Bài 5:
a: Thay x=64 vào A, ta được:
\(A=\dfrac{2+8}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)
b: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
c: \(\dfrac{A}{B}>\dfrac{3}{2}\)
=>\(\dfrac{2+\sqrt{x}}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{3}{2}>0\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)
=>\(\dfrac{2\left(\sqrt{x}+1\right)-3\sqrt{x}}{2\sqrt{x}}>0\)
=>\(-\sqrt{x}+2>0\)
=>\(-\sqrt{x}>-2\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
Kết hợp ĐKXĐ, ta được: 0<x<4
Bài 6:
a:Thay x=9 vào A, ta được:
\(A=\dfrac{3+4}{3-1}=\dfrac{7}{2}\)
b: \(B=\dfrac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\dfrac{2}{\sqrt{x}+3}\)
\(=\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{2}{\sqrt{x}+3}\)
\(=\dfrac{3\sqrt{x}+1-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}-1}\)
c: \(\dfrac{A}{B}>=\dfrac{x}{4}+5\)
=>\(\dfrac{\sqrt{x}+4}{\sqrt{x}-1}:\dfrac{1}{\sqrt{x}-1}>=\dfrac{x+20}{4}\)
=>\(\sqrt{x}+4>=\dfrac{x+20}{4}\)
=>\(4\sqrt{x}+16>=x+20\)
=>\(x+20-4\sqrt{x}-16< =0\)
=>\(x-4\sqrt{x}+4< =0\)
=>\(\left(\sqrt{x}-2\right)^2< =0\)
=>\(\sqrt{x}-2=0\)
=>x=4(nhận)
