Question

\(A=\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x}{x+2}\right)\)

a) Rút gọn

b)Tính giá trị của A khi giá trị tuyệt đối của x = \(\dfrac{1}{2}\)

c)Với giá trị nào của x thì A=2

d)Với giá trị nào của x thì A<b

e)TÌm giá trị của x để A có giá trị nguyên

Answer 1

ĐKXĐ \(x\ne2;x\ne-2;x\ne0\)\(A=\left(\dfrac{x^3}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x+2+\dfrac{10-x}{x+2}\right)\)\(=\left(\dfrac{x^2}{x\left(x+2\right)\left(x-2\right)}\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x}{x+2}\right)\)\(=\left(\dfrac{3x^2}{3x\left(x-2\right)\left(x+2\right)}-\dfrac{6.x\left(x+2\right)}{3x\left(x-2\right)\left(x+2\right)}+\dfrac{3x\left(x-2\right)}{3x\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{\left(x-2\right)\left(x+2\right)^23x}{3x\left(x-2\right)\left(x+2\right)}+\dfrac{\left(10-x\right)3x\left(x-2\right)}{3x\left(x-2\right)\left(x+2\right)}\right)\)\(=\dfrac{3x^2-6x^2-12x+3x-6}{3x\left(x^2-4\right)}.\dfrac{3x\left(x^2-4\right)}{3x^4+12x^3+12x^2-6x^3-24x^2-24x}=\dfrac{-3x^2-9x-6}{3x^4+6x^3-12x^2-24x}=\dfrac{-3\left(x^2+6x+2\right)}{-3\left(-x^4-2x^3+4x^2+8x\right)}=\dfrac{x^2+6x+2}{-x^4-2x^3+4x^2+8x}\)

Answer 2

a, \(A=\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x}{x+2}\right)\)

\(=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left[\dfrac{\left(x-2\right)\left(x+2\right)+10-x}{\left(x+2\right)}\right]\)

\(=\left(\dfrac{x}{x^2-4}+\dfrac{2x+4+2-x}{\left(2-x\right)\left(x+2\right)}\right)\left(\dfrac{x+2}{x^2-4+10-x}\right)\)

\(=\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}+\dfrac{x+6}{\left(2-x\right)\left(x+2\right)}\right]\left(\dfrac{x+2}{x^2-x+6}\right)\)

\(=\left[\dfrac{-x}{\left(2-x\right)\left(x+2\right)}+\dfrac{x+6}{\left(2-x\right)\left(x+2\right)}\right]\left(\dfrac{x+2}{x^2-x+6}\right)\)

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

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