Bài 1:Với x>0 cho các biểu thức:
\(A=\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1};B=\dfrac{\sqrt{x}}{x+\sqrt{x}};P=\dfrac{A}{B}\)
a,Rút gọn và tính giá trị của P khi x=4
b,Tìm các giá trị thực của x để A\(\le\)3B
c,So sánh B với 1
d,Tìm x thỏa mãn \(P\sqrt{x}+\left(2\sqrt{5}-1\right)\sqrt{x}=3x-2\sqrt{x-4}+3\)
\(•A=\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\\ •B=\dfrac{\sqrt{x}}{x+\sqrt{x}}=\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
a) \(P=\dfrac{A}{B}=\dfrac{\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}}{\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b) thay x=4 vào P, ta được:
\(\dfrac{4+\sqrt{4}+1}{\sqrt{4}}=\dfrac{4+2+1}{2}=\dfrac{7}{2}\)
c)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}+1}\)
ta có: \(\sqrt{x}\ge0\)
\(\Leftrightarrow\sqrt{x}+1\ge1\\ \Rightarrow\dfrac{1}{\sqrt{x}+1}\le\dfrac{1}{1}=1\)
vậy B<1
d) DKXĐ : x >=0
\(P\sqrt{x}+\left(2\sqrt{5}-1\right)\sqrt{x}=3x-2\sqrt{x-4}+3\\ \Leftrightarrow x+\sqrt{x}+1+2\sqrt{5x}-\sqrt{x}=3x-2\sqrt{x-4}+3\\x-\sqrt{x-4}-\sqrt{5x}+1=0\\ x+1=\sqrt{x-4}+\sqrt{5x}\\ \left(x+1\right)^2=\left(\sqrt{x-4}+\sqrt{5x}\right)^2\\ x^2+2x+1=x-4+5x+2\sqrt{\left(x-4\right)5x}\\ x^2-4x+5=2\sqrt{5x^2-20x}\\ \left(x^2-4x+5\right)^2=\left(2\sqrt{5x^2-20x}\right)^2\\ x^4-8x^3+6x^2+40x+25=0\\ \left(x-5\right)^3\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
vậy x=5
