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24 tháng 12 2021

b: \(=\dfrac{7x-42-x^2+36}{x\left(x-6\right)}=\dfrac{-x^2+7x-6}{x\left(x-6\right)}=\dfrac{-x+1}{x}\)

24 tháng 12 2021

\(\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x-3\right)}=\dfrac{x\left(x+3\right)-3\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+3x-3x-9}{x\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x}\)

13 tháng 12 2018

\(\frac{x^2+3x+9}{2x+10}.\frac{x+5}{x^3-27}\)

\(=\frac{x^2+3x+9}{2\left(x+5\right)}.\frac{x+5}{\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\frac{\left(x+5\right)\left(x^2+3x+9\right)}{2\left(x+5\right)\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\frac{1}{2\left(x-3\right)}\)

\(\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\left(\frac{x^2-36}{x^2+1}\right)\)

\(=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\left[\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\right]\)

\(=\frac{\left(6x+1\right)\left(x+6\right)+\left(6x-1\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{12x^2+12}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{12\left(x^2+1\right).\left(x-6\right)\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)\left(x^2+1\right)}\)

\(=\frac{12}{x}\)

24 tháng 6 2017

Phân thức đại số

Phân thức đại số

28 tháng 6 2017

Phép trừ các phân thức đại số

14 tháng 12 2018

a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)

14 tháng 12 2018

thanks

28 tháng 6 2017

Phép cộng các phân thức đại số

Phép cộng các phân thức đại số

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

b: \(=\dfrac{x\left(x+3\right)}{\left(x+3\right)^2}+\dfrac{3}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x}{x+3}+\dfrac{3}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x^2-3x+3x+9-6x}{\left(x-3\right)\left(x+3\right)}\)

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

c: \(=\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+9x-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x^2+9x-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x-3}\)

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

21 tháng 4 2017

Giải bài 35 trang 50 Toán 8 Tập 1 | Giải bài tập Toán 8

18 tháng 11 2018

a) \(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}=\dfrac{\left(x-3\right)^2-x^2+9}{x\left(x-3\right)}=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}=\dfrac{18-6x}{x\left(x-3\right)}=\dfrac{-6\left(x-3\right)}{x\left(x-3\right)}=\dfrac{-6}{x}\)b) \(\dfrac{1}{x-2}-\dfrac{6x}{x^3-8}+\dfrac{x-2}{x^2+2x+4}=\dfrac{x^2+2x+4+\left(x-2\right)^2-6x}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{x^2-4x+4+\left(x-2\right)^2}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{\left(x-2\right)^2+\left(x-2\right)^2}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2\left(x-2\right)^2}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2\left(x-2\right)}{x^2+2x+4}\)

30 tháng 11 2018

\(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}=\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x\left(x-3\right)}=\dfrac{\left(x-3\right).\left(x-3\right)}{x.\left(x-3\right)}-\dfrac{x.x}{\left(x-3\right).x}+\dfrac{9}{x.\left(x-3\right)}=\dfrac{x^2-6x+9}{x.\left(x-3\right)}-\dfrac{x^2}{x.\left(x-3\right)}+\dfrac{9}{x.\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x.\left(x-3\right)}-\dfrac{x^2}{x.\left(x-3\right)}+\dfrac{9}{x.\left(x-3\right)}=\dfrac{\left(x-3\right)^2-x^2+9}{x.\left(x-3\right)}\)

17 tháng 7 2017

Nguyễn Huy Tú :v

17 tháng 7 2017

a,\(\dfrac{3}{x-3}\) - \(\dfrac{6x}{9-x^2}\) + \(\dfrac{x}{x+3}\) (*)

đkxđ: x khác 3, x khác -3

(*) \(\dfrac{3(x+3)}{\left(x-3\right).\left(x+3\right)}\)- \(\dfrac{6x}{\left(x-3\right).\left(x+3\right)}\) + \(\dfrac{x\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}\)

=>3x+9 -6x + x2+3x

<=>x2 + 3x-6x+3x + 9

<=>x2 +9

<=>(x-3).(x+3)

a: \(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)

\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)

\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)

\(=\dfrac{3}{x-3}\cdot\dfrac{-\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{-3}{x-3}\)

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

\(=\dfrac{x+1}{x+2}\cdot\dfrac{\left(x+3\right)^2}{\left(x+2\right)\left(x+1\right)}=\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2}\)

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

\(=\dfrac{2\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x+y\right)^2}{x^2+y^2}\cdot\dfrac{1}{2}\)

\(=\dfrac{\left(x+y\right)}{x-y}\)