Question

[SONGOKU DRAGON BALL]

a,(x-4).(x+3)=0

b,(x^2+16).(x^2-16)=0

c,(x^2+10).(x-3)<0

d,(-x^2-9).(2-x)<0

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

Lam nhung cau cac ban biet nhe

Ai lam dung minh tick tick tick cho

Answer 1

[Kiệt Nguyễn]

a) \(\left(x-4\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)

Vậy \(x\in\left\{-3;4\right\}\)

b)\(\left(x^2+16\right)\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+16=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{-16}\\x=\sqrt{16}=4\end{cases}}\)

Vậy \(x=4\)

Answer 2

[Đặng Tú Phương]

\(\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)

\(\left(x^2+16\right)\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+16=0\\x^2-16=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=-16\left(loại\right)\\x^2=16\end{cases}}\Rightarrow x=\left(\pm4\right)^2\)

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

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

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