Ta có: \(\left(3x-7\right)^{2005}=\left(3x-7\right)^{2003}\)

\(\Leftrightarrow\left(3x-7\right)^{2005}-\left(3x-7\right)^{2003}=0\)

\(\Leftrightarrow\left(3x-7\right)^{2003}\left[\left(3x-7\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(3x-7\right)^{2003}=0\\\left(3x-7\right)^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x\in\left\{\frac{8}{3};2\right\}\end{cases}}\)

Vậy \(x\in\left\{\frac{7}{3};\frac{8}{3};2\right\}\)

\(\left(3x-7\right)^{2005}=\left(3x-7\right)^{2003}\)

\(\Rightarrow\left(3x-7\right)^{2005}-\left(3x-7\right)^{2003}=0\)

\(\Leftrightarrow\left(3x-7\right)^{2003}[\left(3x-7\right)^2-1]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(3x-7\right)^{2003}=0\\\left(3x-7\right)^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x-7=0\\3x-7=1\end{cases}}\)hoặc \(3x-7=-1\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{8}{3}\end{cases}}\)hoặc \(x=2\)

Vậy ...............................

(3x - 7)²⁰⁰⁷ = (3x - 7)²⁰⁰⁵

=> (3x - 7)²⁰⁰⁷ - (3x - 7)²⁰⁰⁵ = 0

=> (3x - 7)²⁰⁰⁵ [(3x - 7)² - 1] = 0

=> (3x - 7)²⁰⁰⁵ = 0 hoặc (3x - 7)² - 1 = 0

+) (3x - 7)²⁰⁰⁵ = 0

=> 3x - 7 = 0

=> 3x = 7

=> x = 7/3

+) (3x - 7)² - 1 = 0

=> (3x - 7)² = 1

=> 3x - 7 = 1  => 3x = 8 => x = 8/3

     3x - 7 = -1 => 3x = 6 => x = 2

Vậy: x \(\in\){-7/3;8/3;2

3x-7=1=>x=2\(\frac{2}{3}\)

3x-7=0=>x=2\(\frac{1}{3}\)

a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)

\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)

\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow x+8-12+20x=0\)

\(\Leftrightarrow21x-4=0\)

\(\Leftrightarrow21x=4\)

\(\Leftrightarrow x=\dfrac{4}{21}\)

Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)

Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!

a) 

PT \(\Leftrightarrow \frac{4x+2}{12}-\frac{3x-6}{12}=\frac{12-8x}{12}-\frac{12x}{12}\)

\(\Leftrightarrow 4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow 21x=4\Leftrightarrow x=\frac{4}{21}\)

b) 

PT \(\Leftrightarrow \frac{30x+15}{20}-\frac{100}{20}-\frac{6x+4}{20}=\frac{24x-12}{20}\)

\(\Leftrightarrow 30x+15-100-6x-4=24x-12\Leftrightarrow -89=-12\) (vô lý)

Vậy pt vô nghiệm.