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

\(A=x+13+\dfrac{36}{x}=\left(x+\dfrac{36}{x}\right)+13\ge2\sqrt{\dfrac{x.36}{x}}+13=12+13=25.\text{ Dấu }"="\text{ xảy ra khi: }x=\dfrac{36}{x}\text{ hay: }x=6\)

Ta có: \(A=\dfrac{x^2+13x+36}{x}=\dfrac{25x+x^2-12x+36}{x}\) \(=\dfrac{25x+\left(x-6\right)^2}{x}=25+\dfrac{\left(x-6\right)^2}{x}\ge25\)

Dấu bằng xảy ra \(\Leftrightarrow x=6\)

 Vậy \(Min_A=25\) khi \(x=6\)

Áp dụng BĐT Cauchy : 

\(\frac{\left(x+16\right)\left(x+9\right)}{x}=\frac{x^2+25x+144}{x}=x+\frac{144}{x}+25\ge2\sqrt{x.\frac{144}{x}}+25=49\)

Đẳng thức xảy ra khi \(x=12\)

Vậy ...............................................

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x=12 ban nhe

k mik nha

\(A=\frac{\left(x+4\right)\left(x+9\right)}{x}=\frac{x^2+13x+36}{x}\)

Dễ thấy \(x\ne0\) do \(x\) là mẫu nên ta có: 

\(A=x+13+\frac{36}{x}\). Do \(x>0\) nên ta áp dụng BĐT AM-GM:

\(x+\frac{36}{x}\ge2\sqrt{x\cdot\frac{36}{x}}=2\sqrt{36}=12\)

\(\Rightarrow A\ge13+12=25\)

Đẳng thức xảy ra khi \(x=\frac{36}{x}\Rightarrow x^2=36\Rightarrow x=6\left(x>0\right)\)

\(A=\frac{\left(x+4\right)\left(x+9\right)}{x}\left(x>0\right)\)

\(\Leftrightarrow Ax=x^2+13x+36\)

\(\Leftrightarrow x^2+x\left(13-A\right)+36=0\left(1\right)\)

Để pt(1) có nghiệm \(\Leftrightarrow\Delta\ge0\)

\(\Leftrightarrow\left(13-A\right)^2-4\cdot36\ge0\)

\(\Leftrightarrow\left(13-A\right)^2-12^2\ge0\)

\(\Leftrightarrow\left(13-A-12\right)\left(13-A+12\right)\ge0\)

\(\Leftrightarrow\left(1-A\right)\left(25-A\right)\ge0\)

\(\Leftrightarrow\left[\begin{matrix}A\le1\\A\ge25\end{matrix}\right.\)

Với A=25 ta tìm được x=6

Vậy GTNN của A là 25 khi x=6

a) \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)

\(P=\left[\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\left[\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{-4x-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(P=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}-3\right)}\)

\(P=\dfrac{-4\sqrt{x}\cdot\sqrt{x}}{-\left(\sqrt{x}-3\right)}\)

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

b) \(P=\dfrac{4x}{\sqrt{x}-3}\)

\(P=4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\)

Theo BĐT côsi ta có:

\(P\ge\sqrt{\dfrac{4\left(\sqrt{x}-3\right)\cdot36}{\sqrt{x}-3}}+24=36\)

Vậy: \(P_{min}=36\Leftrightarrow x=36\) 

\(A=\frac{\left(4x+1\right)\left(4+x\right)}{x}=\frac{4x^2+17x+4}{x}=4x+17+\frac{4}{x}=17+4\left(x+\frac{1}{x}\right)\)

Bạn tự chứng minh được \(x>0\Rightarrow x+\frac{1}{x}\ge2\)

Khi đó \(A\ge17+4.2=17+8=25\)

Dấu "=" xảy ra khi: \(x=\frac{1}{x}\Rightarrow x^2=1\Rightarrow x=1\left(x>0\right)\)

Vậy \(A_{min}=25\Leftrightarrow x=1\) (x > 0)