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 = \(\dfrac{x^2+3}{-x}-x+2\)

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

Các bạn giúp mình giải bài này nha

tìm GTLN,GTNN của biểu thức

\(\sqrt{x+3}\)+\(\sqrt{5-x}\)

a)

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

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

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

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

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

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

B = \(\frac{x^2+3+x^2-2x}{x}\)

B = \(\frac{2x^2-2x+3}{x}\)

B = \(2\left(x-1\right)+3\)

b) để B nguyên thì B  \(\ge\)0

<=> 2 ( x - 1 ) + 3  \(\ge\)0

<=> 2x - 2 + 3  \(\ge\)0

<=> 2x + 1  \(\ge\)0

<=>   x  \(\ge\)\(\frac{-1}{2}\)

k mình nhé bạn

ĐKXĐ: \(x\ne0\)

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

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

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

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

Ta có bảng xét dấu:

Với \(x< 0,y=\frac{x^2+3}{-x}+2-x=\frac{2x^2-2x+3}{-x}\)

Với \(0< x\le2,y=\frac{x^2+3}{x}+2-x=\frac{2x+3}{x}\)

Với \(x>2,y=\frac{x^2+3}{x}+x-2=\frac{2x^2-2x+3}{x}\)

- Ta thấy ngay, với cả ba trường hợp thì \(y\in Z\Leftrightarrow x\in U\left(3\right)=\left\{-3;-1;1;3\right\}\)

Bài 1:

\(\sqrt{24+8\sqrt{15}-\sqrt{9-4\sqrt{5}}}\)

\(=\sqrt{24+8\sqrt{15}-\left(\sqrt{5}-2\right)}\)

\(=\sqrt{26+8\sqrt{15}-\sqrt{5}}\)

Bài 2:

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

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

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

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

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

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

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

wrecking ball sai rồi \(\frac{\sqrt{\left(x^2+3\right)^2}}{x}=\frac{trituyetdoix^2+3}{x}\) bằng