đk: x khác 0

A = \(\sqrt{\dfrac{x^4-6x^2+9+12x^2}{x^2}}+\sqrt{x^2+4x+4-8x}\)

= \(\sqrt{\dfrac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}\)

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

= \(\dfrac{x^2+3}{\left|x\right|}+\left|x-2\right|\)

TH1: x \(\ge2\)

A = \(\dfrac{x^2+3}{x}+x-2\)

= \(\dfrac{x^2+3+x^2-2x}{x}=\dfrac{2x^2-2x+3}{x}\)

TH2: \(0< x< 2\)

A = \(\dfrac{x^2+3}{x}-x+2\)

= \(\dfrac{x^2+3-x^2+2x}{x}=\dfrac{2x+3}{x}\)

TH3: x < 0

A = \(\dfrac{x^2+3}{-x}-x+2\)

= \(\dfrac{-x^2-3}{x}-x+2=\dfrac{-x^2-3-x^2+2x}{x}=\dfrac{-2x^2+2x-3}{x}\)

Điều kiện: x khác 0

\(=\sqrt{\frac{x^4-6x^2+9+12x^2}{x^2}}+\sqrt{x^2+4x+4-8x}\)

\(=\sqrt{\frac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}=\sqrt{\frac{\left(x^2+3\right)^2}{x^2}}+\sqrt{\left(x-2\right)^2}\)

\(=\left|\frac{x^2+3}{x}\right|+\left|x-2\right|=\frac{x^2+3}{\left|x\right|}+\left|x-2\right|\)

\(\sqrt{\frac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\)

=\(\frac{\sqrt{x^4-6x+9+12x^2}}{\sqrt{x^2}}+\sqrt{x^2+4x+4-8x}\)

=\(\frac{\sqrt{x^4+6x+9}}{x}+\sqrt{x^2-4x+4}\)

=\(\frac{\sqrt{\left(x^2+3\right)^2}}{x}+\sqrt{\left(x-2\right)^2}\)

=\(\frac{\sqrt{\left(x^2+3\right)^2}}{x}+\left|x-2\right|\)

=\(\frac{x^2+3}{x}+\left|x-2\right|\)

TH1: x\(\ge\)2 =>|x-2|=x-2

=>\(\frac{x^2+3}{x}+\left|x-2\right|\)

=\(\frac{x^2+3}{x}+x-2\)

=\(\frac{x^2+3}{x}+\frac{x^2-2x}{x}=\frac{2x^2-2x+3}{x}\)

TH2:x\(\le\)2 =>|x-2|=2-x

=>\(\frac{x^2+3}{x}+\left|x-2\right|\)

=\(\frac{x^2+3}{x}+2-x\)

=\(\frac{x^2+3}{x}+\frac{2x-x^2}{x}=\frac{2x+3}{x}\)

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

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

b: \(m\left(\sqrt{x}-3\right)\cdot B>x+1\)

=>\(-4xm\left(3\sqrt{x}-2\right)>\left(\sqrt{x}+2\right)\cdot\left(x+1\right)\)

=>\(-12m\cdot x\sqrt{x}+8xm>x\sqrt{x}+2x+\sqrt{x}+2\)

=>\(x\sqrt{x}\left(-12m-1\right)+x\left(8m-2\right)-\sqrt{x}-2>0\)

Để BPT luôn đúng thì m<-0,3

`(\sqrt(3x^2-12x+12)-x+2)/(x-2)`

`=(\sqrt(3(x^2-4x+4))-(x-2))/(x-2)`

`=(\sqrt(3(x-2)^2)) -(x-2))/(x-2)`

`=(\sqrt3. (x-2) - (x-2))/(x-2)`

`=( (\sqrt3-1) (x-2))/(x-2)`

`=\sqrt3-1`

`=>` C.

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

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

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

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

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

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: \(P_{min}=36\Leftrightarrow x=36\) 

rút gọn R ?

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

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

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

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

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

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

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

\(=\dfrac{8x-8\sqrt{x}+8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}-2-\sqrt{x}+2}\)

\(=\dfrac{16x-8\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{2\sqrt{x}}{\sqrt{x}}\)

\(=\dfrac{2\left(16-8\sqrt{x}\right)}{\sqrt{x}+2}\)

\(=\dfrac{32-16\sqrt{x}}{\sqrt{x}+2}\)