`(x^2 - 7x + 6)(x - 7) < 0`
Chia hai trường hợp
\(\left\{ \begin{array}{l}x^2 - 7x + 6 < 0\\x - 7 > 0\end{array} \right.\)
`<=>` \(\left\{ \begin{array}{l}x < 1\\x > 6\\x < 7\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x < 1\\6 < x < 7\end{array} \right.\)
\(\left\{ \begin{array}{l}x^2 - 7x + 6 > 0\\x - 7 < 0\end{array} \right.\)
`<=>` \(\left\{ \begin{array}{l}1 < x < 6\\x > 7\end{array} \right.\)
`=>` loại
`(x + 7)/(-2x^2 + 3x - 7) > 0`
`<=> (x + 7)/(-2(x^2 - \dfrac{3}{2}x - \dfrac{7}{2})) > 0`
`=> x + 7 < 0`
`<=> x < -7`
`(4x^2 + 3x - 1)(-3x^2 - x + 4) > 0`
Dễ dàng nhận thấy `4x^2 + 3x - 1 > 0`
`=> -3x^2 - x + 4 > 0`
`=> -4/3 < x < 1`
`2/(x - 1) <= 5/(2x - 1)` `(ĐK: x ne 1; x ne 1/2)`
`<=> 2/(x - 1) - 5/(2x - 1) <= 0`
`<=> (4x - 2 - 5x + 5)/((x - 1)(2x - 1)) <= 0`
`<=> (3 - x)/(2x^2 - 3x + 1) <= 0`
Xét 2 trường hợp
\(\left\{ \begin{array}{l}3 - x <= 0\\2x^2 - 3x + 1 > 0\end{array} \right.\)
`<=>` \(\left\{ \begin{array}{l}x >= 3\\x < \dfrac{1}{2}\\x > 1\end{array} \right.\)
`=> x >= 3`
\(\left\{ \begin{array}{l}3 - x >= 0\\2x^2 - 3x + 1 < 0\end{array} \right.\)
`<=>` \(\left\{ \begin{array}{l}x <= 3\\\dfrac{1}{2} < x < 1\end{array} \right.\)
`=> 1/2 < x < 1`
Bài 5:
`(2x^2 - x)/(1 - 2x) >= -1` `(ĐK: x ne 1/2)`
`<=> (x(2x - 1))/(1 - 2x) >= -1`
`<=> (x(1 - 2x))/(1 - 2x) <= 1`
`<=> x <= 1`
