Question

Cho biểu thức:

Answer 1

a,\(ĐKXĐ:\left\{{}\begin{matrix}2x+12\ne0\\x\ne0\\2x\left(x+6\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-6\\x\ne0\\2x\left(x+6\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-6\\x\ne0\end{matrix}\right.\)

b, \(P=\dfrac{x^2+2x}{2x+12}+\dfrac{x-6}{x}+\dfrac{108-6x}{2x\left(x+6\right)}\)

\(\Leftrightarrow P=\dfrac{x\left(x^2+2x\right)}{2x\left(x+6\right)}+\dfrac{2\left(x-6\right)\left(x+6\right)}{2x\left(x+6\right)}+\dfrac{108-6x}{2x\left(x+6\right)}\)

\(\Leftrightarrow P=\dfrac{x^3+2x^2}{2x\left(x+6\right)}+\dfrac{2\left(x^2-36\right)}{2x\left(x+6\right)}+\dfrac{108-6x}{2x\left(x+6\right)}\)

\(\Leftrightarrow P=\dfrac{x^3+2x^2}{2x\left(x+6\right)}+\dfrac{2x^2-72}{2x\left(x+6\right)}+\dfrac{108-6x}{2x\left(x+6\right)}\)

\(\Leftrightarrow P=\dfrac{x^3+2x^2+2x^2-72+108-6x}{2x\left(x+6\right)}\)

\(\Leftrightarrow P=\dfrac{x^3+4x^2-6x+36}{2x\left(x+6\right)}\)

\(\Leftrightarrow P=\dfrac{x^3+6x^2-2x^2-12x+6x+36}{2x\left(x+6\right)}\)

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

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

\(\Leftrightarrow P=\dfrac{x^2-2x+6}{2x}\)

\(c,P=\dfrac{3}{2}\Leftrightarrow\dfrac{x^2-2x+6}{2x}=\dfrac{3}{2}\\ \Rightarrow2x^2-4x+12=6x\\ \Leftrightarrow2x^2-10x+12=0\\ \Leftrightarrow x^2-5x+6=0\\ \Leftrightarrow x^2-2x-3x+6=0\\ \Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{2;3\right\}\)

\(d,P=-\dfrac{9}{2}\\ \Leftrightarrow\dfrac{x^2-2x+6}{2x}=-\dfrac{9}{2}\\ \Rightarrow2x^2-4x+12=-18x\\ \Leftrightarrow2x^2+14x+12=0\\ \Leftrightarrow x^2+7x+6=0\\ \Leftrightarrow x^2+6x+x+6=0\\ \Leftrightarrow x\left(x+6\right)+\left(x+6\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\left(ktm\right)\end{matrix}\right.\)

Vậy x=-1

\(e,P=1\\ \Leftrightarrow\dfrac{x^2-2x+6}{2x}=1\\ \Rightarrow x^2-2x+6=2x\\ \Leftrightarrow x^2-4x+6=0\\ \Leftrightarrow\left(x^2-4x+4\right)+2=0\\ \Leftrightarrow\left(x-2\right)^2+2=0\left(vô.lí\right)\)

vậy không có x thỏa mãn đề bài