\(...=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\) \(\left(x\ge0;x\ne4\right)\)

\(=\dfrac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

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

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(...=\left(\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-4\dfrac{\sqrt{6}}{2}\right)\left(\dfrac{3\sqrt{6}}{3}-2\sqrt{3}-\sqrt{6}\right)\) (Trục căn)

\(=\left(\dfrac{9\sqrt{6}}{6}+\dfrac{4\sqrt{6}}{6}-\dfrac{12\sqrt{6}}{6}\right)\left(\sqrt{6}-2\sqrt{3}-\sqrt{6}\right)\)

\(=\dfrac{\sqrt{6}}{6}.\left(-2\sqrt{3}\right)=\dfrac{1}{\sqrt{6}}.\left(-2\sqrt{3}\right)=-\sqrt{2}\)

\(\widehat{mOy}+\widehat{xOn}=70^o\)

mà \(\widehat{mOy}=\widehat{xOn}\left(đối.đỉnh\right)\)

\(\Rightarrow\widehat{mOy}=70^o:2=35^o\)

mà \(\widehat{mOy}+\widehat{mOx}=180^o\left(kề.bù\right)\)

\(\Rightarrow\widehat{mOx}=180^o-35^o=145^o\)

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

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

\(ab=180;BCNN\left(a;b\right)=60\)

\(ab=BCNN\left(a;b\right).UCLN\left(a;b\right)\)

\(\Rightarrow UCLN\left(a;b\right)=\dfrac{180}{60}=3\)

Đặt \(a=3x;b=3y\) \(\left(x;y>0\right)\)

\(\Rightarrow xy=\dfrac{180}{3.3}=20\)

\(\Rightarrow\left(x;y\right)=\left(1;20\right);\left(2;10\right);\left(4;5\right);\left(5;4\right);\left(10;2\right);\left(20;1\right)\)

\(\Rightarrow\left(a;b\right)=\left(3;60\right);\left(6;30\right);\left(12;15\right);\left(15;12\right);\left(30;6\right);\left(60;3\right)\)

Giả sử \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}>\sqrt{\dfrac{\sqrt{x}+1}{\sqrt{x}-1}}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}>\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}>0\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left[\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-1\right]>0\left(1\right)\)

Với \(x>1\Rightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{x}+1}{\sqrt{x}-1}>1>0\\\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-1>0\end{matrix}\right.\)

\(\Rightarrow\left(1\right)\) luôn luôn đúng \(\forall x>1\)

Vậy \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}>\sqrt{\dfrac{\sqrt{x}+1}{\sqrt{x}-1}}\)

Gọi \(x\in Z^+\) là số lượng đĩa CD cần bán để hòa vốn

Theo đề bài ta có :

\(30000x=120000000+15000x\)

\(\Leftrightarrow15000x=120000000\)

\(\Leftrightarrow x=8000\)

Vậy công ty cần bán \(8000\) đĩa CD mỗi tháng để hòa vốn.

\(\overrightarrow{AC}+\overrightarrow{CD}-\overrightarrow{EC}=\overrightarrow{AE}-\overrightarrow{DB}+\overrightarrow{CB}\left(1\right)\)

\(\Leftrightarrow\overrightarrow{AD}+\overrightarrow{DB}=\overrightarrow{AE}+\overrightarrow{EC}+\overrightarrow{CB}\)

\(\Leftrightarrow\overrightarrow{AB}=\overrightarrow{AC}+\overrightarrow{CB}\)

\(\Leftrightarrow\overrightarrow{AB}=\overrightarrow{AB}\left(đúng\right)\)

Vậy \(\left(1\right)đpcm\)

\(y=x^2-9\left|x\right|=\left[{}\begin{matrix}x^2-9x\left(x\ge0\right)\\x^2+9x\left(x< 0\right)\end{matrix}\right.\)

Theo đồ thị ta có hàm số \(y=x^2-9x\) đạt cực tiểu tại \(x=\dfrac{9}{2}\)

\(\Rightarrow y_{min}=\dfrac{81}{4}-9.\dfrac{9}{2}=-\dfrac{81}{4}\)

Đồ thị hàm số \(y=x^2+9x\) sẽ đối xứng qua trục \(Oy\)

\(\Rightarrow\) Để hàm số \(y=x^2-9\left|x\right|\) cắt đường thẳng \(y=m\) tại \(4\) điểm phân biệt

\(\Rightarrow-\dfrac{81}{4}< m< 0\)