a) \(3^{7} \cdot 27^{5} \cdot 81^{3}\)

• Đổi về cơ số 3: \(27 = 3^{3} , \textrm{ }\textrm{ } 81 = 3^{4}\).

\(3^{7} \cdot \left(\right. 3^{3} \left.\right)^{5} \cdot \left(\right. 3^{4} \left.\right)^{3} = 3^{7} \cdot 3^{15} \cdot 3^{12} = 3^{7 + 15 + 12} = 3^{34} .\)

b) \(100^{6} \cdot 1000^{5} \cdot 10 \textrm{ } 000^{3}\)

• Đổi về cơ số 10: \(100 = 10^{2} , \textrm{ }\textrm{ } 1000 = 10^{3} , \textrm{ }\textrm{ } 10 \textrm{ } 000 = 10^{4}\).

\(\left(\right. 10^{2} \left.\right)^{6} \cdot \left(\right. 10^{3} \left.\right)^{5} \cdot \left(\right. 10^{4} \left.\right)^{3} = 10^{12} \cdot 10^{15} \cdot 10^{12} = 10^{39} .\)

c) \(\frac{36^{5}}{18^{5}}\)

• Do cùng số mũ:

\(\left(\left(\right. \frac{36}{18} \left.\right)\right)^{5} = 2^{5} = 32.\)

d) \(24 \cdot 5^{2} + 5^{2} \cdot 5^{3}\)

\(= 24 \cdot 25 + 25 \cdot 125 = 600 + 3125 = 3725.\)

e) \(\frac{125^{4}}{5^{8}}\)

• Đổi về cơ số 5: \(125 = 5^{3}\).

\(\left(\right. 5^{3} \left.\right)^{4} : 5^{8} = 5^{12} : 5^{8} = 5^{12 - 8} = 5^{4} = 625.\)

TÓM lại là a) \(3^{34}\)
b) \(10^{39}\)
c) \(32\)
d) \(3725\)
e) \(625\)

bạn muốn chép đáp án hay sem cách làm///??

a: \(3^7\cdot27^5\cdot81^3\)

\(=3^7\cdot\left(3^3\right)^5\cdot\left(3^4\right)^3\)

\(=3^7\cdot3^{15}\cdot3^{12}=3^{7+15+12}=3^{34}\)

b: \(100^6\cdot1000^5\cdot10000^3\)

\(=\left(10^2\right)^6\cdot\left(10^3\right)^5\cdot\left(10^4\right)^3\)

\(=10^{12}\cdot10^{15}\cdot10^{12}=10^{15+12+12}=10^{39}\)

c: \(36^5:18^5=\left(\frac{36}{18}\right)^5=2^5=32\)

d: \(24\cdot5^2+5^2\cdot5^3\)

\(=5^2\left(24+5^3\right)\)

\(=25\cdot\left(24+125\right)=25\cdot149=3725\)

e: \(125^4:5^8=\left(5^3\right)^4:5^8=5^{12}:5^8=5^{12-8}=5^4\)

a/ 100 : { 2. [ 52 - ( 35 - 8 )]}

= 100 : { 2. [ 52 -  27 ]}

= 100 : { 2. 25 }

= 100 : 50

= 2

b/ 12 : { 390 : [ 500 - ( 125 + 35 . 7 )]}

= 12 : { 390 : [ 500 - ( 125 + 245 )]}

= 12 : { 390 : [ 500 - 370  ]}

= 12 : { 390 : 130 }

= 12 : 3

= 4

c/ 3. 5² - 16 : 2²

= 3 . 25 - 16 : 4

= 75 - 4

= 71

    Bài này dễ mà 

b, \(3737.43-4343.37=\left(37.101\right).43-\left(43.101\right).37=0\)

suy ra B = 0

c, \(D=\frac{2^{12}\left(13+65\right)}{2^{10}.104}+\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{2^{12}.78}{2^{10}.104}+\frac{3^{10}.16}{3^9.2^4}\)

\(=\frac{2^{12}.2.39}{2^{10}.2^3.13}+\frac{3^{10}.2^4}{3^9.2^4}=\frac{39}{13}+3=6\)

a)\(\frac{2}{3}+\frac{3}{4}+\frac{5}{6}\)

\(=\frac{8+9+10}{12}\)

\(=\frac{27}{12}=\frac{9}{4}\)

b)\(\frac{15}{8}-\frac{7}{12}+\frac{5}{6}\)

\(=\frac{45-14+20}{24}\)

\(=\frac{51}{24}=\frac{17}{8}\)

2)

a)\(\frac{2}{5}+\frac{7}{13}+\frac{3}{5}+\frac{1}{7}\)

\(=\frac{2}{5}+\frac{3}{5}+\frac{7}{13}+\frac{1}{7}\)

\(=1+\frac{7}{13}+\frac{1}{7}\)

\(=\frac{20}{13}+\frac{1}{7}\)

\(=\frac{153}{91}\)

Tí tớ trả lời tiếp

b)\(5\frac14+3\frac25-4\frac14\)

=\(\left(5\frac14-4\frac14\right)+3\frac25\)

=\(\left\lbrack\left(5-4\right)+\left(\frac14-\frac14\right)\right\rbrack+\frac{17}{5}\)

=\(1+0+\frac{17}{5}\)

=\(\frac55+\frac{17}{5}\)

=\(\frac{22}{5}\)

a. \(2\cdot125\cdot\left(-5\right)\cdot8\) 

\(=\left(2\cdot\left(-5\right)\right)\cdot\left(125\cdot8\right)\) 

\(=-10\cdot1000\) 

\(=-10000\) 

b. \(56\cdot\left(-35\right)+156\cdot35\) 

\(=-1960+5460\) 

\(=3500\) 

c. \(\frac{11}{15}+\frac{-9}{10}=\frac{22}{30}+\frac{-27}{30}=\frac{-5}{30}=\frac{-1}{6}\) 

d. \(\frac{-4}{5}+\frac{2}{7}+\frac{3}{31}+\frac{5}{7}+\frac{-1}{5}\) 

\(=\left(\frac{-4}{5}+\frac{-1}{5}\right)+\left(\frac{2}{7}+\frac{5}{7}\right)+\frac{3}{31}\) 

\(=-1+1+\frac{3}{31}\) 

\(=0+\frac{3}{31}\) 

\(=\frac{3}{31}\)

B1a)\(11\frac34-\left(6\frac56-4\frac12\right)+1\frac23\)

=\(11\frac34-6\frac56+4\frac12+1\frac23\)

=\(\left(11-6+4+1\right)+\left(\frac34-\frac56+\frac12+\frac23\right)\)

=\(10+\left(\frac{9}{12}-\frac{10}{12}+\frac{6}{12}+\frac{8}{12}\right)\)

=\(10+\left(-\frac{1}{12}+\frac{6}{12}+\frac{8}{12}\right)\)

=10+\(\frac{13}{12}\)

=\(\frac{120}{12}+\frac{13}{12}\)

=\(\frac{133}{12}\)

b)\(2\frac{17}{20}-1\frac{11}{5}+6\frac{9}{20}:3\)

= \(\frac{57}{20}-\frac{16}{5}+\frac{129}{20}\times\frac13\)

=\(\frac{57}{20}-\frac{16}{5}+\frac{129}{60}\)

=\(\frac{171}{60}-\frac{192}{60}+\frac{129}{60}\)

=\(\frac{108}{60}\)

=\(\frac95\)

\(\frac{-3}{14}\) - \(\frac{5}{-14}\)

= \(\frac{-3}{14}+\frac{5}{14}\)

= \(\frac{2}{14}\)

= \(\frac17\)

- \(\frac54\) - \(\frac34\)

= -(\(\frac54+\frac34\))

= - \(\frac84\)

= - 2

\(\frac{15}{6}-\frac{-10}{20}\)

= \(\frac52\) + \(\frac12\)

= \(\frac62\)

= 3

\(\frac{26}{-35}\) - \(\frac{6}{35}\)

= - (\(\frac{26}{35}\) + \(\frac{6}{35}\))

=- \(\frac{32}{35}\)

2.
a) (2x + 1)³ = 125
    (2x + 1)³ = 5³
     2x + 1    = 5
     2x          = 5 - 1
     2x          = 4
       x          = 4:2
       x          = 2
Vậy x = 2
b) 5^x+1 = 5⁴
    x + 1 = 4
    x       = 4 - 1 
    x       = 3
Vây x = 3

a) \(A=1.2+2.3+3.4+...+98.99+99.100\)

\(3A=1.2.3+2.3.4+3.4.3+...+99.100.3\) 

\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)....99.100.\left(101-98\right)\) 

\(3A=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(0.1.2+1.2.3+2.3.4+...+98.99.100\right)\)

\(3A=99.100.101-0.1.2\) 

\(3A=999900-0\)

\(3A=999900\)

\(A=999900:3\)

\(\Rightarrow A=333300\)