2. (2025·合肥期中)對(duì)有理數(shù)a,b定義了一種新的運(yùn)算,叫“乘加法”,記作“$a★b$”,并按照此運(yùn)算寫出了一些式子:$2★3= 5,(-2)★3= -5,2★(-3)= -5,(-2)★(-3)= 5,(-2)★(-2)= 4,2★(-2)= -4,2★0= 2,(-2)★0= 2……$
(1)根據(jù)以上式子特點(diǎn)將“乘加法”法則補(bǔ)充完整:
同號(hào)得
正
,異號(hào)得
負(fù)
,并把絕對(duì)值
相加
;一個(gè)數(shù)與0相“乘加”等于
這個(gè)數(shù)的絕對(duì)值
.
(2)根據(jù)法則計(jì)算:$(-4)★2= $
-6
;$(-\frac {1}{3})★(-3)= $
$3\frac{1}{3}$
.
(3)若括號(hào)的作用與它在有理數(shù)運(yùn)算中的作用相同,請(qǐng)計(jì)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)正負(fù)相加這個(gè)數(shù)的絕對(duì)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】已知有理數(shù)p(p不為0和1),我們把$1-\frac {1}{p}$稱為p的倒數(shù)差,如:3的倒數(shù)差是$1-\frac {1}{3}= \frac {2}{3},-\frac {1}{2}$的倒數(shù)差是1-(-2)= 3.
【理解運(yùn)用】若p= 4,p_{1}是p的倒數(shù)差,p_{2}是p_{1}的倒數(shù)差,p_{3}是p_{2}的倒數(shù)差,…,以此類推.
(1)分別求出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}的值為____.
$\frac{53}{12}$
【拓展提升】
(3)設(shè)有理數(shù)a,b,c(都不為0和1),將一個(gè)數(shù)組(a,b,c)中的數(shù)分別按照材料中“倒數(shù)差”的定義作變換,第1次變換后得到數(shù)組(a_{1},b_{1},c_{1}),第2次變換后得到數(shù)組(a_{2},b_{2},c_{2}),…,第n次變換后得到數(shù)組(a_{n},b_{n},c_{n}).若數(shù)組(a,b,c)確定為$(-3,\frac {1}{2},-1),$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?yàn)閜 = 4,p?是p的倒數(shù)差,p?是p?的倒數(shù)差,p?是p?的倒數(shù)差,所以結(jié)合題意可得p? = 1 - $\frac{1}{p}$ = 1 - $\frac{1}{4}$ = $\frac{3}{4}$,p? = 1 - $\frac{1}{p?}$ = 1 - $\frac{1}{\frac{3}{4}}$ = -$\frac{1}{3}$,p? = 1 - $\frac{1}{p?}$ = 1 - $\frac{1}{-\frac{1}{3}}$ = 4.
(2)$\frac{53}{12}$解析:由p?,p?,p?的值,可得p?,p?,p?三個(gè)值為一組循環(huán)數(shù),后面每三個(gè)數(shù)一組進(jìn)行重復(fù),所以p? + p? + p?和p? + p? + p?及p? + p? + p?得數(shù)均相同,即p? = p? = p? = $\frac{3}{4}$,p? = p? = p? = -$\frac{1}{3}$,p? = p? = p? = 4.所以p? + p? + p? = $\frac{3}{4}$ - $\frac{1}{3}$ + 4 = $\frac{53}{12}$.
(3)因?yàn)?a,b,c)確定為(-3,$\frac{1}{2}$,-1),所以第1次變換后a? = 1 - $\frac{1}{a}$ = 1 - $\frac{1}{-3}$ = $\frac{4}{3}$,b? = 1 - $\frac{1}$ = 1 - $\frac{1}{\frac{1}{2}}$ = -1,c? = 1 - $\frac{1}{c}$ = 1 - $\frac{1}{-1}$ = 2.所以第2次變換后a? = 1 - $\frac{1}{a?}$ = 1 - $\frac{1}{\frac{4}{3}}$ = $\frac{1}{4}$,b? = 1 - $\frac{1}{b?}$ = 1 - $\frac{1}{-1}$ = 2,c? = 1 - $\frac{1}{c?}$ = 1 - $\frac{1}{2}$ = $\frac{1}{2}$.所以第3次變換后a? = 1 - $\frac{1}{a?}$ = 1 - $\frac{1}{\frac{1}{4}}$ = -3,b? = 1 - $\frac{1}{b?}$ = 1 - $\frac{1}{2}$ = $\frac{1}{2}$,c? = 1 - $\frac{1}{c?}$ = 1 - $\frac{1}{\frac{1}{2}}$ = -1.所以同理可得a? = a? = a? = a?? = $\frac{4}{3}$,a? = a? = a? = a?? = $\frac{1}{4}$,a? = a? = a? = a?? = -3,b? = b? = b? = b?? = -1,b? = b? = b? = b?? = 2,b? = b? = b? = b?? = $\frac{1}{2}$,c? = c? = c? = c?? = 2,c? = c? = c? = c?? = $\frac{1}{2}$,c? = c? = c? = c?? = -1.所以a? + b? + c? = a? + b? + c? = a? + b? + c? = a?? + b?? + c?? = $\frac{4}{3}$ - 1 + 2 = $\frac{7}{3}$,a? + b? + c? = a? + b? + c? = a? + b? + c? = a?? + b?? + c?? = $\frac{1}{4}$ + 2 + $\frac{1}{2}$ = $\frac{11}{4}$,a? + b? + c? = a? + b? + c? = a? + b? + c? = a?? + b?? + c?? = -3 + $\frac{1}{2}$ - 1 = -$\frac{7}{2}$.所以a? + b? + c? + a? + b? + c? +... + a?? + b?? + c?? = 4(a? + b? + c?) + 4(a? + b? + c?) + 4(a? + b? + c?) = 4×$\frac{7}{3}$ + 4×$\frac{11}{4}$ + 4×(-$\frac{7}{2}$) = $\frac{19}{3}$.
