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

hỉu j chết liền

Đồ đầu đất .Làm đươc rồi ,khỏi cần làm nữa

hì hì

Ta sẽ xét tính biến thiên của hàm số : 

Ta có \(f\left(x\right)=\left(x^3-3x^2+3x-1\right)+4=\left(x-1\right)^3+4\)

\(f\left(\frac{2017}{2016}\right)-f\left(\frac{2016}{2015}\right)=\left(\frac{2017}{2016}-1\right)^3-\left(\frac{2016}{2015}-1\right)^3\)

\(=\left(\frac{1}{2016}-\frac{1}{2015}\right)\left[\left(\frac{2017}{2016}-1\right)^2+\left(\frac{2016}{2015}-1\right)^2+\left(\frac{2017}{2016}-1\right)\left(\frac{2016}{2015}-1\right)\right]\)

\(=\left(\frac{1}{2016}-\frac{1}{2015}\right)\left(\frac{1}{2016^2}+\frac{1}{2015^2}+\frac{1}{2016}.\frac{1}{2015}\right)< 0\)

\(\Rightarrow f\left(\frac{2017}{2016}\right)-f\left(\frac{2016}{2015}\right)< 0\Rightarrow f\left(\frac{2017}{2016}\right)< f\left(\frac{2016}{2015}\right)\)

Ta sẽ xét tính biến thiên của hàm số : 

Ta có f\left(x\right)=\left(x^3-3x^2+3x-1\right)+4=\left(x-1\right)^3+4f(x)=(x3−3x2+3x−1)+4=(x−1)3+4

f\left(\frac{2017}{2016}\right)-f\left(\frac{2016}{2015}\right)=\left(\frac{2017}{2016}-1\right)^3-\left(\frac{2016}{2015}-1\right)^3f(20162017​)−f(20152016​)=(20162017​−1)3−(20152016​−1)3

=\left(\frac{1}{2016}-\frac{1}{2015}\right)\left[\left(\frac{2017}{2016}-1\right)^2+\left(\frac{2016}{2015}-1\right)^2+\left(\frac{2017}{2016}-1\right)\left(\frac{2016}{2015}-1\right)\right]=(20161​−20151​)[(20162017​−1)2+(20152016​−1)2+(20162017​−1)(20152016​−1)]

=\left(\frac{1}{2016}-\frac{1}{2015}\right)\left(\frac{1}{2016^2}+\frac{1}{2015^2}+\frac{1}{2016}.\frac{1}{2015}\right)&lt; 0=(20161​−20151​)(201621​+201521​+20161​.20151​)<0

\Rightarrow f\left(\frac{2017}{2016}\right)-f\left(\frac{2016}{2015}\right)&lt; 0\Rightarrow f\left(\frac{2017}{2016}\right)&lt; f\left(\frac{2016}{2015}\right)⇒f(20162017​)−f(20152016​)<0⇒f(20162017​)<f(20152016​)

x = 1

b) x = 7/3 hoặc x = 8/3                  

\(\left(3x-7\right)^{2018}=\left(3x-7\right)^{2016}\)

\(\Leftrightarrow\left(3x-7\right)^{2018}-\left(3x-7\right)^{2016}=0\)

\(\Leftrightarrow\left(3x-7\right)^{2016}\left[\left(3x-7\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(3x-7\right)^{2016}=0\\\left(3x-7\right)^2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left(3x-7\right)^{2016}=0\\\left(3x-7\right)^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-7=0\\3x-7=1\\3x-7=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3x=7\\3x=8\\3x=6\end{matrix}\right.\)

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