Bài I:
1: Thay x=4 vào A, ta được:
\(A=\dfrac{4}{2+1}=\dfrac{4}{3}\)
2: \(B=\dfrac{3}{\sqrt{x}+1}+\dfrac{x+5}{x-1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{3}{\sqrt{x}+1}+\dfrac{\left(x+5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)+x+5-\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3\sqrt{x}-3+x-\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
3: P=A*B
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{x}{\sqrt{x}+1}=\dfrac{x}{\sqrt{x}-1}\)
P<=4
=>P-4<=0
=>\(\dfrac{x-4\sqrt{x}+4}{\sqrt{x}-1}< =0\)
=>\(\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}< =0\)
=>\(\sqrt{x}-1< 0\)
=>\(\sqrt{x}< 1\)
=>0<=x<1
Kết hợp ĐKXĐ, ta được: 0<=x<1