Question

tìm x,biết:( 2x+3) .(3x-5)<0

Answer 1

(2x + 3). (3x - 5) < 0

Xét 2 trường hợp:

+ \(\hept{\begin{cases}2x+3>0\\3x-5< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-\frac{3}{2}\\x< \frac{5}{3}\end{cases}}\Rightarrow-\frac{3}{2}< x< \frac{5}{3}}\)   (đúng)

+  \(\hept{\begin{cases}2x+3< 0\\3x-5>0\end{cases}\Rightarrow\hept{\begin{cases}x< -\frac{3}{2}\\x>\frac{5}{3}\end{cases}}}\) (vô lí)

                             Vậy -3/2 < x < 5/3 

Answer 2

\(\left(2x+3\right)\left(3x-5\right)< 0\)

TH1 : \(\hept{\begin{cases}2x+3< 0\\3x-5>0\end{cases}=>\hept{\begin{cases}x< \frac{-3}{2}\\x>\frac{5}{3}\end{cases}}}\)

TH2 : \(\hept{\begin{cases}2x+3>0\\3x-5< 0\end{cases}=>\hept{\begin{cases}x>\frac{-3}{2}\\x< \frac{5}{3}\end{cases}}}\)

Ủng hộ mik nha

Answer 3

\(\left(2x+3\right)\left(3x-5\right)< 0\)

\(\orbr{\begin{cases}\hept{\begin{cases}2x+3>0\\3x-5< 0\end{cases}}\\\hept{\begin{cases}2x+3< 0\\3x-5>0\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}x>-1,5\\x< \frac{5}{3}\end{cases}}\\\hept{\begin{cases}x< -1,5\\x>\frac{5}{3}\end{cases}\left(vn\right)}\end{cases}}\)

\(\Leftrightarrow-1,5< x< \frac{5}{3}\)