Question

[Kim Chi Truong]

tìm x :

a, 3x . ( x +1/5 ) = 0

b, ( x -2) ( 3+x) = 0

c, (x-3) (x +9) > 0

ai onl làm ơn giải cái nhoa!

Answer 1

[Vũ Thị Như Quỳnh]

mình không biets khó quá

Answer 2

[o0o I am a studious person o0o]

\(3x\left(x+\frac{1}{5}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x=0\\x+\frac{1}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}}\)

\(\left(x-2\right)\left(3+x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\3+x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

\(\left(x-3\right)\left(x+9\right)>0\)

Th1 : \(\hept{\begin{cases}x-3>0\\x+9>0\end{cases}\Rightarrow\hept{\begin{cases}x>3\\x>-9\end{cases}\Rightarrow}x>3}\)

Th2 : \(\hept{\begin{cases}x-3< 0\\x+9< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 3\\x< -9\end{cases}\Rightarrow}x< -9}\)