《2025年初中升高中衔接读本南京出版社数学》

4. 若$2x^2-5x+2＜0$,则$\sqrt{4x^2-4x+1}+2|x-2|= $______.

答案： 3

5. 解下列一元二次不等式：
(1)$3x^2+x-2\leq0$；
(2)$4x^2+1＞4x$；
(3)$(2x-1)^2+(x-1)^2\geq x$；
(4)$(x^2-x^3+2x)(x^2-1)(-x^2+4x-8)\leq0$.

答案： （1）原不等式可化为(3x-2)(x+1)≤0，所以原不等式的解集为$\left\lbrace x\mid -1\leqslant x\leqslant \dfrac{2}{3}\right\rbrace$．（2）原不等式可化为$4x^{2}-4x+1＞0$，$(2x-1)^{2}＞0$，所以原不等式的解集为$\left\lbrace x\mid x\neq \dfrac{1}{2}\right\rbrace$．（3）原不等式可化为$5x^{2}-7x+2\geqslant 0$，$(5x-2)(x-1)\geqslant 0$，则$\left\{\begin{array}{l} 5x-2\geqslant 0,\\ x-1\geqslant 0\end{array}\right. $或$\left\{\begin{array}{l} 5x-2\leqslant 0,\\ x-1\leqslant 0.\end{array}\right. $解得$\left\{\begin{array}{l} x\geqslant \dfrac {2}{5},\\ x\geqslant 1\end{array}\right. $或$\left\{\begin{array}{l} x\leqslant \dfrac {2}{5},\\ x\leqslant 1.\end{array}\right. $所以$x\leqslant \dfrac {2}{5}$或$x\geqslant 1$．所以原不等式的解集为$\left\lbrace x\mid x\leqslant \dfrac{2}{5}或x\geqslant 1\right\rbrace$．（4）原不等式可化为$(x^{3}-x^{2}-2x)(x^{2}-1)(x^{2}-4x+8)\leqslant 0$，化简得$x(x-2)(x-1)(x+1)^{2}[(x-2)^{2}+4]\leqslant 0$，所以$x(x-2)(x-1)\leqslant 0$，①$\left\{\begin{array}{l} x\leqslant 0,\\ x-2\geqslant 0,\\ x-1\geqslant 0,\end{array}\right. $此不等式组无解；②$\left\{\begin{array}{l} x\geqslant 0,\\ x-2\geqslant 0,\\ x-1\leqslant 0,\end{array}\right. $此不等式组无解；③$\left\{\begin{array}{l} x\leqslant 0,\\ x-1\leqslant 0,\\ x-2\leqslant 0,\end{array}\right. $解得$x\leqslant 0$；④$\left\{\begin{array}{l} x\geqslant 0,\\ x-2\leqslant 0,\\ x-1\geqslant 0,\end{array}\right. $解得$1\leqslant x\leqslant 2$．综上，原不等式的解集为$\left\lbrace x\mid x\leqslant 0或1\leqslant x\leqslant 2\right\rbrace$．