Tham khảo:

a)      \((4{x^2} - 5):(x - 2) = \dfrac{{4{x^2} - 5}}{{x - 2}} = 4x + 8 + \dfrac{{11}}{{x - 2}}\)

Vậy \( (4{x^2} - 5):(x - 2)= 4x + 8 + \dfrac{{11}}{{x - 2}}\)

b)      \((3{x^3} - 7x + 2):(2{x^2} - 3) = \dfrac{{3{x^3} - 7x + 2}}{{2{x^2} - 3}}\)

Vậy \( (3{x^3} - 7x + 2):(2{x^2} - 3)= \dfrac{3}{2}x + \dfrac{{\dfrac{-5}{2}x + 2}}{{2{x^2} - 3}}\)

a: \(\dfrac{2x^3-5x^2-x+1}{2x+1}\)

\(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}\)

\(=x^2-3x+1\)

b: \(\dfrac{x^3-2x+4}{x+2}\)

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

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

Tham khảo:

a)      \((45{x^5} - 5{x^4} + 10{x^2}):5{x^2}\)\( = 9{x^3} - {x^2} + 2\)

b)      \((9{t^2} - 3{t^4} + 27{t^5}):3t = (27{t^5} - 3{t^4} + 9{t^2}):3t\\=(27t^5):(3t) - (3t^4):(3t)+(9t^2):(3t) = 9{t^4} - {t^3}+3t\)

\(a)(2{y^4} - 13{y^3} + 15{y^2} + 11y - 3):({y^2} - 4y - 3)=2y^2-5y+1\)

b) \((5{x^3} - 3{x^2} + 10):({x^2} + 1)=5x-3+\dfrac{-5x+13}{x^2+1}\)

\(a,=\dfrac{5x}{4y^3}\times\left(\dfrac{-20y}{x^4}\right)=\dfrac{-100xy}{4x^4y^3}=\dfrac{-25}{x^3y^2}\\ b,=\dfrac{\left(x-4\right)\left(x+4\right)}{\left(x+4\right)}\times\dfrac{x}{2\left(x-4\right)}=\dfrac{x}{2}\)

\(c,=\dfrac{2\left(x+3\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\times\dfrac{2\left(x-2\right)}{\left(x+3\right)^3}=\dfrac{4}{\left(x+3\right)^2.\left(x^2+2x+4\right)}\)

a) \(\dfrac{5x}{4y^3}:\left(-\dfrac{x^4}{20y}\right)=\dfrac{5x}{4y^3}\cdot\left(-\dfrac{20y}{x^4}\right)=\dfrac{5\cdot-5}{y^2\cdot x^3}=\dfrac{-25}{x^3y^2}\)

b) \(\dfrac{x^2-16}{x+4}:\dfrac{2x-8}{x}=\left(x-4\right)\cdot\dfrac{x}{2\left(x-4\right)}=\dfrac{x}{2}\)

c) \(\dfrac{2x+6}{x^3-8}:\dfrac{\left(x+3\right)^3}{2x-4}=\dfrac{2\left(x+3\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\cdot\dfrac{2\left(x-2\right)}{\left(x+3\right)^3}=\dfrac{4}{\left(x^2+2x+4\right)\left(x+3\right)^2}\)

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)

Bài 4 là:

\(\dfrac{1}{2.5}\) + \(\dfrac{1}{5.8}\) +....+\(\dfrac{1}{92.95}\) + \(\dfrac{1}{95.98}\)

                  Đúng không bạn

Bạn chỉ nên đưa câu khó lên vì làm cả đề thì ko làm kịp đc

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

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

\(=2x^2+x+1\)

b: \(=\dfrac{x^5+x^2+x^3+1}{x^3+1}=x^2+1\)

c: \(=\dfrac{2x^3-x^2-x+6x^2-3x-3+2x+6}{2x^2-x-1}\)

\(=x+3+\dfrac{2x+6}{2x^2-x-1}\)

d: \(=\dfrac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)

\(=\dfrac{3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5}{3x^2-2x+1}\)

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