Question

Tìm x bít

(3x-7)^2009=(3x-7)^2007

Answer 1

Ta có:(3x-7)²⁰⁰⁹=(3x-7)²⁰⁰⁷

\(\Rightarrow\)(3x-7)²⁰⁰⁹-(3x-7)²⁰⁰⁷=0

​\(\Rightarrow\)​(3x-7)²⁰⁰⁷ [(3x-7)²-1]=0​​

\(\Rightarrow\)\(\orbr{\begin{cases}\left(3x-7\right)^{2007}=0\\\left(3x-7\right)^2-1=0\end{cases}}\)

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

\(\Rightarrow\)\(\orbr{\begin{cases}3x=7\\3x-7=1\end{cases}}\)

​\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\3x=8\end{cases}}\)​

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{8}{3}\end{cases}}\)

Vậy......

Answer 2

\(\left(3x-7\right)^{2009}=\left(3x-7\right)^{2007}\)

\(\Rightarrow\left(3x-7\right)^{2009}-\left(3x-7\right)^{2007}=0\)

\(\Rightarrow\left(3x-7\right)^{2007}\left[\left(3x-7\right)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(3x-7\right)^{2007}=0\\\left(3x-7\right)^2-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x-7=0\\3x-7=\pm1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x\in\left\{\frac{8}{3};2\right\}\end{cases}}\)