a) 3^x = 243

=> 3^x = 35

=> x = 5

b) 2^x.4 = 128

=> 2^x = 32

=> 2^x = 2⁵

=> x = 5

c) 5^x - 15 = 110

=> 5^x = 125

=> 5^x = 5³

=> x = 3

a) 3 mũ 5 = 243

b) 2 mũ 5.4=128

c)5 mũ 3 -15=110

\(a,\left(x+2,8\right)+\left(x+4,5\right)+\left(x+2,7\right)=27,5\\ \Leftrightarrow\left(x+x+x\right)+\left(2,8+4.5+2,7\right)=27,5\\ \Leftrightarrow3x+9=27,5\\ \Leftrightarrow3x=18,5\\ \Leftrightarrow x=\dfrac{37}{6}\\ b,\left(2,4+x\right)+\left(2,6+x\right)+\left(2,8+x\right)+...+\left(4,2+x\right)=150,66\\ \Leftrightarrow10x+\left(2,4+4,2\right):2.10=150,66\\ \Leftrightarrow10x+33=150,66\\ \Leftrightarrow10x=117,66\\ \Leftrightarrow x=11,766\)

Đặt A = -5+5²-5³+5⁴-...-5²⁰¹⁷+5²⁰¹⁸

=>5A = -5²+5³-5⁴+5⁵-...-5²⁰¹⁸+5²⁰¹⁹

=> 5A+A=(-5²+5³-5⁴+5⁵-...+5²⁰¹⁹)+(-5+5²-5³+5⁴-...+5²⁰¹⁸)

=> 6A = 5²⁰¹⁹-5

=> A = \(\dfrac{5^{2019}-5}{6}\) 

Ta có :

B = 1+3²+3⁴+...+3²⁰¹⁸

=> 3²B=3²+3⁴+3⁶+...+3²⁰²⁰

=> 3²B-B=(3²+3⁴+3⁶+...+3²⁰²⁰)-(1+3²+3⁴+...+3²⁰¹⁸)

=> 8B=3²⁰²⁰-1

=> \(B=\dfrac{3^{2020}-1}{8}\)

3^1 019 090

Ta có :

A = 2+2+2²+2³+...+2²⁰¹⁷

=> 2A=4+2²+2³+2⁴+...+2²⁰¹⁸

=> 2A-A=(4+2²+2³+2⁴+...+2²⁰¹⁸)-(2+2+2²+...+2²⁰¹⁷)

=> A=2²⁰¹⁸

giúp mik vs

Ta có :

A = 1+3+3²+3³+...+3²⁰⁰⁰ ( có tất cả 2001 số hạng )

=> A = (1+3+3²)+(3³+3⁴+3⁵)+...+(3¹⁹⁹⁸+3¹⁹⁹⁹+3²⁰⁰⁰) ( có đủ 667 nhóm )

=> A = (1+3+3²)+3³.(1+3+3²)+...+3¹⁹⁹⁸.(1+3+3²)

=> A = 13+3³.13+...+3¹⁹⁹⁸.13

=> A = 13.(1+3³+...+3¹⁹⁹⁸) ⋮ 13

\(20^{3x}=20^{13}:\left(4.5\right)^4\)

\(\Rightarrow20^{3x}=20^{13}:20^4\)

\(\Rightarrow20^{3x}=20^9\)

`=>3x=9`

`=>x=3`

20^3x=20¹³:(4.5)⁴

=> 20^3x=20¹³:20⁴

=> 20^3x=20⁹

=> 3x = 9

=> x = 3

a) 85-(14+3x):5=81

=> (14+3x):5=85-81=4

=> 14+3x=20

=> 3x=6

=> x=2

b) 49+{40-3.[16-81:3³]}

= 49+[40-3.(16-81:27)]

=49+[40-3.(16-3)]

= 49+(40-3.13)

= 49 + (40 - 39)

= 49 + 1 = 50

c) (x-6)²=16

=> (x-6)²=4²

=> x-6=4

=> x=10

d) 3^x+2-3^x=72

=> 2=72 =>  Đề sai

`@`\(85-\left(14+3x\right):5=81\)

\(\Rightarrow\left(14+3x\right):5=4\)

\(\Rightarrow14+3x=20\)

`=>3x=6`

`=>x=2`

`@`\(\left(x-6\right)^2=16\)

`=>(x-6)^2=4^2`

\(\Rightarrow\left[{}\begin{matrix}x-6=4\\x-6=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=10\\x=2\end{matrix}\right.\)

`@`\(3^{x+2}-3^x=72\)

\(\Rightarrow3^x.9-3^x=72\)

\(\Rightarrow3^x\left(9-1\right)=72\)

\(\Rightarrow3^x.8=72\)

\(\Rightarrow3^x=9\)

\(\Rightarrow x=2\)

Đặt \(S=1+2+2^2+...+2^{2012}\)

=> \(2S=2+2^2+2^3+...+2^{2013}\)

\(2S-S=2^{2013}-1=S\)

Biểu thức ban đầu có dạng:

\(\dfrac{2^{2013}-1}{2^{2014}\times2}=\dfrac{2^{2013}-1}{2^{2015}}\)

Đề yêu cầu gì em?

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.100}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}=\dfrac{100}{101}\)