《2025年课堂点睛八年级数学上册北师大版宁夏专版》

1. 已知 $\left\{\begin{array}{l}x = 2,\\ y = 1\end{array} \right.$ 是二元一次方程组 $\left\{\begin{array}{l}mx + ny = 8,\\ nx - my = 1\end{array} \right.$ 的解,则 $m + 3n$ 的值为 9.

答案： 9

2. 已知 $\left\{\begin{array}{l}x = -2,\\ y = 1\end{array} \right.$ 是二元一次方程组 $\left\{\begin{array}{l}mx + ny = 7,\\ ny - mx = 3\end{array} \right.$ 的解,求 $m,n$ 的值.

答案： 解:把$\left\{\begin{array}{l} x=-2,\\ y=1\end{array}\right. $代入$\left\{\begin{array}{l} mx+ny=7,\\ ny-mx=3,\end{array}\right. $得$\left\{\begin{array}{l} -2m+n=7,\\ 2m+n=3,\end{array}\right. $解得$\left\{\begin{array}{l} m=-1,\\ n=5.\end{array}\right. $

3. 如果关于 $x,y$ 的二元一次方程组 $\left\{\begin{array}{l}x - 3y = k + 2,\\ x - y = 4\end{array} \right.$ 的解适合方程 $3x + y = -8$,则 $k$ 的值为 12.

答案： 12

4. 若关于 $x,y$ 的方程组 $\left\{\begin{array}{l}6x - 2y = k + 3,①\\ x + 3y = 6k - 7②\end{array} \right.$ 的解满足 $x - y = 4$,则 $k$ 的值为 -2.

答案： -2

5. 已知方程组 $\left\{\begin{array}{l}x = y + 5,\\ x + y + m = 0\end{array} \right.$ 和方程组 $\left\{\begin{array}{l}2x - y = 5,\\ x + y + m = 0\end{array} \right.$ 有相同的解,则 $m$ 的值是 5.

答案： 5

6. 已知关于 $x,y$ 的方程组 $\left\{\begin{array}{l}3x - y = 5,\\ 4ax + 5by = -22\end{array} \right.$ 和 $\left\{\begin{array}{l}2x + 3y = -4,\\ ax - by = 8\end{array} \right.$ 有相同的解,求 $(-a)^b$ 值.

答案： 解:因为两组方程组有相同的解,所以原方程组可化为
(1)$\left\{\begin{array}{l} 3x-y=5,\\ 2x+3y=-4\end{array}\right. $和
(2)$\left\{\begin{array}{l} 4ax+5by=-22,\\ ax-by=8.\end{array}\right. $解方程组
(1)得$\left\{\begin{array}{l} x=1,\\ y=-2.\end{array}\right. $代入
(2)得$\left\{\begin{array}{l} 4a-10b=-22,\\ a+2b=8,\end{array}\right. $解得$\left\{\begin{array}{l} a=2,\\ b=3.\end{array}\right. $所以$(-a)^{b}=(-2)^{3}=-8.$

7. 已知方程组 $\left\{\begin{array}{l}ax + by = 3,\\ 5x - cy = 1,\end{array} \right.$ 甲正确地解得 $\left\{\begin{array}{l}x = 2,\\ y = 3,\end{array} \right.$ 而乙粗心地把 $c$ 看错了,得 $\left\{\begin{array}{l}x = 3,\\ y = 6,\end{array} \right.$ 则 $a + b + c$ 的值为 5.

答案： 5

8. 甲、乙两人共同解方程组 $\left\{\begin{array}{l}ax + 5y = 15,①\\ 4x - by = -2,②\end{array} \right.$ 由于甲看错了方程①中的 $a$,得到方程组的解为 $\left\{\begin{array}{l}x = -3,\\ y = -1,\end{array} \right.$ 乙看错了方程②中的 $b$,得到方程组的解为 $\left\{\begin{array}{l}x = 5,\\ y = 4,\end{array} \right.$ 试计算 $a^{2027} + \left(-\dfrac{1}{10}b\right)^{2026}$.

答案： 解:根据题意把$\left\{\begin{array}{l} x=-3,\\ y=-1\end{array}\right. $代入$4x-by=-2$得:$-12+b=-2$,解得:$b=10$.把$\left\{\begin{array}{l} x=5,\\ y=4\end{array}\right. $代入$ax+5y=15$得:$5a+20=15$,解得:$a=-1$,则$a^{2027}+(-\frac {1}{10}b)^{2026}=(-1)^{2027}+(-\frac {1}{10}×10)^{2026}=0.$