Question

với x>0,cho các biểu thức

A=\(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\) , B=\(\frac{\sqrt{x}}{x+\sqrt{x}}\)và P=\(\frac{A}{B}\)

a)rút gọn và tính giá trị của P khi x=4

b)tìm các giá trị thực của x để A≤3B

c)so sánh B với 1

d)tìm x thỏa mãn \(P\sqrt{x}+\left(2\sqrt{5}-1\right)\sqrt{x}=3x-2\sqrt{x-4}+3\)

Answer 1

\(x>0\Leftrightarrow\sqrt{x}>0\)

\(A=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{x+\sqrt{x}}\)

\(\Rightarrow P=\frac{A}{B}=A:B=\frac{x+\sqrt{x}+1}{x+\sqrt{x}}:\frac{\sqrt{x}}{x+\sqrt{x}}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)

\(x=4\Rightarrow P=\frac{7}{2}\)

\(A\le3B\Leftrightarrow\frac{x+\sqrt{x}+1}{x+\sqrt{x}}\le\frac{3x}{x+\sqrt{x}}\Leftrightarrow;\frac{\sqrt{x}+1-2x}{x+\sqrt{x}}\le0\left(x+\sqrt{x}>0\right)\Leftrightarrow\sqrt{x}-2x+1\le0\left(x+\sqrt{x}>0\right)\Leftrightarrow2x-\sqrt{x}-1\ge0\Leftrightarrow2x\ge1+\sqrt{x}\)\(+,0< x< 1\Rightarrow0^2< \left(\sqrt{x}\right)^2< 1^2\Rightarrow0< \sqrt{x}< 1\left(\sqrt{x}>0\right);\sqrt{x}-x=\sqrt{x}\left(1-\sqrt{x}\right)>0\left(0< \sqrt{x}< 1\right)\Rightarrow\sqrt{x}>x\Rightarrow2x< \sqrt{x}+x< \sqrt{x}+1\left(x< 1\right)\left(loai\right)\) \(+,x\ge1\Rightarrow\left(\sqrt{x}\right)^2\ge1^2\Rightarrow\sqrt{x}\ge1\left(\sqrt{x}>0\right)\Rightarrow x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)\ge0\left(\sqrt{x}\ge1\right)\Rightarrow x\ge\sqrt{x}\Rightarrow2x\ge\sqrt{x}+x\ge\sqrt{x}+1\left(x\ge1\right)\left(thoaman\right)\)

\(B=1-\frac{x}{x+\sqrt{x}}\Rightarrow B< 1\left(x;\sqrt{x}>0\right)\)