Giá trị của phân thức \(\frac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}\) được xác định với điều kiện ( x + 1 )( 2x - 6 )\(\ne\) 0

<=> 2( x + 1 )( x - 3 ) \(\ne\) 0

<=> x + 1 \(\ne\) 0 và x - 3 \(\ne\) 0

+, x + 1 \(\ne\) 0

<=> x \(\ne\) -1

+, x - 3 \(\ne\) 0

<=> x \(\ne\) 3

Vậy ĐKXĐ : x  \(\ne\) -1; 3

Ta có : \(\frac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}\) 

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

\(=\frac{3x}{2\left(x-3\right)}\) 

Giá trị của biểu thức \(\frac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}\) bằng 1

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

\(\Rightarrow3x=2x-6\)  

\(\Rightarrow3x-2x=-6\)

 \(\Rightarrow x=-6\)

Vậy x = -6

\(x^3-7x-6=0\)

\(x^3-3x^2+3x^2+2x-9x-6=0\)

\(x^2.\left(x-3\right)+3x.\left(x-3\right)+2.\left(x-3\right)=0\)

\(\left(x+3\right).\left(x^2+3x+2\right)=0\Rightarrow\left(x-3\right).\left(x^2+3x+x+2\right)=0\)

\(\Rightarrow\left(x-3\right).\left(x+1\right).\left(x+2\right)=0\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\text{hoặc }x=-2\)

giusp minh voi

a, ĐỂ \(\frac{3x+3}{x^2-1}=\frac{3x+3}{\left(x+1\right)\left(x-1\right)}\)    Xác định 

\(\Rightarrow\left(x+1\right)\left(x-1\right)\ne0\)

\(\Rightarrow\hept{\begin{cases}x+1\ne0\\x-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-1\\x\ne1\end{cases}}}\)

KL : \(x\ne\pm1\)

b , 

\(\frac{3x+3}{x^2-1}\)xác định 

\(\Leftrightarrow x^2-1\ne0\Leftrightarrow x\ne\pm1\)

Vậy điều kiện xác định của \(\frac{3x+3}{x^2-1}\)là \(x\ne\pm1\)

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

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

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

\(\Leftrightarrow3=-2\left(x-1\right)\)

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

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

Vậy \(x=\frac{-1}{2}\)là giá trị cần tìm

Bạn đã ib nhờ mik thì mik làm cho trót vại UwU

\(\frac{1}{x\left(x-y\right)\left(x-z\right)}+\frac{1}{y\left(y-z\right)\left(y-x\right)}+\frac{1}{z\left(z-x\right)\left(z-y\right)}.\)

\(=-\frac{1}{x\left(x-y\right)\left(z-x\right)}-\frac{1}{y\left(y-z\right)\left(x-y\right)}-\frac{1}{z\left(z-x\right)\left(y-z\right)}\)

\(=-\frac{y^2x-yz^2}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}-\frac{xz^2-x^2z}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}-\frac{x^2y-xy^2}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)

\(=\frac{-y^2z+yz^2-xz^2+x^2z-x^2y+xy^2}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)

\(=\frac{-\left(y^2z-x^2z\right)+\left(yz^2-xz^2\right)-\left(x^2y-xy^2\right)}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)

\(=\frac{-z\left(y^2-x^2\right)+z^2\left(y-x\right)-xy\left(x-y\right)}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)

\(=\frac{-z\left(y-x\right)\left(x+y\right)+z^2\left(y-x\right)+xy\left(y-x\right)}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)

\(=\frac{\left(y-x\right)\left[-z\left(x+y\right)+x^2+xy\right]}{xyz\left(x-y\right)\left(z-x\right)\left(y-z\right)}\)

\(=\frac{\left(y-x\right)\left[-z\left(x+y\right)+x^2+xy\right]}{-xyz\left(y-x\right)\left(z-x\right)\left(y-z\right)}\)

\(=-\frac{-z\left(x+y\right)+z^2+xy}{xyz\left(z-x\right)\left(y-z\right)}\)

\(=-\frac{-zx-zy+z^2+xy}{xyz\left(z-x\right)\left(y-z\right)}\)

\(=\frac{-\left(zx-xy\right)-\left(zy-z^2\right)}{xyz\left(z-x\right)\left(y-z\right)}\)

\(=\frac{-x\left(z-y\right)-z\left(y-z\right)}{xyz\left(z-x\right)\left(y-z\right)}\)

\(=\frac{x\left(y-z\right)-z\left(y-z\right)}{xyz\left(z-x\right)\left(y-z\right)}\)

\(=\frac{\left(y-z\right)\left(x-z\right)}{xyz\left(z-x\right)\left(y-z\right)}\)

\(=\frac{x-z}{xyz\left(z-x\right)}\)

\(=\frac{-\left(z-x\right)}{xyz\left(z-x\right)}\)

\(=\frac{-1}{xyz}\)