《2025年一遍过七年级数学上册沪科版》

1 下列方程组中,是三元一次方程组的是( )

A.$\left\{\begin{array}{l} x^{2}-y= 1,\\ y+z= 0,\\ xz= 2\end{array} \right.$
B.$\left\{\begin{array}{l} \frac {1}{x}+y= 1,\\ \frac {1}{y}+z= 2,\\ \frac {1}{z}+x= 6\end{array} \right.$
C.$\left\{\begin{array}{l} a+b+c+d= 1,\\ a-c= 2,\\ b-d= 3\end{array} \right.$
D.$\left\{\begin{array}{l} m+n= 18,\\ n+t= 12,\\ t+m= 0\end{array} \right.$

答案： D A选项,$x^{2}-y=1$与$xz=2$中,$x^{2}$,$xz$的次数为2,故A选项不是三元一次方程组;B选项,$\frac{1}{x}$,$\frac{1}{y}$,$\frac{1}{z}$不是整式,故B选项不是三元一次方程组;C选项,有四个未知数,故C选项不是三元一次方程组.

2 [2024 娄底月考]下列四组数值中,是方程组$\left\{\begin{array}{l} x+2y+z= 0,\\ 2x-y-z= 1,\\ 3x-y-z= 2\end{array} \right.$的解的是( )

A.$\left\{\begin{array}{l} x= 0,\\ y= 1,\\ z= -2\end{array} \right.$
B.$\left\{\begin{array}{l} x= 0,\\ y= 0,\\ z= 1\end{array} \right.$
C.$\left\{\begin{array}{l} x= 0,\\ y= -1,\\ z= 0\end{array} \right.$
D.$\left\{\begin{array}{l} x= 1,\\ y= -2,\\ z= 3\end{array} \right.$

答案： D

3 [2024 泉州期中]解方程组$\left\{\begin{array}{l} 3x-y+2z= 3,\\ 2x+y-5z= 11,\\ 7x+y-5z= 1,\end{array} \right.$若要使运算简便,消元的方法应选取( )

A.先消去 x
B.先消去 y
C.先消去 z
D.以上说法都不对

答案： B y的系数是1或-1,因此先消去y会使运算更简便.

4 一题多解 [2025 晋城期中]若$\left\{\begin{array}{l} x+y= 5,\\ y+z= 6,\\ x+z= 7,\end{array} \right.则x+y+z= $______。

答案： 9 通解 $\left\{\begin{array}{l} x+y=5,\enclose{circle}{1}\\ y+z=6,\enclose{circle}{2}\\ x+z=7,\enclose{circle}{3}\end{array}\right. $①+②+③,得$2(x+y+z)=18$,即$x+y+z=9.$
另解 $\left\{\begin{array}{l} x+y=5,\enclose{circle}{1}\\ y+z=6,\enclose{circle}{2}\\ x+z=7,\enclose{circle}{3}\end{array}\right. $①-②,得$x-z=-1$④,③+④,得$2x=6$,解得$x=3$.把$x=3$代入①,得$y=2$.把$x=3$代入③,得$z=4$,所以$x+y+z=3+2+4=9.$

5 解下列方程组：
(1)$\left\{\begin{array}{l} x-y+z= 0,\\ 4x+2y+z= 3,\\ 25x+5y+z= 60;\end{array} \right.$
(2)$\left\{\begin{array}{l} 3x+2y+5z= 2,\\ x-2y-z= 6,\\ 4x+2y-7z= 30;\end{array} \right.$
(3)$\left\{\begin{array}{l} 3x-y+z= 10,\\ x+2y-z= 6,\\ x+y+2z= 17;\end{array} \right.$
(4)$\left\{\begin{array}{l} 2x+4y-3z= 2,\\ 4x+7y+z= 3,\\ 8x+3y-2z= -5.\end{array} \right.$

答案： 解:
(1)$\left\{\begin{array}{l} x-y+z=0,\enclose{circle}{1}\\ 4x+2y+z=3,\enclose{circle}{2}\\ 25x+5y+z=60,\enclose{circle}{3}\end{array}\right. $
②-①,得$3x+3y=3$,即$x+y=1$,④
③-①,得$24x+6y=60$,即$4x+y=10$,⑤
⑤-④,得$3x=9$,解得$x=3.$
把$x=3$代入④,得$3+y=1$,解得$y=-2.$
把$x=3$,$y=-2$代入①,得$3+2+z=0$,解得$z=-5.$
所以$\left\{\begin{array}{l} x=3,\\ y=-2,\\ z=-5.\end{array}\right. $
(2)$\left\{\begin{array}{l} 3x+2y+5z=2,\enclose{circle}{1}\\ x-2y-z=6,\enclose{circle}{2}\\ 4x+2y-7z=30,\enclose{circle}{3}\end{array}\right. $
①+②,得$4x+4z=8$,即$x+z=2$.④
②+③,得$5x-8z=36$.⑤
④×5-⑤,得$13z=-26$,解得$z=-2.$
把$z=-2$代入④,得$x=4.$
把$x=4$,$z=-2$代入①,得$12+2y-10=2$,解得$y=0.$
所以$\left\{\begin{array}{l} x=4,\\ y=0,\\ z=-2.\end{array}\right. $
(3)$\left\{\begin{array}{l} 3x-y+z=10,\enclose{circle}{1}\\ x+2y-z=6,\enclose{circle}{2}\\ x+y+2z=17,\enclose{circle}{3}\end{array}\right. $
①+②,得$4x+y=16$,④
②×2+③,得$3x+5y=29$,⑤
④×5-⑤,得$17x=51$,解得$x=3.$
把$x=3$代入④,得$y=4.$
将$x=3$,$y=4$代入①,得$9-4+z=10$,解得$z=5,$
所以$\left\{\begin{array}{l} x=3,\\ y=4,\\ z=5.\end{array}\right. $
(4)$\left\{\begin{array}{l} 2x+4y-3z=2,\enclose{circle}{1}\\ 4x+7y+z=3,\enclose{circle}{2}\\ 8x+3y-2z=-5,\enclose{circle}{3}\end{array}\right. $
②-①×2,得$-y+7z=-1$.④
②×2-③,得$11y+4z=11$.⑤
由④得$y=7z+1$.⑥
把⑥代入⑤,得$11(7z+1)+4z=11$,解得$z=0.$
把$z=0$代入⑥,得$y=1.$
把$y=1$,$z=0$代入①,得$2x+4-0=2$,解得$x=-1.$
所以$\left\{\begin{array}{l} x=-1,\\ y=1,\\ z=0.\end{array}\right. $