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 ........
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
Điều kiện : \(\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)
Rút gọn :
\(\frac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}\)
\(=\frac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}\)
\(=\frac{3x}{2\left(x-3\right)}\)
Phân thức = 1 \(\Leftrightarrow3x=2\left(x-3\right)\)
\(\Leftrightarrow3x=2x-6\)
\(\Leftrightarrow x=6\left(tmdk\right)\)
a) DK x khác -1 và x khác 3
b)
P=1 \(\Leftrightarrow3x^2+3x=\left(x+1\right)\left(2x-6\right)\)
\(\Leftrightarrow3x^2+3x=2x^2-4x-6\Leftrightarrow x^2+7x+6=0\)
\(\left(x+\frac{7}{2}\right)^2=\frac{25}{4}=\left(\frac{5}{4}\right)^2\)\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}-\frac{7}{2}=-1\\x=-\frac{5}{2}-\frac{7}{2}=-6\end{cases}}\)
