Question

Giúp mình với!!! Bài này về bất đẳng thức Cauchy ak!!!

1. Cho x > 1 hãy tìm GTNN của:

P=\(\dfrac{x}{\sqrt{x}-1}\)

2. Tìm GTNN của:

B=\(\dfrac{x+15}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}+3}\)

\(\left(x\ge0;x\ne1,x\ne9\right)\)

Answer 1

`1. P = x/(sqrt x-1)`

`= (x-1+1)/(sqrtx-1)`

`= ((sqrt x+1)(sqrt x-1))/(sqrt x-1) +1/(sqrt x-1)`

`= sqrt x+1 + 1/(sqrt x-1)`

`= sqrtx-1 + 1/(sqrt x-1) + 2 >= 4`.

ĐTXR `<=> (sqrtx-1)^2 = 1`.

`<=> x =4` hoặc `x = 0 ( ktm)`.

Vậy Min A `= 4 <=> x= 4`.

Answer 2

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

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

Với x>1\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-1>0\\\dfrac{1}{\sqrt{x}-1}>0\end{matrix}\right.\)

Áp dụng BĐT AM-GM cho 2 số dương \(\sqrt{x}-1\) và \(\dfrac{1}{\sqrt{x}-1}\), ta có:

\(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\ge2\sqrt{(\sqrt{x}-1).\dfrac{1}{\sqrt{x}-1}}=2\)

\(\Rightarrow P\ge2+2=4\)

Dấu = xảy ra khi: \(\sqrt{x}-1=1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)

KL;....

Answer 3

2:

\(B=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}\)

\(=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\)

=>\(B>=2\cdot\sqrt{25}-6=4\)

Dấu = xảy ra khi (căn x+3)^2=25

=>căn x+3=5

=>căn x=2

=>x=4

Answer 4

2) \(B=\dfrac{x+15}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}+3}=\dfrac{x+16}{\sqrt{x}+3}\)

\(=\dfrac{x+3\sqrt{x}-3\sqrt{x}-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\)

\(\ge2\sqrt{(\sqrt{x}+3).\dfrac{25}{\sqrt{x}+3}}-6=2.5-6=4\) (BĐT Cô si cho 2 số \(\sqrt{x}+3\) và \(\dfrac{25}{\sqrt{x}+3}\)dương )

Dấu bằng xảy ra khi \(\left(\sqrt{x}+3\right)^2=25\Leftrightarrow\sqrt{x}+3=5\)

\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\) (t/m)

KL:....