\(\frac{1}{2}+\frac{1}{2}\div x=\frac{1}{3}\)

\(\Leftrightarrow\)\(\frac{1}{2}\div x=\frac{1}{3}-\frac{1}{2}\)

\(\Leftrightarrow\)\(\frac{1}{2}\div x=\left(-\frac{1}{6}\right)\)

\(\Leftrightarrow\)\(x=\frac{1}{2}\div\left(-\frac{1}{6}\right)\)

\(\Leftrightarrow\)\(x=\frac{1}{2}\times\left(-6\right)\)

\(\Leftrightarrow\)\(x=-\frac{6}{2}=-3\)

Ta xét:

\(\left|3x+2\right|\ge0\)

\(\left|2x+5\right|\ge0\)

\(\left|x+1\right|\ge0\)

\(\Rightarrow\left|3x+2\right|+\left|2x+5\right|+\left|x+1\right|\ge0\)

\(\Rightarrow10x\ge0\)

\(\Rightarrow x\ge0\)

Do vậy: \(3x+2+2x+5+x+1=10x\)

\(\Rightarrow x\left(3+2+1\right)+\left(2+5+1\right)=10x\)

\(\Rightarrow6x+8=10x\)

\(\Rightarrow10x-6x=8\)

\(\Rightarrow4x=8\)

\(\Rightarrow x=2\)

Vì 2x = 3y

\(\Rightarrow\frac{x}{3}=\frac{y}{2}\)

ĐẶt \(\frac{x}{3}=\frac{y}{2}=k\)\(\Rightarrow\hept{\begin{cases}x=3k\\y=2k\end{cases}}\)

Mà xy = 54

=> 3k . 2k = 54

=> 6k² = 54

=> k² = 9

=> k = ± 3

\(\Rightarrow\hept{\begin{cases}x=9.\left(\pm3\right)=\pm27\\y=9.\left(\pm2\right)=\pm18\end{cases}}\)

Theo bài ra ta có : 

\(2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\)

Đặt : 

\(\frac{x}{3}=\frac{y}{2}=k\)

\(\Rightarrow\hept{\begin{cases}x=3k\\y=2k\end{cases}}\)

Lại có : 

\(3k.2k=54\)

\(6k^2=54\)

\(k^2=9\)

\(k=\pm3\)

\(\Rightarrow\hept{\begin{cases}x=3k\\y=2k\end{cases}}\Rightarrow\hept{\begin{cases}x=\pm9\\y=\pm6\end{cases}}\)

\(\sqrt{-4}\)

\(=\sqrt{-1.4}\)

\(=\sqrt{4.i^2}\)

\(=2i\)

Căn bậc hai của -4 là:

\(-\sqrt{4}=-2\)

Nếu mà nó nhưu thế này thì là sai nha

\(\sqrt{-4}\Rightarrow\varnothing\)

\(B=\left|x-3\right|+\left|x-5\right|+\left|x-7\right|\)

\(=\left|x-3\right|+\left|7-x\right|+\left|x-5\right|\)

\(\ge\left|x-3+7-x\right|+0\)

\(=4\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x-3\right)\left(7-x\right)\ge0\\x-5=0\end{cases}}\Leftrightarrow x=5\).