`Answer:`

\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+...+\left(x+\frac{1}{729}\right)=\frac{4209}{729}\)

\(\Leftrightarrow\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{3^2}\right)+\left(x+\frac{1}{3^3}\right)+...+\left(x+\frac{1}{3^6}\right)=\frac{4209}{729}\)

\(\Leftrightarrow6x+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\right)=\frac{4209}{729}\text{(*)}\)

Đặt \(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\)

\(\Leftrightarrow3N=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)

\(\Leftrightarrow3N-N=\left(1+\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\right)\)

\(\Leftrightarrow2N=1-\frac{1}{3^6}\)

\(\Leftrightarrow2N=\frac{728}{729}\)

\(\Leftrightarrow N=\frac{364}{729}\)

\(\text{(*)}\Leftrightarrow6x+\frac{364}{729}=\frac{4209}{729}\)

\(\Leftrightarrow6x=\frac{3845}{729}\)

\(\Leftrightarrow x=\frac{3845}{4374}\)

a, không có x :v

b, (2x + 1)³ = 729

=> (2x + 1)³ = 9³

=> 2x + 1 = 9

=> 2x = 8

=> x = 4

vậy_

c, 9⁷ : x = 9⁷

=> x = 9⁷ : 9⁷

=> x = 1 

vậy_

d, x⁵ : 8³ = 64

=> x⁵ : 8³ = 8²

=> x⁵ = 8².8³

=> x⁵ = 8⁵

=> x = 8

vậy_

a) 3^x : 3 = 81

3^x = 27 = 3³

=> x = 3

b) (2x+1)³ = 729 = 9³

=> 2x + 1 = 9

2x = 8

x = 4

c) 9⁷: x = 9⁷

x = 1

d) x⁵ : 8³ = 64 = 8²

x⁵ = 8³.8² = 8⁵

=> x = 8

Hình hiển thị bị lỗi rồi. Bạn nên gõ hẳn đề ra để được hỗ trợ tốt hơn nhé.

d) \(\left|2x-3\right|=x-3\)

TH1: \(\left|2x-3\right|=2x-3\) với \(2x-3\ge0\Leftrightarrow x\ge\dfrac{3}{2}\)

Pt trở thành:

\(2x-3=x-3\) (ĐK: \(x\ge\dfrac{3}{2}\) )

\(\Leftrightarrow2x-x=-3+3\)

\(\Leftrightarrow x=0\left(ktm\right)\)

TH2: \(\left|2x-3\right|=-\left(2x-3\right)\) với \(2x-3< 0\Leftrightarrow x< \dfrac{3}{2}\)

Pt trở thành:

\(-\left(2x-3\right)=x-3\)

\(\Leftrightarrow-2x+3=x-3\)

\(\Leftrightarrow-2x-x=-3-3\)

\(\Leftrightarrow-3x=-6\)

\(\Leftrightarrow x=-\dfrac{6}{-3}=2\left(ktm\right)\)

Vậy Pt vô nghiệm

=>3x-2x^2-3+2x+2x^2+6x-x-3=14

=>10x-6=14

=>x=2

lm chi tiết giúp mình dc ko ạ:")?

a: \(3x^2\left(x^2-2x+5\right)\)

\(=3x^2\cdot x^2-3x^2\cdot2x+3x^2\cdot5\)

\(=3x^4-6x^3+15x^2\)

b: (x+1)(2x-3)

\(=2x^2-3x+2x-3\)

\(=2x^2-x-3\)

c: \(\left(2x^3-3x^2+x+15\right):\left(2x+3\right)\)

\(=\left(2x^3+3x^2-6x^2-9x+10x+15\right):\left(2x+3\right)\)

\(=\frac{x^2\left(2x+3\right)-3x\left(2x+3\right)+5\left(2x+3\right)}{2x+3}=x^2-3x+5\)

3^x+3*3^2x-1*3^x=729

\(\Rightarrow3^{x+3+2x-1+x}=3^6\)

\(\Rightarrow3^{4x+2}=3^6\)

\(\Rightarrow4x+2=6\)

\(\Rightarrow4x=4\)

\(\Rightarrow x=1\)