a) Để \(A=0\)
\(\Leftrightarrow x^2+5\times x=0\)
\(\Leftrightarrow x\times\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy để \(A=0\) thì \(x=0\) hoặc \(x=-5\)
b) Để \(A< 0\)
\(\Leftrightarrow x^2+5\times x< 0\)
\(\Leftrightarrow x\times\left(x+5\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x+5< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x>-5\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x< -5\end{matrix}\right.\left(vôlí\right)\end{matrix}\right.\Leftrightarrow-5< x< 0\) \(\Leftrightarrow x\in\left\{-4;-3;-2;-1\right\}\)
Vậy để \(A< 0\) thì \(x\in\left\{-4;-3;-2;-1\right\}\)
c) Để \(A=-6\)
\(\Leftrightarrow x^2+5\times x=-6\)
\(\Leftrightarrow x\times\left(x+5\right)=-6\)
\(\Leftrightarrow x\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Ta có bảng sau:
| \(x\) | \(-6\) | \(-3\) | \(-2\) | \(-1\) | \(1\) | \(2\) | \(3\) | \(6\) |
| \(x+5\) | \(-1\) | \(2\) | \(3\) | \(4\) | \(6\) | \(7\) | \(8\) | \(11\) |
| \(x\times\left(x+5\right)\) | \(6\) | \(-6\) | \(-6\) | \(-4\) | \(6\) | \(14\) | \(24\) | \(66\) |
Mà \(x\times\left(x+5\right)=-6\)
\(\Rightarrow x\in\left\{-3;-2\right\}\)
Vậy để \(A=-6\) thì \(x\in\left\{-3;-2\right\}\)
Ta có:
a) \(x^2+5x=0\)
\(\Rightarrow6x=0\)
\(\Rightarrow x=0\)
b) \(x^2+5x< 0\)
\(\Rightarrow6x< 0\Rightarrow x< 0\)
c) \(x^2+5x=-6\)
\(\Rightarrow6x=-6\)
\(\Rightarrow x=-1\)
Cho mk sửa lại nhé!
a) \(x^2+5x=0\)
\(\Rightarrow x\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
b) \(x^2+5x< 0\)
\(\Rightarrow x\left(x+5\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}x>0;x+5< 0\\x< 0;x+5>0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>0;x< -5\\x< 0;x>-5\end{matrix}\right.\)
\(\Rightarrow-5< x< 0\)
c) \(x^2+5x=-6\)
\(\Rightarrow x\left(5+x\right)=-6\)
\(\Rightarrow x\) và \(5+x\inƯ\left(6\right)\)
mà \(Ư\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(\Rightarrow x\) và \(5+x\in\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Xét các t/h sau:
_ Nếu \(x=1\) thì \(5+x=-6\)
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