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

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{49\cdot51}\)

\(\Rightarrow A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(\Rightarrow A=\frac{1}{3}-\frac{1}{51}=\frac{17}{51}-\frac{1}{51}=\frac{16}{51}\)

\(B=5\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{100}-\frac{1}{103}\right)\)

\(\Rightarrow B=5\cdot\left(1-\frac{1}{103}\right)=5\cdot\frac{102}{103}=\frac{510}{103}\)

\(C=5\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}\right)\)

\(\Rightarrow C=5\cdot\left(1-\frac{1}{101}\right)=5\cdot\frac{100}{101}=\frac{500}{101}\)

\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)

\(B=\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(B=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{5}{3}\left(1-\frac{1}{103}\right)\)

\(B=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)

\(C=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)

\(C=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

\(C=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(C=\frac{5}{2}\left(1-\frac{1}{101}\right)\)

\(C=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)

sao bn phũ với mk thế:(( đx ko giải lại còn nói thế

Nó dễ mà :(

a) Ta có :

M= \(5+5^2+5^3+...+5^{60}\)

5M= \(5^2+5^3+5^4+...+5^{61}\)

5M - M= \(\left(5^2+5^3+5^4+...+5^{61}\right)\) - \(\left(5+5^2+5^3+...+5^{60}\right)\)

4M= \(5^{61}-5\)

M= \(\dfrac{5^{61}-5}{4}\)

thế lm sao lm đc câu c

cho 5 

______________

_____________
hok tút

\(E⋮5\) Vì có cơ số đều là 5

Bài 1:

a) Ta có: \(\frac{3}{5}+\frac{4}{15}\)

\(=\frac{9}{15}+\frac{4}{15}\)

\(=\frac{13}{15}\)

b) Ta có: \(\frac{-3}{5}+\frac{5}{7}\)

\(=\frac{-21}{35}+\frac{25}{35}=\frac{4}{35}\)

c) Ta có: \(\frac{5}{6}:\frac{-7}{12}\)

\(=\frac{5}{6}\cdot\frac{-12}{7}=\frac{-60}{42}=\frac{-10}{7}\)

d) Ta có: \(\frac{-21}{24}:\frac{-14}{8}\)

\(=\frac{-7}{8}:\frac{-7}{4}\)

\(=\frac{-7}{8}\cdot\frac{4}{-7}=\frac{4}{8}=\frac{1}{2}\)

e) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)

\(=\frac{-3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)

\(=-\frac{3}{5}\cdot2=\frac{-6}{5}\)

f) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)

\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)

\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)

\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)

g) Ta có: \(\frac{4}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{19}+\frac{5}{7}\)

\(=\frac{4}{19}\cdot\frac{-3}{7}+\frac{5}{19}\cdot\frac{-3}{7}+\frac{-3}{7}\cdot\frac{5}{-3}\)

\(=-\frac{3}{7}\left(\frac{4}{19}+\frac{5}{19}+\frac{-5}{3}\right)\)

\(=\frac{-3}{7}\cdot\left(\frac{27}{57}+\frac{-95}{57}\right)\)

\(=\frac{-3}{7}\cdot\frac{-68}{57}=\frac{68}{133}\)

h) Ta có: \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}\)

\(=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{5}{13}\right)\)

\(=\frac{5}{9}\)

a)(25.5-5².2):(5.2)-3
= (25.5-25.2):10-3
= 25.(5-2):10-3
= 25.3:10-3
=75:10-3=7,5-3=4,5
b)(6.5² -13⁷).2-2³.(7+3)(Sai đề)
c)2³-5³ :5² +12.2²
= 8-125:25+12.4
= 8-5+12.4=8-5+48=3+48=51
d)2.[(95+5²:5):2² +180] -2².10²
= 2.[(95+25:5):4+180]-4.100
= 2.[(95+5):4+180]-400
= 2.(100:4+180)-400
= 2. (25+180)-400
= 2. 205-400
= 410-400=10
e)2⁷.2²+5⁴:5³.24-3.2⁵
= 128+625:125.24-3.32
= 128+5.24-96
= 128+120-96
= 248-96=152
f)2.[(7-3³ :3²):2²+99]-100
=2.[(7-27:9):4+99]-100
=2.[(7-3):4+99]-100
=2. (4:4+99)-100
=2.  (1+99)-100
=2.   100-100
= 200-100
=100
Chúc Bạn Học Tốt ^_^

A = 5 + 5² + 5³ + 5⁴ + ... + 5²⁰⁰

5A = 5² + 5³ + 5⁴ + 5⁵ + ... + 5²⁰¹

5A - A = (5² + 5³ + 5⁴ + 5⁵ + ... + 5²⁰¹) - (5 + 5² + 5³ + 5⁴ + ... + 5²⁰⁰)

4A = 5²⁰¹ - 5 < 5²⁰¹

=> A < 5²⁰¹

tối mik giải cho nhé, giờ bận

5⁵ - 5⁴ + 5³

= 5³ ( 25 - 5 + 1 )

= 5³. 21

Mà 21 ⋮ 7 ⇒ 5⁵ - 5⁴ + 5³ ⋮ 7