《2025年绩优学案八年级数学上册北师大版》

6. 若方程组$\begin{cases}3x - y = 4k - 5,\\2x + 6y = k\end{cases} 的解中x + y = 16$,则$k$等于( )

A.15
B.17
C.18
D.16

答案： B

7. 若$\begin{cases}x = 2,\\y = 1\end{cases} 是关于x,y的二元一次方程组\begin{cases}ax + by = 7,\\ax - by = 1\end{cases} $的一组解,则$a + b = $____。

答案： 5

8. 已知方程组$\begin{cases}2x + 5y = - 6,\\ax - by = - 4\end{cases} 与\begin{cases}3x - 5y = 16,\\bx + ay = - 8\end{cases} $方程组的解相同,求$(2a + b)^{2025}$的值。

答案： 解:由题意得$\left\{\begin{array}{l} 2x+5y=-6,\\ 3x-5y=16,\end{array}\right. $解得$\left\{\begin{array}{l} x=2,\\ y=-2,\end{array}\right. $再将$\left\{\begin{array}{l} x=2,\\ y=-2\end{array}\right. $代入另外两个方程中,组成关于a,b 的方程组$\left\{\begin{array}{l} 2a+2b=-4,\\ 2b-2a=-8\end{array}\right. $解得$\left\{\begin{array}{l} a=1,\\ b=-3\end{array}\right. $然后将$a=1,b=-3$代入$(2a+b)^{2025},(2×1-3)^{2025}=(-1)^{2025}=-1$。$(2a+b)^{2025}$的值为-1。

9. 解方程组$\begin{cases}ax + by = 2,\\cx - 7y = 8\end{cases} $时,甲同学正确解得$\begin{cases}x = 3,\\y = 2,\end{cases} 乙同学因把c写错而得到\begin{cases}x = 2,\\y = 1,\end{cases} 求7a + 7b + 3c$的值。

答案： 解:将$\left\{\begin{array}{l} x=3,\\ y=2\end{array}\right. $和$\left\{\begin{array}{l} x=2,\\ y=1\end{array}\right. $代入方程组中的第一个方程$ax+by=2$得关于a,b 的方程组$\left\{\begin{array}{l} 3a+2b=2,\\ 2a+b=2\end{array}\right. $解得$\left\{\begin{array}{l} a=2,\\ b=-2,\end{array}\right. $则$a+b=0,$再将$\left\{\begin{array}{l} x=3,\\ y=2\end{array}\right. $代入方程组中的第二个方程$cx-7y=8$得$3c-14=8$,所以$3c=22$。将$a+b=0$和$3c=22$代入$7a+7b+3c=7(a+b)+3c=0+22=22$。

10. 阅读探索：
解方程组$\begin{cases}(a - 1) + 2(b + 2) = 6,\\2(a - 1) + (b + 2) = 6。\end{cases} $
解：设$a - 1 = x,b + 2 = y原方程组可以化为\begin{cases}x + 2y = 6,\\2x + y = 6,\end{cases} 解得\begin{cases}x = 2,\\y = 2,\end{cases} $即：$\begin{cases}a - 1 = 2,\\b + 2 = 2,\end{cases} 所以\begin{cases}a = 3,\\b = 0,\end{cases} $此种解方程组的方法叫换元法。
(1)运用上述方法解下列方程组$\begin{cases}(\frac{a}{4} - 1) + 2(\frac{b}{5} + 2) = 10,\\2(\frac{a}{4} - 1) + (\frac{b}{5} + 2) = 11;\end{cases} $
(2)已知关于$x,y的方程组\begin{cases}a_1x + b_1y = c_1,\\a_2x + b_2y = c_2\end{cases} 的解为\begin{cases}x = 6,\\y = 7,\end{cases} 求关于m,n的方程组\begin{cases}a_1(m - 2) + b_1(n + 3) = c_1,\\a_2(m - 2) + b_2(n + 3) = c_2\end{cases} $的解。

答案： 解:
(1)设$\frac {a}{4}-1=x,\frac {b}{5}+2=y,$所以原方程组可变为$\left\{\begin{array}{l} x+2y=10,\\ 2x+y=11,\end{array}\right. $解这个方程组得$\left\{\begin{array}{l} x=4,\\ y=3,\end{array}\right. $即$\left\{\begin{array}{l} \frac {a}{4}-1=4,\\ \frac {b}{5}+2=3,\end{array}\right. $所以$\left\{\begin{array}{l} a=20,\\ b=5\end{array}\right. $。
(2)设$\left\{\begin{array}{l} m-2=x,\\ n+3=y,\end{array}\right. $可得$\left\{\begin{array}{l} m-2=6,\\ n+3=7,\end{array}\right. $解得$\left\{\begin{array}{l} m=8,\\ n=4\end{array}\right. $。