a) ĐKXĐ: 

\(\left\{{}\begin{matrix}x^2-9\ne0\\x+3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne-3\end{matrix}\right.\Leftrightarrow x\ne\pm3\) 

b) \(A=\dfrac{x+15}{x^2-9}-\dfrac{2}{x+3}\)

\(A=\dfrac{x+15}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{x+15-2x+6}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{21-x}{\left(x+3\right)\left(x-3\right)}\)

c) Thay x = - 1 vào A ta có: 

\(A=\dfrac{21-\left(-1\right)}{\left(-1+3\right)\left(-1-3\right)}=\dfrac{21+1}{2\cdot-4}=\dfrac{22}{-8}=-\dfrac{11}{4}\)

đưa máy tính mà tính chứ

a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)

b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)

Mik cảm ơn ạ 

\(\frac{9\times8\times5}{6\times4\times15}=\frac{120}{360}=\frac{120:120}{360:120}=\frac{1}{3}\)

Kiểu này ó bạn, tự hiểu đi<3

TL

9x8x5/6x4x15=360/360=1

nha

HT

Bài làm 

\(A=\frac{2x+6}{\left(x-3\right)\left(x-2\right)}=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x-2\right)}\)

\(B=\frac{x^2-9}{x^2-6x+9}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\frac{x+3}{x-3}\)

\(A=\frac{2x+6}{\left(x-3\right)\left(x-2\right)}=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x-2\right)}\)

\(B=\frac{x^2-9}{x^2-6x+9}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\frac{x+3}{x-3}\)

a) \(\dfrac{2\left(x+1\right)^2}{4x\left(x+1\right)}\left(x\ne0;x\ne-1\right)\)

\(=\dfrac{2\left(x+1\right)^2:2\left(x+1\right)}{4x\left(x+1\right):2\left(x+1\right)}\)

\(=\dfrac{x+1}{2x}\)

b) \(\dfrac{\left(8-x\right)\left(-x-2\right)}{\left(x+2\right)^2}\left(x\ne-2\right)\)

\(=\dfrac{-\left(8-x\right)\left(x+2\right)}{\left(x+2\right)^2}\)

\(=\dfrac{-\left(8-x\right)}{x+2}\)

\(=\dfrac{x-8}{x+2}\)

c) \(\dfrac{2\left(x-y\right)}{y-x}\left(x\ne y\right)\)

\(=\dfrac{2\left(x-y\right)}{-\left(x-y\right)}\)

\(=-2\)

d) \(\dfrac{\left(x+2\right)^2}{2x+4}\left(x\ne-2\right)\)

\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)

\(=\dfrac{x+2}{2}\)

ĐKXĐ: \(x\neq0;x\neq-1\)

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

$---$

ĐKXĐ: \(x\neq-2\)

\(\dfrac{(8-x)(-x-2)}{(x+2)^2}=\dfrac{-(8-x)(x+2)}{(x+2)^2}=\dfrac{x-8}{x+2}\)

$---$

ĐKXĐ: \(x\neq y\)

\(\dfrac{2(x-y)}{y-x}=\dfrac{-2(y-x)}{y-x}=-2\)

$---$

ĐKXĐ: \(x\neq-2\)

\(\dfrac{(x+2)^2}{2x+4}=\dfrac{(x+2)^2}{2(x+2)}=\dfrac{x+2}{2}\)

1. = \(\dfrac{x+y}{x-y}\)
2. = \(\dfrac{x}{x+3}\)

\(\frac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{27}+x^{24}+x^{21}+x^{18}+x^{15}+x^{12}+x^9+x^6+x^3+1}=\frac{x^{24}+x^{18}+x^{12}+x^6+1}{x^{24}\left(x^3+1\right)+x^{18}\left(x^3+1\right)+x^{12}\left(x^3+1\right)+x^6\left(x^3+1\right)+\left(x^3+1\right)}\)

=\(\frac{x^{24}+x^{18}+x^{12}+x^6+1}{\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)}=\frac{1}{x^3+1}\)

\(M=\frac{x^8\left(x+1\right)+x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+x+1}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{\left(x^8+x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^8+x^6+x^4+x^2+1}{x-1}\)

M=\(\frac{\left(x^9+x^8\right)\left(x^7+x^6\right)+...+\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)

M=\(\frac{x^8\left(x+1\right)+x^6\left(x+1\right)+...+\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)

M=\(\frac{\left(x+1\right)\left(x^8+x^6+x^4+x^2\right)}{\left(x+1\right)\left(x-1\right)}\)

M=\(\frac{x^8+x^6+x^4+x^2}{x-1}\)