Question

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

a/ ĐKXĐ

b/Tìm giá trị của x để phân thức bằng 1

Answer 1

a/ \(x\ne1;x\ne3\)

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

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

\(\Leftrightarrow\dfrac{3x\cdot\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=1\)

\(\Leftrightarrow\dfrac{3x}{2x-6}=1\)

\(\Leftrightarrow3x=2x-6\)

\(\Leftrightarrow3x-2x=-6\)

\(\Rightarrow x=-6\left(đk:x\ne-1,x\ne-3\right)\)

\(\Rightarrow x=6\)

Vậy \(x=6\)

Answer 2

a) \(ĐKXĐ:\left\{{}\begin{matrix}x+1\ne0\\2x-6\ne0\end{matrix}\right.< =>\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)

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

Ta có: \(P=1\\ < =>\dfrac{3x}{2\left(x-3\right)}=1\\ < =>3x=2\left(x-3\right)\\ < =>3x=2x-6\\ < =>3x-2x=-6\\ < =>x=-6\left(TMĐK\right)\)

Vậy: Để P=1 => x=-6

Answer 3

ĐKXĐ: \(x\ne1,x\ne3\)

\(\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}=\dfrac{3x\left(x+1\right)}{\left(x+1\right)2\left(x+3\right)}=\dfrac{3x}{2\left(x+3\right)}\)Để phân thức = 1 \(\Rightarrow3x=2\left(x+3\right)\Rightarrow3x=2x+3\Rightarrow x=3\)

Answer 4

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

a) ĐKXĐ : \(x\ne-1;x\ne3\)

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

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

\(\Leftrightarrow\dfrac{3x}{2x-6}=1\)

\(\Leftrightarrow3x=2x-6\)

\(\Leftrightarrow x=-6\)

Vạy \(x=-6\) thì phân thức trên bằng 1.

Answer 5

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

a) ĐKXĐ \(x\ne-1;x\ne3\).

b) Để P = 1 :

\(\Leftrightarrow1=\dfrac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}\)

\(\Leftrightarrow3x^2+3x=\left(x+1\right)\left(2x-6\right)\)

\(\Leftrightarrow3x^2+3x=2x^2-4x-6\)

\(\Leftrightarrow x^2+7x+6=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\left(KTMĐK\right)\\x=-6\left(TMĐK\right)\end{matrix}\right.\)

Vậy x = -6 thì GTPT bằng 1.

Answer 6

a) Điều kiện xác định: \(x\ne1;x\ne3\)

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

\(\Leftrightarrow3x^2+3x=\left(x+1\right)\left(2x-6\right)\)

\(\Leftrightarrow3x^2+3x=2x^2-4x-6\)

\(\Leftrightarrow x^2+7x+6=0\)

\(\Leftrightarrow x^2+x+6x+6=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(loại,vì:x\ne-1;-3\right)\\x=-6\left(TM\right)\end{matrix}\right.\)

Vậy \(P=1\) khi \(x=-6\)