\(\left(2x-15\right)^3=\left(2x-15\right)^5\\ \Rightarrow\left(2x-15\right)^2=1\\ \Rightarrow\left[{}\begin{matrix}2x-15=-1\\2x-15=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

giúp mik ik mik tic cho và nhớ cho thêm cả hướng dẫn

ĐKXĐ: \(-5\le x\le3\)

Đặt \(\sqrt{x+5}+\sqrt{3-x}=t>0\Rightarrow t^2=8+2\sqrt{-x^2-2x+15}\)

\(\Rightarrow-2\sqrt{-x^2-2x+15}=8-t^2\) (1)

Pt trở thành:

\(t+8-t^2-2=0\Leftrightarrow-t^2+t+6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(loại\right)\end{matrix}\right.\)

Thế vào (1): \(-2\sqrt{-x^2-2x+15}=-1\)

\(\Leftrightarrow\sqrt{-x^2-2x+15}=\dfrac{1}{2}\)

\(\Leftrightarrow-x^2-2x+15=\dfrac{1}{4}\)

\(\Leftrightarrow...\)

a,x=2

b,x=5

x=4

x=6

a) \(\left(2x+1\right)^3=125\)

\(\Rightarrow\left(2x+1\right)^3=5^3\)

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=5-1\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=4:2\)

\(\Rightarrow x=2\)

Vậy x = 2

b) \(\left(x-5\right)^4=\left(x-5\right)^6\)

\(\Rightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)

\(\Rightarrow\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2=1\end{cases}}\)

TH 1 : \(\left(x-5\right)^4=0\Rightarrow x-5=0\Rightarrow x=5\)

TH 2 : \(\left(x-5\right)^2=1\Rightarrow\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

Vậy  \(x\in\left\{5;6;4\right\}\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2=1\end{cases}}\)

TH 1 : \(\left(2x-15\right)^3=0\Rightarrow2x-15=0\Rightarrow2x=15\Rightarrow x=\frac{15}{2}\)

TH 2 : \(\left(2x-15\right)^2=1\Rightarrow\orbr{\begin{cases}2x-15=1\\2x-15=-1\end{cases}}\Rightarrow\orbr{\begin{cases}2x=16\\2x=14\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=7\end{cases}}\)

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

_Chúc bạn học tốt_

a) 2^x.2^4=128

=>2^x.2^2=2^7

=>2^x=2^7:2^2

=>2^x=2^5

=>x=5

b)x^15=x

=>x^15-x=0

=>x(x^16-x)=0

=>2 trượng hợp:x=0 và x^16-1=0(x^16-1=0 cx 2 th nha)

b),d),e) như nhau nha!

c) dễ rồi

\(a)2^x\cdot4=128\)

\(\Rightarrow2^x=\frac{128}{4}\)

\(\Rightarrow2^x=32\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

\(b)x^{15}=x\)

\(\Rightarrow x^{15}-x=0\)

\(\Rightarrow x(x^{14}-1)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\x^{14}=1\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=1\end{cases}}\)

\(c)(2x+1)^3=125\)

\(\Rightarrow(2x+1)^3=5^3\)

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=5-1\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=4:2=2\)

\(d)(x-5)^4=(x-5)^6\)

\(\Rightarrow(x-5)^6-(x-5)^4=0\)

\(\Rightarrow(x-5)^4\cdot\left[(x-5)^2-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}(x-5)^4=0\\(x-5)^2-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

\(e)(2x-15)^5=(2x-15)^3\)

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

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

\(\Rightarrow\orbr{\begin{cases}(2x-15)^3=0\\(2x-15)^2-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\varnothing\\x=8\end{cases}}\)

Chúc bạn hoc tốt :>

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

\(\Rightarrow\left[\begin{array}{nghiempt}2x-15=0\\2x-15=1\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x=15\\2x=16\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{15}{2}\\x=8\end{array}\right.\)

bai nay thieu  = -1

diều kiện xác định là các mẫu phải khác o; số chia cũng khác o nhé:

ĐK: +)  \(x+5\ne0\Rightarrow x\ne-5\)

+)  \(2x-15\ne0\Rightarrow x\ne\frac{15}{2}\)

+)  \(x^2-25\ne0\Rightarrow\left(x+5\right)\left(x-5\right)\ne0\Rightarrow x\ne\pm5\)

+)  \(1-x\ne0\Rightarrow x\ne1\)

Vậy điều kiện xác đinh của A là : \(x\ne1;x\ne\frac{15}{2};x\ne\pm5\)