Cho biểu thức P =\(\dfrac{x}{x+2}\) +\(\dfrac{2}{x-2}\)+\(\dfrac{2x+4}{4-x^2}\)với x≠2 , x≠ -2
Tìm giá trị của P tại |x+1|=3
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25 tháng 2 2021
a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
b) Ta có: \(B=\left(\dfrac{x-2}{2x-2}+\dfrac{3}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(1-\dfrac{x-3}{x+1}\right)\)
\(=\left(\dfrac{x-1}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(\dfrac{x+1-x-3}{x+1}\right)\)
\(=\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\dfrac{-2}{x+1}\)
\(=\dfrac{x^2-1-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{-2}\)
\(=\dfrac{-2x+2}{2\left(x-1\right)}\cdot\dfrac{-1}{2}\)
\(=\dfrac{-2\left(x-1\right)}{2\left(x-1\right)}\cdot\dfrac{-1}{2}\)
\(=\dfrac{1}{2}\)
Vậy: Khi x=2005 thì \(B=\dfrac{1}{2}\)
\(\dfrac{x}{x+2}+\dfrac{2}{x-2}+\dfrac{2x+4}{4-x^2}\\ =\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x+2x+4-2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x}{x+2}\)
\(\left|x+1\right|=3\\ \left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=2\left(loai\right)\\x=-4\left(tm\right)\end{matrix}\right.\)
với x=-4 thì
\(\dfrac{-4}{-4+2}=\dfrac{-4}{-2}=2\)
\(=>P=\dfrac{x}{x+2}+\dfrac{2}{x-2}+\dfrac{-2x-4}{x^2-4}\)`(x ne +-2)`
\(P=\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{-2x-4}{\left(x+2\right)\left(x-2\right)}\)
\(P=\dfrac{x^2-2x+2x+4-2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2-2x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(P=\dfrac{x}{x+2}\)
`|x+1| =3`
`=>[(x+1=3),(x+1=-3):}`
`=> [(x=3-1=2(ktm) ),(x=-3-1=-4(t/m)):}`
Thay `x=-4` vào `P` ta đc
`P= (-4)/(-4+2) = 2`