Реши

Краткие ответы

Вариант 1

1. в) \(a^5\)
2. в)
3. \(7\)
4. \(1\,000\,000\)
5. \(25\)
6. \(-\dfrac17\)
7. \(x^2;\ 0{,}36\)
8. \(2\)
9. \(37^9>2^{18}\cdot3^{18}\)
10. \(200\)

Вариант 2

1. г) \(b^4\)
2. г)
3. \(5\)
4. \(100\,000\,000\)
5. \(36\)
6. \(\dfrac17\)
7. \(x^2;\ 0{,}49\)
8. \(9\)
9. \(35^6<2^{12}\cdot3^{12}\)
10. \(400\)

Вариант 3

1. в) \(8^5\)
2. а) \(-27\)
3. \(\left(\dfrac23\right)^5\)
4. \(3^{11}\)
5. \(a^{23}\)
6. \(\dfrac1{49}\)
7. \(0{,}0016\)
8. \(-0{,}8999\)
9. \(-0{,}91\)
10. \(64a^4\)

Вариант 4

1. г) \(7^4\)
2. б) \(-32\)
3. \(\left(\dfrac37\right)^4\)
4. \(2^8\)
5. \(a^{38}\)
6. \(\dfrac1{81}\)
7. \(0{,}027\)
8. \(-3{,}0999\)
9. \(-0{,}96\)
10. \(49a^4\)

Вариант 5*

1. б) \(4b^6\)
2. б) \(-0{,}001\)
3. \(-24\)
4. \(-\dfrac{73}{9}\)
5. \(x^3;\ -\dfrac8{27}\)
6. \(3^{8n+3}\)
7. \(-10^9=-1\,000\,000\,000\)
8. Например: \(b^1\cdot b^2\cdot b^9\), \(b^3\cdot b^4\cdot b^5\), \(b^4\cdot b^4\cdot b^4\)
9. \(x\)
10. \(2^{4n+2}\)

Ключевые правила

\[ a^m\cdot a^n=a^{m+n},\qquad a^m:a^n=a^{m-n}, \]

\[ (a^m)^n=a^{mn},\qquad a^n\cdot b^n=(ab)^n. \]

Важно: \(-a^2=-(a^2)\), а не \((-a)^2\).

Вариант 1

2. Проверим равенства:

\[ m^3\cdot m^6=m^9,\quad m^{21}:m^7=m^{14},\quad m^8:m^2=m^6,\quad (m^5)^3=m^{15}. \]

Верно только в).

\[ 8-(-1)^6=8-1=7. \]

\[ 25^3\cdot4^3=(25\cdot4)^3=100^3=1\,000\,000. \]

\[ 5^7\cdot5^5:5^{10}=5^{7+5-10}=5^2=25. \]

\[ -8^2\cdot\frac1{56}+1^5 =-\frac{64}{56}+1 =-\frac87+1 =-\frac17. \]

\[ \frac{(x^7)^2}{(x^4\cdot x)^2\cdot x^2} =\frac{x^{14}}{x^{10}\cdot x^2} =x^2. \]

При \(x=0{,}6\):

\[ x^2=0{,}6^2=0{,}36. \]

\[ \frac{32^3\cdot8^2}{16^5} =\frac{(2^5)^3(2^3)^2}{(2^4)^5} =2^{15+6-20}=2. \]

\[ 2^{18}\cdot3^{18}=6^{18}=(6^2)^9=36^9. \]

Так как \(37>36\), то

\[ 37^9>36^9. \]

\[ a:b=\frac{18\cdot10^{n+3}}{9\cdot10^{n+1}} =2\cdot10^2=200. \]

Вариант 2

\[ k^4\cdot k^3=k^7,\quad k^{24}:k^8=k^{16},\quad (k^4)^3=k^{12},\quad k^5\cdot k^2=k^7. \]

Верно только г).

\[ 6-(-1)^8=6-1=5. \]

\[ 20^4\cdot5^4=(20\cdot5)^4=100^4=10^8. \]

\[ 6^8\cdot6^6:6^{12}=6^{8+6-12}=6^2=36. \]

\[ -6^2\cdot\frac1{42}+1^6 =-\frac{36}{42}+1 =-\frac67+1 =\frac17. \]

\[ \frac{x^5(x^4\cdot x)^3}{(x^2)^9} =\frac{x^5\cdot x^{15}}{x^{18}} =x^2. \]

При \(x=0{,}7\):

\[ x^2=0{,}49. \]

\[ \frac{27^2\cdot9^4}{81^3} =\frac{(3^3)^2(3^2)^4}{(3^4)^3} =3^{6+8-12}=3^2=9. \]

\[ 2^{12}\cdot3^{12}=6^{12}=(6^2)^6=36^6. \]

Так как \(35<36\), то

\[ 35^6<36^6. \]

\[ a:b=\frac{24\cdot10^{n+4}}{6\cdot10^{n+2}} =4\cdot10^2=400. \]

Вариант 3

3. Пять одинаковых множителей дают показатель \(5\):

\[ \frac23\cdot\frac23\cdot\frac23\cdot\frac23\cdot\frac23 =\left(\frac23\right)^5. \]

\[ 3\cdot3^{10}=3^{1+10}=3^{11}. \]

\[ (a^9)^3:a^4=a^{27}:a^4=a^{23}. \]

\[ \left(\frac17\right)^{12}:\left(\frac17\right)^{10} =\left(\frac17\right)^2=\frac1{49}. \]

\[ 1{,}6^4\cdot\left(\frac18\right)^4 =\left(\frac{1{,}6}{8}\right)^4 =(0{,}2)^4=0{,}0016. \]

\[ (-0{,}1)^4-(-0{,}1)-1 =0{,}0001+0{,}1-1 =-0{,}8999. \]

\[ -10^4\cdot(-0{,}01)^2+0{,}3^2 =-10000\cdot0{,}0001+0{,}09 =-1+0{,}09=-0{,}91. \]

\[ 64^{2n+1}=(8^2)^{2n+1} =8^{4n+2} =8^2(8^n)^4 =64a^4. \]

Вариант 4

\[ \frac37\cdot\frac37\cdot\frac37\cdot\frac37 =\left(\frac37\right)^4. \]

\[ 2\cdot2^7=2^{1+7}=2^8. \]

\[ (a^8)^5:a^2=a^{40}:a^2=a^{38}. \]

\[ \left(\frac19\right)^{13}:\left(\frac19\right)^{11} =\left(\frac19\right)^2=\frac1{81}. \]

\[ 1{,}8^3\cdot\left(\frac16\right)^3 =\left(\frac{1{,}8}{6}\right)^3 =(0{,}3)^3=0{,}027. \]

\[ (-0{,}1)^4+(-0{,}1)-3 =0{,}0001-0{,}1-3 =-3{,}0999. \]

\[ -10^6\cdot(-0{,}001)^2+0{,}2^2 =-1\,000\,000\cdot0{,}000001+0{,}04 =-1+0{,}04=-0{,}96. \]

\[ 49^{2n+1}=(7^2)^{2n+1} =7^{4n+2} =7^2(7^n)^4 =49a^4. \]

Вариант 5*

\[ (2b^3)^2=2^2(b^3)^2=4b^6. \]

\[ (-0{,}1)^3=-0{,}001. \]

\[ -a^2+1=-5^2+1=-25+1=-24. \]

\[ -2^3-\left(-\frac13\right)^2 =-8-\frac19 =-\frac{72}{9}-\frac19 =-\frac{73}{9}. \]

\[ \frac{(x^3\cdot x^6)^4}{x^{33}} =\frac{(x^9)^4}{x^{33}} =\frac{x^{36}}{x^{33}} =x^3. \]

При \(x=-\dfrac23\):

\[ x^3=\left(-\frac23\right)^3=-\frac8{27}. \]

\[ 3^{8n}\cdot27=3^{8n}\cdot3^3=3^{8n+3}. \]

\[ \frac{(-5)^9\cdot14^9}{7^9} =\left(\frac{-5\cdot14}{7}\right)^9 =(-10)^9=-10^9. \]

8. Показатели при умножении складываются:

\[ 1+2+9=12,\qquad 3+4+5=12,\qquad 4+4+4=12. \]

\[ x^{2n+1}\cdot x^{2n+2}:(x^{2n+1})^2 =x^{4n+3}:x^{4n+2}=x. \]

\[ 2\cdot16^n=2^{4n+1}, \]

\[ 2^n\cdot8^n=2^n\cdot2^{3n}=2^{4n}, \]

\[ 4^{2n}=2^{4n}. \]

Тогда

\[ 2^{4n+1}+2^{4n}+2^{4n} =2^{4n}(2+1+1) =2^{4n}\cdot4 =2^{4n+2}. \]