Bài 2:
a: Thay x=1 vào A, ta được:
\(A=\dfrac{1+2}{1+1}=\dfrac{3}{2}\)
b: \(B=\dfrac{4}{\sqrt{x}-2}+\dfrac{16}{4-x}\)
\(=\dfrac{4}{\sqrt{x}-2}-\dfrac{16}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4\left(\sqrt{x}+2\right)-16}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{4}{\sqrt{x}+2}\)
c: Đặt P=A*B
\(=\dfrac{4}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=\dfrac{4}{\sqrt{x}+1}\)
Để P>1 thì P-1>0
=>\(\dfrac{4}{\sqrt{x}+1}-1>0\)
=>\(\dfrac{4-\sqrt{x}-1}{\sqrt{x}+1}>0\)
=>\(3-\sqrt{x}>0\)
=>\(\sqrt{x}< 3\)
=>0<=x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< 9\\x\ne4\end{matrix}\right.\)
mà x chẵn
nên \(x\in\left\{0;2;6;8\right\}\)
