7.(2025沈阳)为了能用平方差公式计算$(a + 3b + c)(a - 3b + c)$,下列变形正确的是()
A.$[a - (3b + c)]^2$
B.$[(a - 3b) + c][(a - 3b) - c]$
C.$[(a + c) + 3b][(a + c) - 3b]$
D.$[a + (3b - c)]^2$

7. C

8.已知$(a - 2b - c)\cdot M = (a - c)^2 - 4b^2$,则M表示的代数式是______.

8.$a+2b-c$

9.已知$(a^2 + b^2 + 3)(a^2 + b^2 - 3) = 7$,$ab = 2$,则$(a + b)^2$的值为______.

9.8

10.(教材变式)计算:
(1)$(a + 2b - 3c)(a - 2b + 3c)$; (2)$(2m + n - 3p)(2m + 3p - n)$;
(3)$(3a - 2b + 5)^2$; (4)$(2a - b - 3)^2$.

10. 解:(1)原式$=[a+(2b-3c)][a-$
$(2b-3c)]$
$=a^{2}-(2b-3c)^{2}$
$=a^{2}-(4b^{2}-12bc+9c^{2})$
$=a^{2}-4b^{2}+12bc-9c^{2}$;
(2)原式$=[2m+(n-3p)][2m-$
$(n-3p)]$
$=(2m)^{2}-(n-3p)^{2}$
$=4m^{2}-(n^{2}-6np+9p^{2})$
$=4m^{2}-n^{2}+6np-9p^{2}$;
(3)原式$=(3a-2b)^{2}+10(3a-$
$2b)+25$
$=9a^{2}+4b^{2}-12ab+30a-$
$20b+25$;
(4)原式$=[2a-(b+3)]^{2}$
$=4a^{2}-4a(b+3)+(b+3)^{2}$
$=4a^{2}-4ab-12a+b^{2}+6b+9$.

11.计算:$(a + 2b - 3c)^2 - (a - 2b + 3c)^2$.

11. 解:原式$=(a^{2}+4b^{2}+9c^{2}+4ab-$
$6ac-12bc)-(a^{2}+4b^{2}$
$+9c^{2}-4ab+6ac-$
$12bc)$
$=a^{2}+4b^{2}+9c^{2}+4ab-$
$6ac-12bc-a^{2}-4b^{2}-$
$9c^{2}+4ab-6ac+12bc$
$=8ab-12ac$.

12.如图,将几个正方形与小长方形拼成一个边长为$(a + b + c)$的正方形.
(1)它的面积表示了一个恒等关系:$(a + b + c)^2 = $______;
(2)若实数$a$,$b$,$c满足a + b + c = 11$,$ab + bc + ac = 36$,求$a^2 + b^2 + c^2$的值.

12. 解:(1)$a^{2}+b^{2}+c^{2}+2ab+2ac+$
$2bc$;
(2)$\because (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+$
$2ab+2ac+2bc$,
$\therefore 11^{2}=a^{2}+b^{2}+c^{2}+2×36$,
$\therefore a^{2}+b^{2}+c^{2}=49$.