Question

tìm x\(\in\)Z

a, \((x^2+3)\left(x+7\right)=0\)

b,\(\left(x^2-1\right)\left(x^2-4\right)=0 \)

c,\(\left(5-x^2\right)\left(x^2-1\right)\ge0\)

Answer 1

\(\left(x^2+3\right)\left(x+7\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+3=0\\x+7=0\end{cases}}\)

\(Dễ,thấy:x^2+3>0\Rightarrow x+7=0\Rightarrow x=-7\)

\(\text{Vậy: x=(-7)}\)

Mấy câu khác tương tự nhé :v

Answer 2

\(\left(x^2+3\right)\left(x+7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+3=0\\x+7=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=-3\left(loại\right)\\x=0-7\end{cases}}\)

\(\Leftrightarrow x=-7\)

\(\left(x^2-1\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-1=0\\x^2-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=1\\x^2=4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm1\\x=\pm2\end{cases}}\)