Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 = \(\dfrac{x^2+3}{-x}-x+2\)
= \(\dfrac{-x^2-3}{x}-x+2=\dfrac{-x^2-3-x^2+2x}{x}=\dfrac{-2x^2+2x-3}{x}\)
\(a.A=\sqrt{\dfrac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}=\sqrt{\dfrac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}=\left|\dfrac{x^2+3}{x}\right|+\left|x-2\right|=\left|x+\dfrac{3}{x}\right|+\left|x-2\right|\left(x\ne0\right)\)
\(b.\) Để : \(A\in Z\Leftrightarrow\left(x+\dfrac{3}{x}\right)\in Z\Leftrightarrow x\in\left\{\pm1;\pm3\right\}\)
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
=>\(\frac{x^2+3}{x}+\left|x-2\right|\)
=\(\frac{x^2+3}{x}+2-x\)
=\(\frac{x^2+3}{x}+\frac{2x-x^2}{x}=\frac{2x+3}{x}\)
\(M=\dfrac{4x-8\sqrt{x}+8x}{x-4}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{12x-8\sqrt{x}}{x-4}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}\left(3\sqrt{x}-2\right)}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{3-\sqrt{x}}=\dfrac{4x\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(3-\sqrt{x}\right)}\)
ĐKXĐ: x>0, x≠0;x≠4
\(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)=\left(\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)\sqrt{x}}{\left(\sqrt{x}+2\right)\left(3-\sqrt{x}\right)}=\dfrac{4x}{\sqrt{x}-3}\)
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: \(P_{min}=36\Leftrightarrow x=36\)