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a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2}{x^2-4}\)

12 tháng 12 2021

a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-2x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)

8 tháng 12 2021

a) \(A=\dfrac{x+2+x^2-2x+1}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-x+1}{\left(x-2\right)\left(x+2\right)}\)

a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2}{x^2-4}\)

a: ĐKXĐ:\(x\notin\left\{2;0\right\}\)

b: \(C=\left(\dfrac{x\left(2-x\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\left(\dfrac{2-x^2+x}{x^2}\right)\)

\(=\dfrac{-x^3+4x^2-4x-4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{x^2}\)

\(=\dfrac{x\left(x^2+4\right)}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}=\dfrac{x+1}{2x}\)

c: Thay x=2017 vào C, ta được:

\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)

8 tháng 12 2021

a)B =  \(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{7x+3}{9-x^2}\left(ĐK:x\ne\pm3\right)\)

\(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{7x+3}{x^2-9}\)

\(\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-7x-3}{\left(x+3\right)\left(x-3\right)}\)

\(\dfrac{3x^2-9x}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x}{x+3}\)

b) \(\left|2x+1\right|=7< =>\left[{}\begin{matrix}2x+1=7< =>x=3\left(L\right)\\2x+1=-7< =>x=-4\left(C\right)\end{matrix}\right.\)

Thay x = -4 vào B, ta có:

B = \(\dfrac{-4.3}{-4+3}=12\)

c) Để B = \(\dfrac{-3}{5}\)

<=> \(\dfrac{3x}{x+3}=\dfrac{-3}{5}< =>\dfrac{3x}{x+3}+\dfrac{3}{5}=0\)

<=> \(\dfrac{15x+3x+9}{5\left(x+3\right)}=0< =>x=\dfrac{-1}{2}\left(TM\right)\)

d) Để B nguyên <=> \(\dfrac{3x}{x+3}\) nguyên

<=> \(3-\dfrac{9}{x+3}\) nguyên <=> \(9⋮x+3\)

x+3-9-3-1139
x-12(C)-6(C)-4(C)-2(C)0(C)6(C)

 

29 tháng 10 2018

\(P=\frac{2\left(x-2\right)\left(x+2\right)}{x^2+x+5}.\frac{5\left(x^2+x+5\right)}{\left(x-4\right)\left(x+3\right)}.\frac{\left(x-1\right)\left(x-4\right)}{10\left(x-2\right)\left(x+2\right)}=\frac{x-1}{x+3}\)

ĐK: \(x\ne\left\{4;-3;1;2;-2\right\}\)

b, \(P\in Z\Rightarrow\frac{x-1}{x+3}\in Z\Rightarrow x-1⋮\left(x+3\right)\Rightarrow-4⋮\left(x+3\right)\Rightarrow\left(x+3\right)\in\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow x\in\left\{-7;-5;-4;-2;-1;1\right\}\)

\(\Rightarrow P\in\left\{2;3;5;-3;-1;0\right\}\)