Violympic toán 8

PN

Thực hiện phép tính:

a) \(\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)\)

b, \(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)

c, \(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)

d,\(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)

TM
6 tháng 1 2019 lúc 9:24
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NT
17 tháng 12 2022 lúc 14:21

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

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

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

a: \(=\dfrac{x+1}{x+2}:\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+3\right)^2}\)

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

b: \(=\dfrac{8}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{\left(x-1\right)\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+3\right)}\)

\(=\dfrac{8+2x^2-2+x^3+3x-x^2-3}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}\)

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

c: \(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{4y^2+4xy}{2\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{4y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2y}{\left(x-y\right)}\)

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