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theo cách tính tổng (bn có thể xem lại ở toán 7 hay 6 j đấy) thì bt trên bằng 1/x - 1/(x+5)
từ đó tính tiếp nha bn
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: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
a/ \(\left(2x+3\right)\left(x-5\right)-\left(x-1\right)^2+x\left(7-x\right)\)
\(=2x^2-2x-15-x^2+2x-1+7x-x^2\)
\(=7x-16\)
1.(2x+3).(x-5)+2x(3-x)+x-10
=2x^2 -10x+3x-15+6x-2x^2+x-10
=2x-25
2.(-x-2)3+(2x-4).(x2+2x+4)-x2.(x-6)
=-x^3+6x^2-12x-8+2x^3+4x^2+8x-4x^2+8x-16-x^3+6x^2
a/\(\left(x-1\right)\left(x^5+x^4+x^3+x^2+x+1\right).\)
\(=\left(x-1\right)\left[\left(x^5+x^4+x^3\right)+\left(x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left[x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(=\left(x^2-1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
Câu b/ quên làm ạ :> Bù nè
b/ \(2\left(3x-1\right)\left(2x+5\right)-\left(4x-1\right)\left(3x-2\right)\)
\(=2\left(6x^2+15x-2x-5\right)-\left(12x^2-8x-3x+2\right)\)
\(=2\left(6x^2+13x-5\right)-\left(12x^2-11x+2\right)\)
\(=12x^2+26x-10-\left(12x^2-11x+2\right)\)
\(=12x^2+26x-10-12x^2+11x-2\)
\(=37x-12\)
a: \(=\dfrac{5}{2x^2y}+\dfrac{2}{3xy}-\dfrac{y}{x^3}\)
\(=\dfrac{5\cdot3\cdot x}{6x^3y}+\dfrac{2\cdot2\cdot x^2}{6x^3y}-\dfrac{6y^2}{6x^3y}\)
\(=\dfrac{15x+4x^2-6y^2}{6x^3y}\)
b: \(=\dfrac{2x-7+3x+5}{10x-4}=\dfrac{5x-2}{10x-4}=\dfrac{1}{2}\)
c: \(=\dfrac{x^4-1-x^4+3x^2}{x^2-1}=\dfrac{3x^2-1}{x^2-1}\)
a: \(=\dfrac{5\left(x+2\right)}{10xy^2}\cdot\dfrac{12x}{x+2}=\dfrac{60x}{10xy^2}=\dfrac{6}{y^2}\)
b: \(=\dfrac{x-4}{3x-1}\cdot\dfrac{3\left(3x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x+4}\)
c: \(=\dfrac{2\left(2x+1\right)}{\left(x+4\right)^2}\cdot\dfrac{\left(x+4\right)}{3\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{3\left(x+3\right)\left(x+4\right)}\)
d: \(=\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\cdot\dfrac{x+1}{x-1}=\dfrac{5}{3}\)