\(2x-\frac{1}{3}+2=0\)

\(2x-\frac{1}{3}=0-2\)

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

\(x=\frac{-5}{3}:2\)

\(x=\frac{-5}{3}.\frac{1}{2}\)

\(x=\frac{-5}{6}\)

~HT~

2x-1/3x+2=0

=>5/3x+2=0

<=>5/3x=-2

<=>x=-6/5

\(\frac{2x-1}{3x+2}=0\left(ĐK:x\ne\frac{-2}{3}\right)\)

\(\Rightarrow2x-1=0\)

\(\Rightarrow2x=1\)

\(\Rightarrow x=\frac{1}{2}\)

Quy đồng :

\(x=\frac{-22}{55};y=\frac{-15}{55}\)

Vậy \(x< y\)

~HT~

\(C=\frac{y^2-1}{y^2}\left(ĐK:y\ne0\right)\)

Ta có: \(C>0\Rightarrow\frac{y^2-1}{y^2}>0\)

Mà \(y^2>0\Rightarrow y^2-1>0\Rightarrow y^2>1\Rightarrow\orbr{\begin{cases}y< -1\\y>1\end{cases}}\)

Đề C có giá trị dương thì:

       \(\frac{y^2-1}{y^2}>0\left(y\in Q^+\right)\) ( Vì \(y^2\ge0;y\ne0\)

\(\rightarrow y^2-1\ge0\)

 Nhưng vì C lờn hơn 0 nên:

\(\rightarrow y^2-1>0\rightarrow y^2>1\rightarrow\orbr{\begin{cases}y>1\\y< -1\end{cases}}\)

Ta có :

\(\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)

\(\Rightarrow\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\)

\(\Rightarrow\left(x+2\right).\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)

\(\Rightarrow x+2=0\left(\text{do }\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ne0\right)\)

\(\Rightarrow x=-2\)

\(2^{225}=\left(2^3\right)^{75}=8^{75}\)

\(3^{150}=\left(3^2\right)^{75}=9^{75}\)

\(8^{75}< 9^{75}\Rightarrow2^{225}< 3^{150}\)