Question

Cho biểu thức \(A=\dfrac{x+3}{\sqrt{x}+3}\) và \(B=\left(\dfrac{x+3\sqrt{x}-2}{x-9}-\dfrac{1}{\sqrt{x}+3}\right).\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\) với \(x\ge0,x\ne9\)
a) Tính giá trị của A khi \(x=16\)
b) Rút gọn biểu thức B
c) Cho \(P=\dfrac{A}{B}\) . Tìm GTNN của P 

Answer 1

a: Thay x=16 vào A, ta được: \(A=\dfrac{16+3}{\sqrt{16}+3}=\dfrac{19}{4+3}=\dfrac{19}{7}\)

b: \(B=\left(\dfrac{x+3\sqrt{x}-2}{x-9}-\dfrac{1}{\sqrt{x}+3}\right)\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\left(\dfrac{x+3\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{1}{\sqrt{x}+3}\right)\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{x+3\sqrt{x}-2-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

c: \(P=A:B=\dfrac{x+3}{\sqrt{x}+3}:\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{x+3}{\sqrt{x}+1}=\dfrac{x-1+4}{\sqrt{x}+1}\)

\(=\sqrt{x}-1+\dfrac{4}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\)

=>\(P>=2\cdot\sqrt{\left(\sqrt{x}+1\right)\cdot\dfrac{4}{\sqrt{x}+1}}-2=2\cdot2-2=2\)

Dấu '=' xảy ra khi \(\left(\sqrt{x}+1\right)^2=4\)

=>\(\sqrt{x}+1=2\)

=>x=1(nhận)

Answer 2

a)

Thay x = 16 TMĐK vào A ta có:

\(A=\dfrac{16+3}{\sqrt{16}+3}=\dfrac{19}{4+3}=\dfrac{19}{7}\)

b) \(B=\left[\dfrac{x+3\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(B=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(B=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

c) \(P=\dfrac{A}{B}=\dfrac{x+3}{\sqrt{x}+3}\div\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{x+3}{\sqrt{x}+1}\)

\(P=\dfrac{x+3}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}+1}+\dfrac{4}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{4}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\)

Vì \(x\ge0\Rightarrow\sqrt{x}+1>0\Rightarrow\dfrac{4}{\sqrt{x}+1}>0\)

Áp dụng BĐT Cauchy ta có:

\(\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}\ge2\sqrt{\left(\sqrt{x}+1\right)\left(\dfrac{4}{\sqrt{x}+1}\right)}=2\sqrt{4}=4\)

\(\Rightarrow\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}+1=\dfrac{4}{\sqrt{x}+1}\Leftrightarrow\left(\sqrt{x}+1\right)^2=4\Leftrightarrow x=1\)

Vậy Min P = 2 \(\Leftrightarrow x=1\)