《2025年一本计算题满分训练七年级数学全一册人教版》

1.已知关于x,y的二元一次方程组$\left\{\begin{array}{l} 2x+y= 4,\\ 4x+3y= 2m+2\end{array} \right. 的解满足x+y= 2$,求实数m的值.

答案： 1.解:根据题意,得$\left\{\begin{array}{l} 2x+y=4,\\ x+y=２,\end{array}\right.$解得$\left\{\begin{array}{l} x=2,\\ y=0,\end{array}\right.$所以$4×2+3×0=2m+2$,解得$m=3.$

2.【一题多解】若方程组$\left\{\begin{array}{l} 2x+3y= m,\\ 3x+5y= m+2\end{array} \right. 的解满足x-y= 5$,求m的值.

答案： 2.一题多解 解:方法1:直接求解法先解方程组,用含m的式子表示x,y,再代入$x-y=5$,得到关于m的方程.$\left\{\begin{array}{l} 2x+3y=m,①\\ ３x+5y=m+2.②\end{array}\right.$$②×2-①×3$,得$y=-m+4.$把$y=-m+4$代入①,得$2x-3m+12=m$,解得$x=2m-6.$
∴这个方程组的解是$\left\{\begin{array}{l} x=2m-6,\\ y=-m+4.\end{array}\right.$$\because x-y=5,$$\therefore 2m-6-(-m+4)=5$,解得$m=5.$方法2:组合法先求x,y的值,再求参.通过对原方程组的两个方程进行一定的整体运算消去参数m.$\left\{\begin{array}{l} 2x+3y=m,①\\ 3x+5y=m+2.②\end{array}\right.$$②-①$,得$x+2y=2.$$\therefore \left\{\begin{array}{l} x+2y=2,\\ x-y=5,\end{array}\right.$解得$\left\{\begin{array}{l} x=4,\\ y=-1.\end{array}\right.$将$\left\{\begin{array}{l} x=4,\\ y=-1\end{array}\right.$代入①,得$m=5.$

3.已知关于x,y的方程组$\left\{\begin{array}{l} 3x-2y= 11+8k,①\\ 2x+7y= -2k-7②\end{array} \right. 的解满足x+y= 2k$,求k的值.

答案： 3.方法:整体代入法解:$②+①$,得$5x+5y=6k+4,$$\therefore 5(x+y)=6k+4,\therefore x+y=\frac {6k+4}{5}.$
∵关于x,y的方程组$\left\{\begin{array}{l} 3x-2y=11+8k,①\\ 2x+7y=-2k-7②\end{array}\right.$的解满足$x+y=2k,$$\therefore 2k=\frac {6k+4}{5},$$\therefore k=1.$

4.【一题多解】已知关于x,y的方程组$\left\{\begin{array}{l} mx+ny= 9,\\ 5ny-3mx= 5\end{array} \right. 的解为\left\{\begin{array}{l} x= 3,\\ y= 2,\end{array} \right. 求m-2n$的值.

答案： 4.一题多解 解:$\because \left\{\begin{array}{l} x=3,\\ y=2\end{array}\right.$是方程组$\left\{\begin{array}{l} mx+ny=9,\\ 5ny-3mx=5\end{array}\right.$的解，$\therefore \left\{\begin{array}{l} 3m+2n=9,①\\ 10n-9m=5.②\end{array}\right.$方法1:$①×3+②$,得$16n=32$,解得$n=2.$把$n=2$代入①,得$m=\frac {5}{3}.$故$m-2n=\frac {5}{3}-4=-\frac {7}{3}.$方法2:由①,得$2n=9-3m$.③把③代入②,得$5×(9-3m)-9m=5$,解得$m=\frac {5}{3}.$把$m=\frac {5}{3}$代入③,得$2n=9-3×\frac {5}{3}=4.$故$m-2n=\frac {5}{3}-4=-\frac {7}{3}.$

5.【易错】当m取何整数时,关于x,y的二元一次方程组$\left\{\begin{array}{l} 2x-my= 6,①\\ x-3y= 0②\end{array} \right. $的解是正整数?

答案： 5.方法:枚举法解:由②,得$x=3y$.③把③代入①,得$6y-my=6$,解得$y=\frac {6}{6-m}.$
∵方程组的解是正整数,$\therefore 6-m=1$或$6-m=6$或$6-m=2$或$6-m=3,$$\therefore m=5$或$m=0$或$m=4$或$m=3.$故m取5或0或4或3时,方程组的解是正整数.