a)\(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)

b)\(\frac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=10\)\(\Leftrightarrow\frac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=10\)

\(\Leftrightarrow\frac{3x}{2x-6}=10\)\(\Leftrightarrow3x=10\left(2x-6\right)\)

\(\Leftrightarrow3x=20x-60\)\(\Leftrightarrow17x=60\Leftrightarrow x=\frac{60}{17}\)

a )\(\left[\begin{array}{nghiempt}x+1\ne0\\2x-3\ne0\end{array}\right.\)

\(ĐKXĐ:x\ne-1,x\ne\frac{3}{2}\)

b ) \(A=\frac{2x^2-3x}{\left(x+1\right)\left(2x-3\right)}=\frac{x\left(2x-3\right)}{\left(x+1\right)\left(2x-3\right)}=\frac{x}{x+1}\)

Để \(A=3\) thì :

 \(\frac{x}{x+1}=3\Leftrightarrow x=3x+3\Leftrightarrow x-3x=3\Leftrightarrow-2x=3\Leftrightarrow x=-\frac{3}{2}\)

Chúc bạn học tốt

a) ĐKXĐ:\(x\ne-1,x\ne\frac{3}{2}\)

b)\(A=\frac{2x^2-3x}{\left(x+1\right)\left(2x-3\right)}=\frac{x\left(2x-3\right)}{\left(x+1\right)\left(2x-3\right)}=\frac{x}{x+1}\)

để A = 3 thì \(\frac{x}{x+1}=3\Leftrightarrow x=3x+3\Leftrightarrow x-3x=3\Leftrightarrow-2x=3\Leftrightarrow x=\frac{-3}{2}\)

DKXD : \(x+1\ne0\Rightarrow x\ne-1,2x-3\ne0\Rightarrow2x\ne3\Rightarrow x\ne\frac{3}{2}\)

\(A=\frac{2x^2-3x}{\left(x+1\right)\left(2x-3\right)}=3\Rightarrow A==\frac{2x^2-3x}{\left(x+1\right)\left(2x-3\right)}=\frac{3.\left(\left(x+1\right)\left(2x-3\right)\right)}{\left(x+1\right)\left(2x-3\right)}\)

\(\Rightarrow A=\frac{2x^2-3x}{\left(x+1\right)\left(2x-3\right)}=\frac{3.\left(2x^2-3x-2x+3\right)}{\left(x+1\right)\left(2x-3\right)}\Rightarrow A=\frac{2x^2-3x}{\left(x+1\right)\left(2x-3\right)}=\frac{6x^2-9x-6x+9}{\left(x+1\right)\left(2x-3\right)}\)\(\Rightarrow A=2x^2-3x=6x^2-15x+9\Rightarrow A=0=4x^2-12x+9\Rightarrow A=0=\left(2x-3\right)^2\)

\(\Rightarrow2x-3=0\Rightarrow x=\frac{3}{2}\left(TMDKXD\right)\)

t i c k cho mình 1 cái nha mình bị trừ 50đ ùi hic hic ủng hộ nhé

chết lộn dấu rồi nha

a) P xác định  <=> \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)

b)\(P=\frac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=1\Leftrightarrow3x^2+3x=\left(x+1\right)\left(2x-6\right)\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)\left(2x-6\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-2x+6\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)

Vì \(x\ne-1\Leftrightarrow x+1\ne0\Rightarrow x+6=0\Leftrightarrow x=-6\)

Vậy ........

ĐKXĐ: x khac -1 và x khac 3

a, ĐKXĐ: x\(\ne\)-1, x\(\ne\)3

b,ta có: P =\(\frac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}\)=\(\frac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}\)\(=\frac{3x}{2x-6}\)

Để P = 1

=>\(\frac{3x}{2x-6}=1\)

=> 2x - 6 = 3x

=> 2x - 3x = 6

=> -x =6

=>x = -6

\(a,ĐKXĐ:\\ \left[{}\begin{matrix}x+1\ne0\\2x-6\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\\ b,P=0\\ \Leftrightarrow\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=0\\ \Leftrightarrow\dfrac{3x\left(x+1\right)}{3\left(x+1\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x}{x-2}=0\\ \Leftrightarrow x=0\left(TM\right)\)

Vậy tại X=0 thì P=0

a) Để P xác định thì: \(\left[{}\begin{matrix}x+1\ne0\\2x-6\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)

b) \(P=\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=\dfrac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=\dfrac{3x}{2x-6}\)

Để \(P=0\) thì: \(\dfrac{3x}{2x-6}=0\)

\(\Leftrightarrow3x=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

a) ĐKXĐ \(x +1\ne0=>x\ne-1;2x-6\ne0=>x\ne3\)

b) ta có 

\(P=\frac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=\frac{3x}{2x-6}\)

để P = 1 thì \(\frac{3x}{2x-6}=1= >3x=2x-6\)

                                                        \(< =>3x-2x=-6=>x=-6\)

a,ĐK:  \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

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

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

c, Với x = 4 thỏa mãn ĐKXĐ thì

\(A=\frac{-3}{4-3}=-3\)

d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)

\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)

Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)