Question

(x-2).(x+14)=0

(x-2).(x+4)=0

(7-x).(x+19)=0

(x-3)(x-5)<0

2x^2-3=29

-6x-(-7)=25

46-(x-11)=-48

Answer 1

\(1.\left(x-2\right).\left(x+14\right)=0\)

<=>\(\orbr{\begin{cases}x-2=0\\x+14=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=2\\x=-14\end{cases}}\)

Answer 2

\(2.\left(x-2\right).\left(x+4\right)=0\)

<=>\(\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)

Answer 3

\(\left(x-2\right)\left(x+14\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+14=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-14\end{cases}}}\)

vậy x= 2 hoặc x=-14

\(\left(x-2\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)

vậy x=2 hoặc x=-4

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

nên x-3 và x-5 khác dấu

\(th1\Leftrightarrow\orbr{\begin{cases}x-3< 0\\x-5>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 3\\x>5\end{cases}}\Leftrightarrow5< x< 3}\left(vl\right)\)

\(th2\Leftrightarrow\orbr{\begin{cases}x-3>0\\x-5< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x>3\\x< 5\end{cases}}\Leftrightarrow3< x< 5}\left(tm\right)\)

vậy với 3<x<5 

thì x=4