题目内容
11.已知关于x、y的二元一次方程组$\left\{\begin{array}{l}{3x+5y=6}\\{3x+ky=10}\end{array}\right.$给出下列结论:①当k=5时,此方程组无解;
②若此方程组的解也是方程6x+15y=16的解,则k=10;
③无论整数k取何值,此方程组一定无整数解(x、y均为整数),
其中正确的是①②③(填序号).
分析 ①将k=5代入,得到方程组得$\left\{\begin{array}{l}{3x+5y=6}\\{3x+5y=10}\end{array}\right.$,求解即可作出判断;
②解方程组$\left\{\begin{array}{l}{3x+5y=6}\\{3x+10y=10}\end{array}\right.$得$\left\{\begin{array}{l}{x=\frac{2}{3}}\\{y=\frac{4}{5}}\end{array}\right.$,把$\left\{\begin{array}{l}{x=\frac{2}{3}}\\{y=\frac{4}{5}}\end{array}\right.$代入6x+15y=16,即可做出判断;
③解方程组$\left\{\begin{array}{l}{3x+5y=6}\\{3x+ky=10}\end{array}\right.$得$\left\{\begin{array}{l}{x=2-\frac{20}{3k-15}}\\{y=\frac{4}{k-5}}\end{array}\right.$,根据k为整数即可作出判断.
解答 解:∵当k=5时,方程组为$\left\{\begin{array}{l}{3x+5y=6}\\{3x+5y=10}\end{array}\right.$,此时方程组无解;∴①正确;
∵解方程组$\left\{\begin{array}{l}{3x+5y=6}\\{3x+10y=10}\end{array}\right.$得$\left\{\begin{array}{l}{x=\frac{2}{3}}\\{y=\frac{4}{5}}\end{array}\right.$,把$\left\{\begin{array}{l}{x=\frac{2}{3}}\\{y=\frac{4}{5}}\end{array}\right.$代入6x+15y=16,方程左右两边相等,∴②正确;
∵解方程组$\left\{\begin{array}{l}{3x+5y=6}\\{3x+ky=10}\end{array}\right.$得$\left\{\begin{array}{l}{x=2-\frac{20}{3k-15}}\\{y=\frac{4}{k-5}}\end{array}\right.$,
又∵k为整数,
∴x、y不能均为整数,∴③正确.
故答案为:①②③.
点评 此题考查了二元一次方程组的解,方程组的解即为能使方程组中两方程成立的未知数的值.
| A. | 底与边不相等的等腰三角形 | B. | 等边三角形 | ||
| C. | 钝角三角形 | D. | 直角三角形 |
| A. | a2•a3=a6 | B. | a6÷a2=a3 | C. | (a2)3=a6 | D. | a6-a2=a4 |