Question

8．若關(guān)于x的不等式組$\left\{\begin{array}{l}{x-3≥a}\\{b-x≥-2}\end{array}\right.$的解集為x=2b+a+1，求使得關(guān)于x的不等式組$\left\{\begin{array}{l}{x-k≥a}\\{x+k＜b}\end{array}\right.$無解的k的取值范圍．

Answer 1

分析 先求得不等式組$\left\{\begin{array}{l}{x-3≥a}\\{b-x≥-2}\end{array}\right.$的解集，根據(jù)題意求得a、b的值，即可得出不等式組$\left\{\begin{array}{l}{x-k≥0}\\{x+k＜1}\end{array}\right.$，求得不等式的解集，根據(jù)題意得出k≥1-k，即可求得k的取值．

解答 解：$\left\{\begin{array}{l}{x-3≥a①}\\{b-x≥-2②}\end{array}\right.$
由①得，x≥3+a，
由②得，x≤b+2，
∵關(guān)于x的不等式組$\left\{\begin{array}{l}{x-3≥a}\\{b-x≥-2}\end{array}\right.$的解集為x=2b+a+1，
∴$\left\{\begin{array}{l}{3+a=2b+a+1}\\{b+2=2b+a+1}\end{array}\right.$
解得$\left\{\begin{array}{l}{a=0}\\{b=1}\end{array}\right.$
∴$\left\{\begin{array}{l}{x-k≥0}\\{x+k＜1}\end{array}\right.$，
∴解x-k≥0得，x≥k，
解x+k＜1得，x＜1-k，
∵關(guān)于x的不等式組$\left\{\begin{array}{l}{x-k≥a}\\{x+k＜b}\end{array}\right.$無解，
∴k≥1-k，
解得k≥$\frac{1}{2}$．

點(diǎn)評 主要考查了一元一次不等式組的解定義，解此類題是要先用字母a，b表示出不等式組的解集，然后再根據(jù)已知解集，對應(yīng)得到相等或不等關(guān)系，解關(guān)于字母a，b的方程組或不等式即可求解．