5)
ĐK: \(x\ne\pm3\)
Biểu thức trở thành:
\(\dfrac{2x}{9-x^2}-\dfrac{\left(x+3\right)}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)}{\left(x+3\right)\left(3-x\right)}\\ =\dfrac{2x-x-3-3+x}{9-x^2}\\ =\dfrac{2x-6}{9-x^2}\\ =\dfrac{2\left(x-3\right)}{\left(3-x\right)\left(3+x\right)}\\ =-\dfrac{2\left(x-3\right)}{\left(x-3\right)\left(3+x\right)}\\ -\dfrac{2}{3+x}\)
6)
ĐK: \(x;y\ne0\)
\(\dfrac{8y}{3x^2}.\dfrac{9x^2}{4y^2}\\ =\dfrac{2.4y.3x^2.3}{3x^2.4y.y}\\ =\dfrac{6}{y}\)
7)
ĐK: \(x\ne-3\)
Biểu thức trở thành:
\(\dfrac{x\left(3+x\right)}{x^2+x+1}.\dfrac{3\left(x^3-1\right)}{x+3}\\ =\dfrac{x\left(3+x\right).3.\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(3+x\right)}\\ =3x^2-x\)
8)
ĐK: \(x\ne\pm4\)
Biểu thức trở thành:
\(\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}.-\dfrac{2\left(x-4\right)}{\left(x+5\right)\left(x+5\right)}\\ =-\dfrac{2\left(x+5\right).2.\left(x-4\right)}{\left(x-4\right)\left(x^2+4x+16\right)\left(x+5\right)\left(x+5\right)}\\ =-\dfrac{4}{\left(x+5\right)\left(x^2+4x+16\right)}\\ =-\dfrac{4}{x^3+4x^2+16x+5x^2+20x+80}\\ =-\dfrac{4}{x^3+9x^2+36x+80}\)
`HaNa♬D`
\(9,\left(\dfrac{1}{x^3+1}-\dfrac{1}{x+1}-\dfrac{1}{x^2-x+1}\right):\dfrac{1}{x^3+1}\left(x\ne1\right)\\ =\left(\dfrac{1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{1}{x+1}-\dfrac{1}{x^2-x+1}\right)\cdot\left(x^3+1\right)\\=\left(\dfrac{1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}\right)\cdot\left(x^3+1\right)\\=\dfrac{1-x^2+x-1-x-1}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\left(x^3+1\right)\\=\dfrac{\left(-x^2-1\right)\left(x^3+1\right)}{\left(x+1\right)\left(x^2-1+1\right)}\\ =-x^2-1\)
