Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(P=\frac{3x^2+3x}{\left(x+1\right)\left(3x-6\right)}\left(ĐKXĐ:x\ne-1;2\right)\)
b) \(P=\frac{3x\left(x+1\right)}{\left(x+1\right)\left(3x-6\right)}\)
\(P=\frac{3x}{3x-6}\)
Khi \(x=3\Leftrightarrow P=\frac{3\times3}{3\times3-6}\)
\(\Leftrightarrow P=3\)
c) Để P = 1 thì \(\frac{3x}{3x-6}=1\)
\(\Leftrightarrow3x=3x-6\)
\(\Leftrightarrow-6x=-6\)
\(\Leftrightarrow x=1\)
d) Ta có : \(P>2\Leftrightarrow\frac{3x}{3x-6}>2\)
\(\Leftrightarrow3x>2\left(3x-6\right)\)
\(\Leftrightarrow3x>6x-12\)
\(\Leftrightarrow-3x>-12\)
\(\Leftrightarrow x< 4\)
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) 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:\(x\ne-1,x\ne\frac{3}{2}\)
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
\(a,ĐKXĐ:\\ \left[{}\begin{matrix}x+1\ne0\\2x-6\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\\ b,P=0\\ \Leftrightarrow\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=0\\ \Leftrightarrow\dfrac{3x\left(x+1\right)}{3\left(x+1\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x}{x-2}=0\\ \Leftrightarrow x=0\left(TM\right)\)
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, \(x\ne-1;3\)
b, Ta có : \(P=\frac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=1\)
\(\Leftrightarrow\frac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=1\Leftrightarrow\frac{3x}{2\left(x-3\right)}=1\)
\(\Leftrightarrow3x=2x-6\Leftrightarrow x=-6\)