2. (2025·合肥期中)对有理数a,b定义了一种新的运算,叫“乘加法”,记作“$a★b$”,并按照此运算写出了一些式子:$2★3= 5,(-2)★3= -5,2★(-3)= -5,(-2)★(-3)= 5,(-2)★(-2)= 4,2★(-2)= -4,2★0= 2,(-2)★0= 2……$
(1)根据以上式子特点将“乘加法”法则补充完整:
同号得
正
,异号得
负
,并把绝对值
相加
;一个数与0相“乘加”等于
这个数的绝对值
.
(2)根据法则计算:$(-4)★2= $
-6
;$(-\frac {1}{3})★(-3)= $
$3\frac{1}{3}$
.
(3)若括号的作用与它在有理数运算中的作用相同,请计算:$[6★(-1)]★[(-1)★\frac {1}{2}]$.
[6★(-1)]★[(-1)★$\frac{1}{2}$]=[-(|6|+|-1|)]★[-(|-1|+|$\frac{1}{2}$|)]=(-7)★(-1$\frac{1}{2}$)=|-7|+|-1$\frac{1}{2}$|=8$\frac{1}{2}$.
答案：(1)正负相加这个数的绝对值
(2)-6 3$\frac{1}{3}$解析:(-4)★2=-(|-4|+|2|)=-6,(-$\frac{1}{3}$)★(-3)=| -$\frac{1}{3}$|+| -3|=3$\frac{1}{3}$.
(3)[6★(-1)]★[(-1)★$\frac{1}{2}$]=[-(|6|+|-1|)]★[-(|-1|+|$\frac{1}{2}$|)]=(-7)★(-1$\frac{1}{2}$)=|-7|+|-1$\frac{1}{2}$|=8$\frac{1}{2}$.
3. (2025·亳州期中)【初步感知】已知有理数p(p不为0和1),我们把$1-\frac {1}{p}$称为p的倒数差,如:3的倒数差是$1-\frac {1}{3}= \frac {2}{3},-\frac {1}{2}$的倒数差是1-(-2)= 3.
【理解运用】若p= 4,p_{1}是p的倒数差,p_{2}是p_{1}的倒数差,p_{3}是p_{2}的倒数差,…,以此类推.
(1)分别求出p_{1},p_{2},p_{3}的值.
$p_{1}=\frac{3}{4}，$$p_{2}=-\frac{1}{3}，$p_{3}=4
(2)p_{7}+p_{8}+p_{9}的值为____.
$\frac{53}{12}$
【拓展提升】
(3)设有理数a,b,c(都不为0和1),将一个数组(a,b,c)中的数分别按照材料中“倒数差”的定义作变换,第1次变换后得到数组(a_{1},b_{1},c_{1}),第2次变换后得到数组(a_{2},b_{2},c_{2}),…,第n次变换后得到数组(a_{n},b_{n},c_{n}).若数组(a,b,c)确定为$(-3,\frac {1}{2},-1),$求a_{1}+b_{1}+c_{1}+a_{2}+b_{2}+c_{2}+... +a_{12}+b_{12}+c_{12}的值.
$\frac{19}{3}$
答案：(1)因为p = 4,p₁是p的倒数差,p₂是p₁的倒数差,p₃是p₂的倒数差,所以结合题意可得p₁ = 1 - $\frac{1}{p}$ = 1 - $\frac{1}{4}$ = $\frac{3}{4}$,p₂ = 1 - $\frac{1}{p₁}$ = 1 - $\frac{1}{\frac{3}{4}}$ = -$\frac{1}{3}$,p₃ = 1 - $\frac{1}{p₂}$ = 1 - $\frac{1}{-\frac{1}{3}}$ = 4.
(2)$\frac{53}{12}$解析:由p₁,p₂,p₃的值,可得p₁,p₂,p₃三个值为一组循环数,后面每三个数一组进行重复,所以p₁ + p₂ + p₃和p₄ + p₅ + p₆及p₇ + p₈ + p₉得数均相同,即p₁ = p₄ = p₇ = $\frac{3}{4}$,p₂ = p₅ = p₈ = -$\frac{1}{3}$,p₃ = p₆ = p₉ = 4.所以p₇ + p₈ + p₉ = $\frac{3}{4}$ - $\frac{1}{3}$ + 4 = $\frac{53}{12}$.
(3)因为(a,b,c)确定为(-3,$\frac{1}{2}$,-1),所以第1次变换后a₁ = 1 - $\frac{1}{a}$ = 1 - $\frac{1}{-3}$ = $\frac{4}{3}$,b₁ = 1 - $\frac{1}{b}$ = 1 - $\frac{1}{\frac{1}{2}}$ = -1,c₁ = 1 - $\frac{1}{c}$ = 1 - $\frac{1}{-1}$ = 2.所以第2次变换后a₂ = 1 - $\frac{1}{a₁}$ = 1 - $\frac{1}{\frac{4}{3}}$ = $\frac{1}{4}$,b₂ = 1 - $\frac{1}{b₁}$ = 1 - $\frac{1}{-1}$ = 2,c₂ = 1 - $\frac{1}{c₁}$ = 1 - $\frac{1}{2}$ = $\frac{1}{2}$.所以第3次变换后a₃ = 1 - $\frac{1}{a₂}$ = 1 - $\frac{1}{\frac{1}{4}}$ = -3,b₃ = 1 - $\frac{1}{b₂}$ = 1 - $\frac{1}{2}$ = $\frac{1}{2}$,c₃ = 1 - $\frac{1}{c₂}$ = 1 - $\frac{1}{\frac{1}{2}}$ = -1.所以同理可得a₁ = a₄ = a₇ = a₁₀ = $\frac{4}{3}$,a₂ = a₅ = a₈ = a₁₁ = $\frac{1}{4}$,a₃ = a₆ = a₉ = a₁₂ = -3,b₁ = b₄ = b₇ = b₁₀ = -1,b₂ = b₅ = b₈ = b₁₁ = 2,b₃ = b₆ = b₉ = b₁₂ = $\frac{1}{2}$,c₁ = c₄ = c₇ = c₁₀ = 2,c₂ = c₅ = c₈ = c₁₁ = $\frac{1}{2}$,c₃ = c₆ = c₉ = c₁₂ = -1.所以a₁ + b₁ + c₁ = a₄ + b₄ + c₄ = a₇ + b₇ + c₇ = a₁₀ + b₁₀ + c₁₀ = $\frac{4}{3}$ - 1 + 2 = $\frac{7}{3}$,a₂ + b₂ + c₂ = a₅ + b₅ + c₅ = a₈ + b₈ + c₈ = a₁₁ + b₁₁ + c₁₁ = $\frac{1}{4}$ + 2 + $\frac{1}{2}$ = $\frac{11}{4}$,a₃ + b₃ + c₃ = a₆ + b₆ + c₆ = a₉ + b₉ + c₉ = a₁₂ + b₁₂ + c₁₂ = -3 + $\frac{1}{2}$ - 1 = -$\frac{7}{2}$.所以a₁ + b₁ + c₁ + a₂ + b₂ + c₂ +... + a₁₂ + b₁₂ + c₁₂ = 4(a₁ + b₁ + c₁) + 4(a₂ + b₂ + c₂) + 4(a₃ + b₃ + c₃) = 4×$\frac{7}{3}$ + 4×$\frac{11}{4}$ + 4×(-$\frac{7}{2}$) = $\frac{19}{3}$.
