Question

Cho biểu thức :

B = \((\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}):\frac{\sqrt{x}}{x+\sqrt{x}}\) (đk: x>0)

a, Rút gọn B

b, Tìm giá trị của B biết x=4

c, Tìm giá trị nhỏ nhất của B

Answer 1

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

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

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

\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)

Answer 2

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

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

\(x=4\Rightarrow B=\frac{4+2+1}{2}=\frac{7}{2}\)

\(B=\sqrt{x}+\frac{1}{\sqrt{x}}+1\ge2\sqrt{\frac{\sqrt{x}}{\sqrt{x}}}+1=3\)

\(B_{min}=3\) khi \(x=1\)

Answer 3

đk x>0

thay x=4(tm) vào B ta có

\(B=\frac{4+\sqrt{4}+1}{\sqrt{4}}=\frac{7}{2}\)

KL....