\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\)

\(=\dfrac{x+1+x-18+x+2}{x-5}\)

\(=\dfrac{3x-15}{x-5}\)

\(=\dfrac{3\left(x-5\right)}{x-5}\)

\(=\dfrac{3}{1}\)

\(=3\)

\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\\ =\dfrac{x+1+x-18+x+2}{x-5}\\ =\dfrac{\left(x+x+x\right)+\left(1-18+2\right)}{x-5}\\ =\dfrac{3x-15}{x-5}=\dfrac{3\left(x-5\right)}{x-5}=3\)

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

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

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

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

\(=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{13\cdot15}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}=\dfrac{14}{15}\)

\(3\left(x+1\right)^2+2\left(x-3\right)^3-5\left(x-5\right)\left(x+5\right)\)

\(=3\left(x^2+2x+1\right)+2\left(x^2-6x+9\right)-5\left(x^2-25\right)\)

\(=3x^2+6x+3+2x^2-12x+18-5x^2+125\)

\(=-6x+146\)

c: \(=\dfrac{\left(x-5\right)\left(x+5\right)}{3x+4}\cdot\dfrac{-5}{x-5}=\dfrac{-5\left(x+5\right)}{3x+4}\)

a: \(\left(x+5\right)\left(x+1\right)-x^2\)

\(=x^2+6x+5-x^2\)

=6x+5

\(\dfrac{3}{2x^2+y}+\dfrac{5}{xy^2+}+\dfrac{x}{y^3}\)

=\(\dfrac{3xy^5}{xy^2.y^3\left(2x^2+y\right)+}+\dfrac{10y^3x^2+5y^4}{xy^2.y^3\left(2x^2+y\right)}+\dfrac{2x^4y^2+x^2y^3}{xy^2.y^3\left(2x^2+y\right)}\)

=\(\dfrac{3xy^5+10y^3x^2+5y^4+2x^4y^2+x^2y^3}{xy^5\left(2x^2+y\right)}\)

=\(\dfrac{3xy^5+11y^3x^2+5y^4+2x^4y^2}{xy^5\left(2x^2+y\right)}\)

   ủa đáp án cứ sao sao:<

a: \(\dfrac{4-x^2}{x-3}+\dfrac{2x-2x^2}{3-x}+\dfrac{5-4x}{x-3}\)

\(=\dfrac{4-x^2-2x+2x^2+5-4x}{x-3}=\dfrac{x^2-6x+9}{x-3}\)

=(x-3)^2/(x-3)

=x-3

b: \(\dfrac{2}{x+2}+\dfrac{-4}{2-x}+\dfrac{5x+2}{4-x^2}\)

\(=\dfrac{2}{x+2}-\dfrac{4}{x-2}-\dfrac{5x+2}{x^2-4}\)

\(=\dfrac{2x-4-4x-8-5x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-7x-14}{\left(x-2\right)\left(x+2\right)}\)

=-7(x+2)/(x-2)(x+2)

=-7/(x-2)

a) \(x\left(x^2+4x+5\right)-x^2\left(x+4\right)\)

\(=x^3+4x^2+5x-x^3-4x^2\)

\(=5x\)

b) \(\left(x-2\right)^2+\left(3-x\right)\left(x-1\right)\)

\(=x^2-4x+4+3x-3-x^2+x\)

\(=1\)

c) \(\left(x+2\right)^3-x\left(x^2+6x+12\right)\)

\(=x^3+6x^2+12x+8-x^3-6x^2-12x\)

\(=8\)

\(\left(5x^4-x^3+3x^2\right)\)

5x⁴-x³+3x²