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21 tháng 4 2017

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

31 tháng 8 2020

Bài làm:

đk: \(x\ne-3;x\ne1\)

Ta có: \(\frac{x^2+6x+9}{1-x}\cdot\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}\)

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

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

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

31 tháng 8 2020

\(ĐKXĐ:\hept{\begin{cases}x\ne-3\\x\ne1\end{cases}}\)

\(\frac{x^2+6x+9}{1-x}.\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}=\frac{-\left(x+3\right)^2}{x-1}.\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}=\frac{-\left(x-1\right)^2}{2\left(x+3\right)}\)

9 tháng 12 2018

\(\left(\frac{1}{x^2-9}+\frac{2}{3-x}+\frac{3}{x+3}\right)\div\frac{x-14}{x+3}\)

\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2}{x-3}+\frac{3}{x+3}\right)\div\frac{x-14}{x+3}\)

\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right)\div\frac{x-14}{x+3}\)

\(=\left(\frac{1-2x-6+3x-9}{\left(x+3\right)\left(x-3\right)}\right).\frac{x+3}{x-14}\)

\(=\frac{x-14}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{x-14}=\frac{1}{x-3}\)

14 tháng 12 2018

\(\left(\frac{1}{x^2-9}+\frac{2}{3-x}+\frac{3}{x+3}\right):\frac{x-14}{x+3}\)

\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2}{x-3}+\frac{3}{x+3}\right):\frac{x-14}{x+3}\)

\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right).\frac{x+3}{x-14}\)

\(=\left(\frac{1-2x-6+3x-9}{\left(x-3\right)\left(x+3\right)}\right).\frac{x+3}{x-14}=\frac{x-14}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{x-14}\)

\(=\frac{1}{x-3}\)

22 tháng 11 2017

 = (x+1).(x+3)-(1-x).(x-3)+2x.(1-x)/(x-3).(x+3)

 = x^2+4x+3+x^2-4x+3+2x-2x^2/(x+3).(x-3)

 = 2x+6/(x+3).(x-3) = 2.(x+3)/(x+3).(x-3) = 2/x-3

k mk nha

22 tháng 11 2017

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

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

\(=\)\(\frac{x^2+4x+3}{x^2-9}\)\(-\)\(\frac{4x-x^2-3}{x^2-9}\)\(+\)\(\frac{2x-2x^2}{x^2-9}\)

\(=\)\(\frac{x^2+4x+3-4x+x^2+3+2x-2x^2}{x^2-9}\)\(=\)\(\frac{6+2x}{\left(x-3\right)\left(x+3\right)}\)\(=\)\(\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\)\(\frac{2}{x-3}\)

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

\(=\dfrac{x+2}{x+2}=1\)

b: \(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\left(\dfrac{x+1-2x+2}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x+2}\right)\)

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

\(=\dfrac{\left(x+2\right)\left(x-1\right)\left(x+1\right)}{2\left(x+5\right)}\cdot\dfrac{-\left(x^2-x-6\right)+x^2-1}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}\)

\(=\dfrac{-x^2+x+6+x^2-1}{2\left(x+5\right)}=\dfrac{x+5}{2\left(x+5\right)}=\dfrac{1}{2}\)

2 tháng 7 2021

\(1,\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\)

\(x-2\sqrt{x}-3\sqrt{x}+6\)

\(x-5\sqrt{x}+6\)

\(2,\left(x+2\right)\left(x-3\right)+x\left(x+1\right)\)

\(x^2+2x-3x-6+x^2+x\)

\(2x^2-6\)

3 tháng 3 2020

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)

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

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

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

\(=\frac{-3}{x-3}\)