(3x - 7)²⁰¹⁵ = (3x - 7)²⁰¹⁷

(3x - 7)²⁰¹⁷ - (3x - 7)²⁰¹⁵ = 0

(3x - 7)²⁰¹⁷[(3x - 7)² - 1] = 0

=> (3x - 7)²⁰¹⁷ = 0 hoặc (3x - 7)² = 1

=> 3x - 7 = 0 hoặc 3x - 7 = ± 1

=> x = 7/3 hoặc x = { 8/3 ; 2 }

Vậy x = { 2; 7/3; 8/3 }

\(y\left(y^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}y=0\\y^2-1=0\end{cases}}\)

(3x - 7)^2015 = (3x - 7)^2017

(3x - 7)^2017 - (3x - 7)^2015 = 0

(3x - 7)^2017 [(3x - 7)^2 - 1] = 0

=> (3x - 7)^2017 = 0 hoặc (3x - 7)^2 = 1

=> 3x - 7 = 0 hoặc 3x - 7 = ± 1

=> x = 7/3 hoặc x = { 8/3 ; 2 }

Vậy x = { 2; 7/3; 8/3 }

1) x (x-2016) + 2015 (2016-x) = 0

 x (x-2016) - 2015 (x- 2016) = 0

(x-2015)(x-2016) =0

\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)

Vậy x= 2015; 2016

2) -5x (x-15) + (15-x) = 0

-5x (x-15) - (x-15) =0

(-5x -1) (x-15) =0

\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)

Vậy x= -1/5; 15

3) 3x (3x-7) - (7-3x) =0

3x(3x-7) + (3x -7) =0

(3x+1) (3x-7) =0

\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)

Vậy x= -1/3 ; 7/3

\(\left(3x-7\right)^{2015}=\left(3x-7\right)^{2017}\)

\(\Leftrightarrow\left(3x-7\right)^{2017}-\left(3x-7\right)^{2015}=0\)

\(\Leftrightarrow\left(3x-7\right)^2=0\)

\(\Leftrightarrow3x-7=0\)

\(\Leftrightarrow3x=7\Leftrightarrow x=\frac{7}{3}\)

Vậy \(x=\frac{7}{3}\)

\(\left(3x-7\right)^{2015}=\left(3x-7\right)^{2017}\)

\(\Rightarrow\left(3x-7\right)^{2017}-\left(3x-7\right)^{2015}=0\)

\(\Rightarrow\left(3x-7\right)^{2015}\left[\left(3x-7\right)^2-1\right]=0\)

\(\Rightarrow\left(3x-7\right)^{2015}=0\) hoặc \(\left(3x-7\right)^2-1=0\)

+) \(\left(3x-7\right)^{2015}=0\Rightarrow3x-7=0\Rightarrow x=\frac{7}{3}\)

+) \(\left(3x-7\right)^2-1=0\Rightarrow\left(3x-7\right)^2=1\)

\(\Rightarrow\left[\begin{matrix}3x-7=1\\3x-7=-1\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=\frac{8}{3}\\x=2\end{matrix}\right.\)

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

thiếu 1 trường hợp nữa

(3x-7)²⁰¹⁵=(3x-1)²⁰¹⁷ =1

mk chịu 

nhưng kb với mk nha

co ai ket ban voi minh ko

Ta có:\(\hept{\begin{cases}\left(3x-33\right)^{2014}\ge0\\\left|y-7\right|^{2015}\ge0\end{cases}}\)\(\Rightarrow\left(3x-33\right)^{2014}+\left|y-7\right|^{2015}\ge0\)

Kết hợp với giả thiết chỉ có \(\left(3x-33\right)^{2014}+\left|y-7\right|^{2015}=0\) đúng

\(\Rightarrow\hept{\begin{cases}3x-33=0\\y-7=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=11\\y=7\end{cases}}\)

Vậy...................

\(\left(3x-33\right)^{2014}-\left(\left|y-7\right|\right)^{2015}\le0\)

Ta có \(\left(3x-33\right)^{2014}\ge0\)với mọi gt \(x\in R\)

và \(\left(\left|y-7\right|\right)^{2015}\ge0\)với mọi gt \(x\in R\)

=> \(\left(3x-33\right)^{2014}-\left(\left|y-7\right|\right)^{2015}\ge0\)với mọi gt \(x\in R\)

Mà \(\left(3x-33\right)^{2014}-\left(\left|y-7\right|\right)^{2015}\le0\)

=> \(\left(3x-33\right)^{2014}-\left(\left|y-7\right|\right)^{2015}=0\)

=> \(\hept{\begin{cases}\left(3x-33\right)^{2014}=0\\\left(\left|y-7\right|\right)^{2015}=0\end{cases}}\)=> \(\hept{\begin{cases}3x-33=0\\y-7=0\end{cases}}\)=> \(\hept{\begin{cases}3x=33\\y=7\end{cases}}\)=> \(\hept{\begin{cases}x=11\\y=7\end{cases}}\)

Ta có:\(\left(3x-33\right)^{2014}\ge0,\left|y-7\right|^{2015}\ge0\Rightarrow\left(3x-33\right)^{2014}+\left|y-7\right|^{2015}\ge0\)

Mà VP\(\le0\)

\(\Rightarrow\left(3x-33\right)^{2014}+\left|y-7\right|^{2015}=0\)

\(\Leftrightarrow\left(3x-33\right)^{2014}=0\Leftrightarrow3x-33=0\Leftrightarrow3x=33\Leftrightarrow x=11\)

\(\Leftrightarrow\left|y-7\right|^{2015}=0\Leftrightarrow\left|y-7\right|=0\Leftrightarrow y-7=0\Leftrightarrow y=7\)

Vậy x=11;y=7

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