1. 求同时满足$a+b+c=9,2a-b+c=5$，且$b≥ c≥0$的$a$的最大整数值及最小整数值.

1. 依题意，$\begin{cases} a+b+c=9, &① \\ 2a-b+c=5, &② \end{cases}$ ①-②,得$-a+2b=4$,即$a=2b-4$ ③,将③代入①,得$(2b-4)+b+c=9$,即$3b+c=13$,$\therefore c=13-3b$ ④.$\because b≥ c≥0$,即$c≥0$,$\therefore 13-3b≥0$,即$b≤ \frac{13}{3}$.$\because b≥ c≥0$,即$b≥ c$,$\therefore b≥ 13-3b$,即$4b≥13$,$\therefore b≥ \frac{13}{4}$,$\therefore \frac{13}{4}≤ b≤ \frac{13}{3}$.又$a=2b-4$,$\therefore 2× \frac{13}{4}-4≤ a≤ 2× \frac{13}{3}-4$,即$\frac{5}{2}≤ a≤ \frac{14}{3}$,$\therefore a$的最大整数值是4,最小整数值是3.

2. 已知由小到大的10个正整数$a_1,a_2,a_3,···,a_{10}$的和是2000，求$a_5$的最大值及此时$a_{10}$的值.

2. 设$a_1,a_2,a_3,a_4$为1,2,3,4,$\therefore a_5+a_6+a_7+\dots+a_{10}=2000-(1+2+3+4)=1990$.$\because a_6≥ a_5+1$,$a_7≥ a_5+2$,$a_8≥ a_5+3$,$a_9≥ a_5+4$,$a_{10}≥ a_5+5$,$\therefore a_5+a_6+a_7+\dots+a_{10}≥ 6a_5+15$,$\therefore 6a_5+15≤ 1990$,解得$a_5≤ 329\frac{1}{6}$,$\therefore a_5$最大能取329,那么可得$a_6,a_7,a_8,a_9$只能分别取330,331,332,333,那么$a_{10}$只能取335.但若$a_1,a_2,a_3,a_4$取1,2,3,5,那么$a_5,\dots,a_9$同样可以取上述值,此时$a_{10}=334$.综上,此时$a_{10}$的值为334或335.

3. 若$2a+b=12$，其中$a≥0$，$b≥0$，又$P=3a+2b$.试确定$P$的最小值和最大值.

3. $\because 2a+b=12$,$\therefore b=12-2a$,$\therefore P=3a+2b=3a+2(12-2a)=24-a$.$\because a≥0$,$b≥0$,即$a≥0$且$12-2a≥0$,$\therefore 0≤ a≤6$,$\therefore 18≤ 24-a≤24$,$\therefore P$的最小值为18,最大值为24.

4. 已知$x,y,z$为非负实数，且满足$x+y+z=30,3x+y-z=50$.求$u=5x+4y+2z$的取值范围.

4. 将已知的两个等式联立成方程组$\begin{cases} x+y+z=30, &① \\ 3x+y-z=50, &② \end{cases}$ ①+②得$4x+2y=80$,$y=40-2x$.将$y=40-2x$代入①,可解得$z=x-10$,$\therefore u=5x+4y+2z=5x+4(40-2x)+2(x-10)=-x+140$.$\because x,y,z$均为非负实数,$\therefore \begin{cases} x≥0, \\ 40-2x≥0, \\ x-10≥0, \end{cases}$解得$10≤ x≤20$,$\therefore 120≤ -x+140≤130$,即$120≤ u≤130$.