Question

[Nguyễn Đình An]

(x-3/2)*(2x+1)>0

Answer 1

[Trí Tiên]

\(\left(x-\frac{3}{2}\right)\left(2x+1\right)>0\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{2}>0\\2x+1>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>\frac{3}{2}\\x>-\frac{1}{2}\end{cases}}\Leftrightarrow x>\frac{3}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{2}< 0\\2x+1< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< \frac{3}{2}\\x< -\frac{1}{2}\end{cases}}\Leftrightarrow x< -\frac{1}{2}\)

Answer 2

[Tạ Đức Hoàng Anh]

C₁: Ta có: \(\left(x-\frac{3}{2}\right).\left(2x+1\right)>0\)

+ \(\hept{\begin{cases}x-\frac{3}{2}>0\\2x+1>0\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}x>\frac{3}{2}\\x>-\frac{1}{2}\end{cases}}\)\(\Rightarrow\)\(x>\frac{3}{2}\)

+ \(\hept{\begin{cases}x-\frac{3}{2}< 0\\2x+1< 0\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}x< \frac{3}{2}\\x< -\frac{1}{2}\end{cases}}\)\(\Rightarrow\)\(x< -\frac{1}{2}\)

 Vậy \(x>\frac{3}{2}\)hoặc  \(x< -\frac{1}{2}\)

C₂: Ta có bảng xét dấu:

Vì \(\left(x-\frac{3}{2}\right).\left(2x+1\right)>0\)\(\Rightarrow\)\(x>\frac{3}{2}\)hoặc  \(x< -\frac{1}{2}\)

Vậy \(x>\frac{3}{2}\)hoặc  \(x< -\frac{1}{2}\)

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