1.  x≥1 <=> \(\frac{1}{x}\le1\Leftrightarrow\frac{1}{x}+1\le2\Leftrightarrow A\le2\Rightarrow MaxA=2\Leftrightarrow x=1\)

2. Áp dụng bđt cosi cho x>0. ta có: \(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\Leftrightarrow P\ge2\Rightarrow MinP=2\Leftrightarrow x=\frac{1}{x}\Leftrightarrow x=1\)

3: \(A=\frac{x^2+x+4}{x+1}=\frac{\left(x^2+2x+1\right)-\left(x+1\right)+4}{x+1}=x+1-1+\frac{4}{x+1}\)

áp dụng cosi cho 2 số dương ta có: \(x+1+\frac{4}{x+1}\ge2\sqrt{x+1.\frac{4}{x+1}}=2\Leftrightarrow A+1\ge2\Rightarrow A\ge3\Rightarrow MinA=3\Leftrightarrow x+1=\frac{4}{x+1}\Leftrightarrow x=1\)

x=1 nhe nhap minh di ma ket ban voi minh nhe

1. Ta có : \(A=\frac{\left(x+4\right)\left(x+9\right)}{x}=\frac{x^2+13x+36}{x}=x+\frac{36}{x}+13\)

Áp dụng bđt Cauchy : \(x+\frac{36}{x}\ge2\sqrt{x.\frac{36}{x}}=12\)

\(\Rightarrow A\ge25\)

Vậy Min A = 25 \(\Leftrightarrow\begin{cases}x>0\\x=\frac{36}{x}\end{cases}\) \(\Leftrightarrow x=6\)

2. \(B=\frac{\left(x+100\right)^2}{x}=\frac{x^2+200x+100^2}{x}=x+\frac{100^2}{x}+200\)

Áp dụng bđt Cauchy : \(x+\frac{100^2}{x}\ge2\sqrt{x.\frac{100^2}{x}}=200\)

\(\Rightarrow B\ge400\)

Vậy Min B = 400 \(\Leftrightarrow\begin{cases}x>0\\x=\frac{100^2}{x}\end{cases}\) \(\Leftrightarrow x=100\)

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2

Lời giải:
1. Áp dụng BĐT Cô-si

$G=\frac{x^2}{x-1}=\frac{(x^2-1)+1}{x-1}=x+1+\frac{1}{x-1}$

$=(x-1)+\frac{1}{x-1}+2$
$\geq 2\sqrt{(x-1).\frac{1}{x-1}}+2=2+2=4$ 

Vậy $G_{\min}=4$. Giá trị này đạt tại $x-1=\frac{1}{x-1}$

$\Leftrightarrow x=0$ hoặc $x=2$

2.

Áp dụng BĐT Cô-si:

$H=x+\frac{1}{x}=(\frac{x}{4}+\frac{1}{x})+\frac{3}{4}x$

$\geq 2\sqrt{\frac{x}{4}.\frac{1}{x}}+\frac{3}{4}x$
$=1+\frac{3}{4}x\geq 1+\frac{3}{4}.2=\frac{5}{2}$ (do $x\geq 2$)

Vậy $H_{\min}=\frac{5}{2}$. Giá trị này đạt tại $x=2$
 

3.

Áp dụng BĐT Cô-si:

$K=x^2+\frac{1}{x}=(\frac{x^2}{54}+\frac{1}{2x}+\frac{1}{2x})+\frac{53}{54}x^2$

$\geq 3\sqrt[3]{\frac{x^2}{54}.\frac{1}{2x}.\frac{1}{2x}}+\frac{53}{54}x^2$
$=\frac{1}{2}+\frac{53}{54}x^2\geq \frac{1}{2}+\frac{53}{54}.3^2=\frac{28}{3}$ (do $x\geq 3$)

Vậy $K_{\min}=\frac{28}{3}$ khi $x=3$

\(A=\frac{x^2}{x-1}=\frac{x^2-x+1}{x-1}+1\)1

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

\(=x-1+\frac{1}{x-1}+2\)

\(\ge2\sqrt{\frac{x-1}{x-1}}+2=4\)

Dấu '=' xảy ra khi (x-1)^2 = 1 -> x=2