfrac{\sqrt{x}+7}{\sqrt{x}+2}\)
\(P=\dfrac{\sqrt{x}+2+5}{\sqrt{x}+2}=1+\dfrac{5}{\sqrt{x}+2}\)
P có giá trị nguyên khi:
\(\dfrac{5}{\sqrt{x}+2}\) nguyên
⇒ 5 ⋮ \(\sqrt{x}+2\)
⇒ \(\sqrt{x}+2\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
Mà: \(\sqrt{x}+2\ge2\)
⇒ \(\sqrt{x}+2=5\)
⇒ \(x=9\left(tm\right)\)
