《2025年时习之暑假衔接七年级数学人教版》

4.已知实数$a,b满足方程组\begin{cases}3a + 2b = 7,\\2a + 3b = 8,\end{cases} 则(a + b)(a - b)$的值是 ( )

A.3
B.-3
C.4
D.-4

答案： B

5.若$a + 4b = 6,3a + 2b = 4$,则$a - b$的值为____.

答案： -1

6.若关于$x,y的二元一次方程组\begin{cases}x + y = 5k,\\x - 2y = - k\end{cases} 的解也是二元一次方程2x - y = 8$的解,则$k$的值为____.

答案： 2

7.在一家水果店,小明买了$1kg$苹果,$4kg$西瓜,$2kg$橙子,共付$27.2$元;小惠买了$2kg$苹果,$6kg$西瓜,$2kg$橙子,共付$32.4$元.则买$1kg西瓜和1kg$橙子需付____元.

答案： 11

8.用整体思想解方程组:
(1)$\begin{cases}x + 2(x + 2y) = 4,\\x + 2y = 1;\end{cases} $
(2)$\begin{cases}2x - 3y - 2 = 0,\frac{2x - 3y + 5}{7} + 2y = 9;\end{cases} $
(3)$\begin{cases}\frac{x - 3y}{3} - \frac{1}{3} = 1,\\2x - \frac{x - 3y}{4} = 5.\end{cases} $

答案：
(1)$\left\{\begin{array}{l} x=2,\\ y=-\frac {1}{2}.\end{array}\right. $
(2)$\left\{\begin{array}{l} x=7,\\ y=4.\end{array}\right. $
(3)$\left\{\begin{array}{l} x=3,\\ y=-\frac {1}{3}.\end{array}\right. $

9.若方程组$\begin{cases}m + 5n = 1 + 2a,\\6m + 2n = 3 - a\end{cases} 的解满足m + n = 3$,求$a$的值.

答案： $\left\{\begin{array}{l} m+5n=1+2a\enclose{circle} {1},\\ 6m+2n=3-a\enclose{circle} {2},\end{array}\right. $①+②,得$7(m+n)=a+4,\therefore m+n=\frac {a+4}{7}.\because m+n=3,\therefore \frac {a+4}{7}=3$,解得$a=17.\therefore a$的值为 17.

10.阅读材料:善于思考的小明同学在解方程组$\begin{cases}3(m + 5) - 2(n + 3) = - 1,\\3(m + 5) + 2(n + 3) = 7\end{cases} $时,采用了一种“整体换元”的解法:
解:把$m + 5,n + 3$看成一个整体,设$m + 5 = x,n + 3 = y$,
原方程组可化为$\begin{cases}3x - 2y = - 1,\\3x + 2y = 7,\end{cases} $
解得$\begin{cases}x = 1,\\y = 2.\end{cases} $ $\therefore \begin{cases}m + 5 = 1,\\n + 3 = 2.\end{cases} $
$\therefore原方程组的解为\begin{cases}m = - 4,\\n = - 1.\end{cases} $
请仿照小明同学的方法,用“整体换元”法解方程组$\begin{cases}3(x + y) - 4(x - y) = 5,\frac{x + y}{2} + \frac{x - y}{6} = 0.\end{cases} $

答案： 设$x+y=m,x-y=n$,原方程组可化为$\left\{\begin{array}{l} 3m-4n=5,\\ \frac {m}{2}+\frac {n}{6}=0,\end{array}\right. $即$\left\{\begin{array}{l} 3m-4n=5\enclose{circle} {1},\\ 3m+n=0\enclose{circle} {2},\end{array}\right. $②-①,得$n=-1$.把$n=-1$代入②,得$m=\frac {1}{3},\therefore \left\{\begin{array}{l} n=-1,\\ m=\frac {1}{3}.\end{array}\right. \therefore \left\{\begin{array}{l} x+y=\frac {1}{3},\\ x-y=-1,\end{array}\right. $解得$\left\{\begin{array}{l} x=-\frac {1}{3},\\ y=\frac {2}{3}.\end{array}\right. $