a) Ta có: (x-1)(x-4)>0
\(\Leftrightarrow\left[{}\begin{matrix}x-4>0\\x-1< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\)
b) Ta có: (x-6)(x-7)<0
\(\Leftrightarrow\left\{{}\begin{matrix}x-6>0\\x-7< 0\end{matrix}\right.\Leftrightarrow6< x< 7\)
c) Ta có: \(\left(x-1\right)\left(x-2\right)\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x-2\le0\end{matrix}\right.\Leftrightarrow1\le x\le2\)
d) Ta có: \(\left(x-2\right)\left(x-\dfrac{2}{3}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2\ge0\\x-\dfrac{2}{3}\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le\dfrac{2}{3}\end{matrix}\right.\)
a) Ta có: (x-1)(x-4)>0
⇔[x−4>0x−1<0⇔[x>4x<1⇔[x−4>0x−1<0⇔[x>4x<1
b) Ta có: (x-6)(x-7)<0
⇔{x−6>0x−7<0⇔6<x<7⇔{x−6>0x−7<0⇔6<x<7
c) Ta có: (x−1)(x−2)≤0(x−1)(x−2)≤0
⇔{x−1≥0x−2≤0⇔1≤x≤2⇔{x−1≥0x−2≤0⇔1≤x≤2
d) Ta có: (x−2)(x−23)≥0(x−2)(x−23)≥0
⇔⎡⎣x−2≥0x−23≤0⇔⎡⎣x≥2x≤23
