đặt 2x-15=t

=>t⁵=t³

=>t⁵-t³=0

=>t³(t²-1)=0

=>t³(t-1)(t+1)=0

=>t³=0;t-1=0 hoặc t+1=0

=>t=-1;0;1

=>2x=14;15;16

=>x=7;7,5;8

vậy x=7;7,5;8

Ta có : 

\(\left(2x-15\right)^3=\left(2x-15\right)^5\)

\(\Leftrightarrow\)\(\left(2x-15\right)^3=\left(2x-15\right)^3.\left(2x-15\right)^2\)

\(\Leftrightarrow\)\(\left(2x-15\right)^2=1\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-15=1\\2x-15=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=16\\2x=14\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{16}{2}\\x=\frac{14}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=7\end{cases}}}\)

Vậy \(x=7\) hoặc \(x=8\)

Chúc bạn học tốt ~ 

1.

b) \(3^x+3^{x+2}=2430\)

\(\Rightarrow3^x.1+3^x.3^2=2430\)

\(\Rightarrow3^x.\left(1+3^2\right)=2430\)

\(\Rightarrow3^x.10=2430\)

\(\Rightarrow3^x=2430:10\)

\(\Rightarrow3^x=243\)

\(\Rightarrow3^x=3^5\)

\(\Rightarrow x=5\)

Vậy \(x=5.\)

c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=\pm1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=15:2\\2x-15=1\\2x-15=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\2x=16\\2x=14\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{15}{2};8;7\right\}.\)

Chúc bạn học tốt!

a/ (x - 3)(x + 5) = 0

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

b/ |9 - 2x| + x + 3 = 2x + 15

=> |9 - 2x| = 2x + 15 - 3 - x

=> |9 - 2x| = x + 12

\(\Rightarrow\orbr{\begin{cases}9-2x=x+12\\9-2x=-x-12\end{cases}}\Rightarrow\orbr{\begin{cases}-2x-x=12-9\\-2x+x=-12-9\end{cases}}\Rightarrow\orbr{\begin{cases}-3x=3\\-x=-21\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=21\end{cases}}\)

c/ TH1: Nếu x > 1 thì x + 1 + x - 1 = 4

=> 2x = 4

=> x = 2

TH2: Nếu x < 1 thì x + 1 + x - 1 = -4

=> 2x = -4

=> x = -2

TH3: Nếu x = 1 thì 1 + 1 + 1 - 1 = 4 (vô lí)

Vậy x = 2 hoặc x = -2

 Câu c thì mình không chắc cho lắm, không biết có đúng không nữa. ._.

cam on nhe!

a) 4.(18- 5x) - 12.( 3x-7) =15.(2x-16) - 6.(x+14)

72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

-20x - 36x - 30x + 6x = -240 - 84 - 72 -84 

-80x = -480

x= 6

4.(18 - 5x) - 12.(3x - 7) = 15.(2x - 16) - 6.(x+14)
4.18 - 4.5x - 12.3x + 12.7 = 15.2x - 15.16 - 6x - 6.14
72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
72 + 84 - 20x - 36x = 30x - 6x - 240 - 84
156 - 56x               = 24x - 324
156                        = 24x - 324 + 56x
156                       = 80x - 324
80x - 324               = 156
80x                       = 156 + 324
80x                       = 480
x                           = 480:80

x                           = 6
câu b giải tương tự  

a) x = 0 hoặc x = 1

b) x = 8

c) x = 0 hoặc x = 1