Chubby Bear Cô chứ?

Đk: \(x\ge-2\)

PT \(\Leftrightarrow\) \(x\left(12-4\sqrt{x+2}\right)+3x^2-20x-7=0\)

\(\Leftrightarrow x.\dfrac{144-16\left(x+2\right)}{12+4\sqrt{x+2}}+\left(x-7\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\dfrac{-4x\left(x-7\right)}{3+\sqrt{x+2}}+\left(x-7\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\\left(3+\sqrt{x+2}\right)\left(3x+1\right)=4x\end{matrix}\right.\)

Đặt \(u=\sqrt{x+2}\Leftrightarrow x=u^2-2\left(u\ge0\right)\)

PT (2) \(\Leftrightarrow\left(3+u\right)\left(3u^2-5\right)=4\left(u^2-2\right)\)

\(\Leftrightarrow9u^2-15+3u^3-5u=4u^2-8\)

\(\Leftrightarrow3u^3+5u^2-5u-7=0\) \(\Leftrightarrow u=\dfrac{-1+\sqrt{22}}{3}\)

\(\Leftrightarrow x=\dfrac{5-2\sqrt{22}}{9}\)

Vậy...

6B.
a)\(\sqrt[3]{4-2x}\ge4\Leftrightarrow4-2x\ge64\)
\(\Leftrightarrow2x\le-60\Leftrightarrow x\le-30\)
Vậy...

b) \(\sqrt[3]{-x^3-3x^2+6x-10}< -x-1\)

\(\Leftrightarrow-x^3-3x^2+6x-10< -\left(x+1\right)^3\)

\(\Leftrightarrow-x^3-3x^2+6x-10< -x^3-3x^2-3x-1\)

\(\Leftrightarrow9x< 9\Leftrightarrow x< 1\)
Vậy...
7A.
a) \(\sqrt[3]{2x+1}=3\Leftrightarrow2x+1=27\Leftrightarrow x=13\)
Vậy...
b) \(\sqrt[3]{5+x}-x=5\)

\(\Leftrightarrow5+x=\left(5+x\right)^3\) \(\Leftrightarrow\left[{}\begin{matrix}5+x=0\\\left(5+x\right)^2=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-4\\x=-6\end{matrix}\right.\)

Vậy...

7B.

a) PT \(\Leftrightarrow2-3x=-8\Leftrightarrow x=\dfrac{10}{3}\)

b) PT \(\Leftrightarrow x-1=\left(x-1\right)^3\) 

\(\Leftrightarrow\left(x-1\right)\left[1-\left(x-1\right)^2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\x=2\end{matrix}\right.\)

Vậy...

a. ĐK: \(x\ne\pm2\)
\(M=\left[\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x+7}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{3-x+x-2}{x-2}\)

\(=\dfrac{x^2+2x-\left(x^2-2x+x-2\right)-2x-7}{\left(x-2\right)\left(x+2\right)}.\left(x-2\right)\)

\(=\dfrac{x-5}{x+2}\)

b. \(\dfrac{x-5}{x+2}< 1\Leftrightarrow\dfrac{x-5}{x+2}-1< 0\)

\(\Leftrightarrow\dfrac{-7}{x+2}< 0\Leftrightarrow x+2>0\)

\(\Leftrightarrow x>-2\)
Vậy \(x>-2,x\ne2\)

\(2\sqrt{3}\left(\sqrt{12}-\sqrt{27}-3\sqrt{2}\right)+6\sqrt{6}\)

\(=2\sqrt{3}\left(\sqrt{3.2^2}-\sqrt{3.3^2}-3\sqrt{2}\right)+6\sqrt{6}\)

\(=2\sqrt{3}\left(2\sqrt{3}-3\sqrt{3}-3\sqrt{2}\right)+6\sqrt{6}\)

\(=\left(2\sqrt{3}\right)^2-6.\left(\sqrt{3}\right)^2-6\sqrt{6}+6\sqrt{6}\)

\(=-6\)