\(\left(3x-7\right)^{2015}=\left(3x-7\right)^{2017}\Rightarrow\left(3x-7\right)^{2017}-\left(3x-7\right)^{2015}=0\Leftrightarrow\left(3x-7\right)^{2015}\left[\left(3x-7\right)^2-1\right]=0\Leftrightarrow\orbr{\begin{cases}3x-7=0\\\left(3x-7\right)^2=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}3x=7\\3x-7=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{1+7}{3}=\frac{8}{3}\end{cases}}\)
Vậy phương trình có hai nghiệm là \(x=\frac{7}{3}\)và \(x=\frac{8}{3}\)

Vì \(\left(3x-7\right)^{2015}=\left(3x-7\right)^{2017}\) =>3x-7=0 hoặc 3x-7=1

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

(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 }

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

(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 }

a) [ 3x-1] + 4x - 3 = 7

 3x - 1  + 4x = 7 + 3 = 10

( 3 + 4 )x - 1 =10

7x - 1 = 10

7x=11

x=11/7

Câu tiếp theo làm tương tự nhé =))

) [ 3x-1] + 4x - 3 = 7

 3x - 1  + 4x = 7 + 3 = 10

( 3 + 4 )x - 1 =10

7x - 1 = 10

7x=11

x=11/7

( 3x/7 + 1 ) : (-4 ) = -1/28

 3x/7 + 1           = -1/28 x (-4 ) 

(3x/7 + 1           = 1/7

3x/7                  = 1/7  - 1

3x/7                  = -6/7

Suy ra 3x = -6

            x = -6 : 3

            x = -2

=>3x/7+1=-1/28x(-4)

=>3x/7+1=1/7

=>3x/7=1/7-1

=>3x/7=-6/7

=>3x=6

=>x=6:3=2

k mình đi

Ta có : (3x/7+1) : (-4) = -1/28

3x/7+1 = -1/28 . (-4)

3x/8 = 1/7

Do đó 3x.7/56 = 8/56

Suy ra 3x.7=8

Suy ra 3x =8 : 7

Suy ra 3x = 8/7

Suy ra x = 8/7 : 3

Do đó x = 8/21

\(\frac{5x+7}{4}+\frac{3x+5}{8}>\frac{9x+4}{5}\)

\(\frac{10\cdot\left(5x+7\right)}{40}+\frac{5\cdot\left(3x+5\right)}{40}>\frac{8\cdot\left(9x+4\right)}{40}\)

10.(5x + 7) + 5.(3x + 5) > 8.(9x + 4)

10.(5x + 7) + 5.(3x + 5) - 8.(9x + 4) > 0

50x + 70 + 15x + 25 - 72x - 32 > 0

- 7x + 63 > 0

- 7.(x - 9) > 0

\(\Rightarrow x-9

Bài 1: 

a) \(x^3-16x=x\left(x-4\right)\left(x+4\right)\)

b) \(3x^2+3y^2-6xy-12=3\left(x^2-2xy+y^2-4\right)=3\left(x-y-2\right)\left(x-y+2\right)\)

c) \(x^2+6x+5=\left(x+1\right)\left(x+5\right)\)

d) \(x^4+x^3+2x^2+x+1=\left(x^2+x+1\right)\left(x^2+1\right)\)

Bài 2: 

a) Ta có: \(\left(x+6\right)^2=144\)

\(\Leftrightarrow\left[{}\begin{matrix}x+6=12\\x+6=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-18\end{matrix}\right.\)

b) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

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

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

c) Ta có: \(2x^2-x-6=0\)

\(\Leftrightarrow2x^2-4x+3x-6=0\)

\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Ta có : |2x - 1| + 1 = x 

=> |2x - 1| = x - 1

\(\Leftrightarrow\orbr{\begin{cases}2x-1=x-1\\2x-1=1-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-x=-1+1\\2x+x=1+1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x=2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)