\(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)

\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)

\(=\left(6-5-3\right)+\left(-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)

\(=-2+0-\frac{1}{2}\)

\(=-\frac{4}{2}-\frac{1}{2}\)

\(=-\frac{5}{2}\)

cam on ban

1-2+2^2 các bạn nha

Ta có: \(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}\)

\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6}=\frac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^6}\)

\(=\frac{3^4\cdot2}{3^6}=\frac{2}{3^2}=\frac29\)

Ta có: \(\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)

\(=\frac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\frac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\frac{-6}{9}=5\cdot\frac{-2}{3}=-\frac{10}{3}\)

Ta có: \(A=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)

\(=\frac29+\frac{10}{3}=\frac29+\frac{30}{9}=\frac{32}{9}\)

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) \(A=1.2+2.3+3.4+...+29.30\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4\left(5-2\right)+...+29.30\left(31-28\right)\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+29.30.31-28.29.30\)

\(\Rightarrow3A=29.30.31\)

\(\Rightarrow A=29.30.31:3\)

\(\Rightarrow A=29.10.31\)

\(\Rightarrow A=8990\)

3A= 1.2.3+2.3.4+3.4.3 +......+ 29.30.3

3A= 1.2. ﴾3 ‐ 0﴿ + 2.3.﴾4 ‐ 1﴿ +3.4. ﴾5 ‐ 2﴿....... . 29.30. ﴾31 ‐ 28﴿

3A = ﴾1.2.3 + 2.3.4 + 3.4.5 +...... +18.20.21﴿ ‐ ﴾0.1.2 + 1.2.3 + 2.3.4 +.......+ 18.19.20﴿

3A = 29.30.31 ‐ 0.1.2

3A =26970‐0

3A= 26970

 A=26970:3

A = 8990.

Vậy A=8990

a, \(5S=5^2+5^3+...+5^{2017}\)

\(5S-S=5^{2017}-5\)

\(S=\frac{5^{2017}-5}{4}\)

b,\(3S=3^2+3^3+...+3^{101}\)

\(3S-S=3^{101}-3\)

\(S=\frac{3^{101}-3}{2}\)

c, \(3S=3-3^2+3^3-...-3^{2016}\)

\(3S+S=1-3^{2016}\)

\(4S=1-3^{2016}\)

\(S=\frac{1-3^{2016}}{4}\)

b, 3S = 3^2+3^3+.....+3^101

2S=3S-S=(3^3+3^3+.....+3^101)-(3+3^2+....+3^100) = 3^101-3

=> S = (3^101-3)/2

Tk mk nha

\(a)\) \(S=1+2+2^2+2^3+...+2^{2017}\)

\(2S=2+2^2+2^3+2^4+...+2^{2018}\)

\(2S-S=\left(2+2^2+2^3+2^4+...+2^{2018}\right)-\left(1+2+2^2+2^3+...+2^{2017}\right)\)

\(S=2^{2018}-1\)

\(b)\) \(S=3+3^2+3^3+...+3^{2017}\)

\(3S=3^2+3^3+3^4+...+3^{2018}\)

\(3S-S=\left(3^2+3^3+3^4+...+3^{2018}\right)-\left(3+3^2+3^3+...+3^{2017}\right)\)

\(2S=3^{2018}-3\)

\(S=\frac{3^{2018}-3}{2}\)

\(c)\) \(S=4+4^2+4^3+...+4^{2017}\)

\(4S=4^2+4^3+4^4+...+4^{2018}\)

\(4S-S=\left(4^2+4^3+4^4+...+4^{2018}\right)-\left(4+4^2+4^3+...+4^{2017}\right)\)

\(3S=4^{2018}-4\)

\(S=\frac{4^{2018}-4}{3}\)

\(d)\) \(S=5+5^2+5^3+...+5^{2017}\)

\(5S=5^2+5^3+5^4+...+5^{2018}\)

\(5S-S=\left(5^2+5^3+5^4+...+5^{2018}\right)-\left(5+5^2+5^3+...+5^{2017}\right)\)

\(4S=5^{2018}-5\)

\(S=\frac{5^{2018}-5}{2}\)

Chúc em học tốt ~ 

Tks anh ạ 

 đặt A = (cái trên )

2A=1+2^2+...+2^101

-

A=1+2+....+2^100

------------------------------

A= 2^101 - 1 

B = 5+5^2+......+5^99

5B=5^2+5^3+....+5^100

-

B = 5+5^2+......+5^99

-----------------------------------

4B= 5^100-5

B=(5^100 - 5)/4

học tốt nha

tổng quát cho bạn luôn

A=n+n^2 + ....+ n^n

nA= n^2 + n^3 +....+n^(n+1)

- 

A=n+n^2 + ....+ n^n

------------------------------------------

(n-1)A = n^(n+1) - n

A= (n^(n+1) - n) / (n-1)

ok

tuy nhiên một vài trường hợp(như câu B) thôi nha còn lại cũng na ná như thế

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

\(A=3+\frac{1}{2}-\frac{2}{3}-2+\frac{2}{3}-\frac{5}{2}-5+\frac{5}{2}-\frac{4}{3}\)

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

\(A=-4+\frac{1}{2}-1-\frac{1}{3}\)

\(A=-5+\frac{1}{2}-\frac{1}{3}\)

\(A=-5+\frac{1}{6}\)

\(A=-4\frac{5}{6}\)

tích đúng mình làm cho