Đáp án:
b, Ta có :
`B = (x - 2)/(x - 6) > 0`
<=> \(\left[ \begin{array}{l}x - 2 > 0 ; x - 6 > 0\\x - 2 < 0 ; x - 6 < 0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x > 2 ; x > 6\\x < 2 ; x < 6\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x > 6\\x < 2\end{array} \right.\)
a, Ta có :
`A = x^2 - x - 2/3x + 2/3`
`= x(x - 1) - 2/3(x - 1)`
`= (x - 2/3)(x - 1)`
Để `A > 0`
`<=> (x - 2/3)(x - 1) > 0`
<=> \(\left[ \begin{array}{l}x - 2/3 > 0 ; x - 1 > 0\\x - 2/3 < 0 ; x - 1 < 0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x > 2/3 ; x > 1\\x < 2/3 ; x < 1\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x > 1\\x < 2/3\end{array} \right.\)
Giải thích các bước giải:
