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4 tháng 7 2019

\(=x^3-\left(\frac{1}{2}\right)^3=x^3-\frac{1}{8}\)

4 tháng 7 2019

\(\left(x-\frac{1}{2}\right)\left(x^2+\frac{1}{2}x+\frac{1}{4}\right)\)

\(=x.\left(x^2+\frac{1}{2}x+\frac{1}{4}\right)-\frac{1}{2}.\left(x^2+\frac{1}{2}x+\frac{1}{4}\right)\)

\(=x^3+\frac{1}{2}x^2+\frac{1}{4}x-\frac{1}{2}x^2-\frac{1}{4}x-\frac{1}{8}\)

\(=x^3-\frac{1}{8}\)

Làm bằng cách hằng đẳng thức còn rắc rối hơn thà nhân lên luôn.

2 tháng 7 2019

\(\left[\frac{2}{3x}-\frac{2}{x+1}\left(\frac{x+1}{3x}-x-1\right)\right]:\frac{x-1}{x}\)

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

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

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

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

16 tháng 12 2020

\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-3}+...+\frac{1}{x-4}-\frac{1}{x-5}\)

\(=\frac{1}{x}-\frac{1}{x-5}=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}=\frac{-5}{x\left(x-5\right)}\)

16 tháng 12 2020

\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+...+\frac{1}{x-4}-\frac{1}{x-5}\)

\(=\frac{1}{x}-\frac{1}{x-5}\)

\(=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}\)

\(=\frac{x-5-x}{x\left(x-5\right)}\)

\(=-\frac{5}{x\left(x-5\right)}\)

24 tháng 11 2019

b) \(\left[\frac{2}{3x}-\frac{2}{x+1}.\left(\frac{x+1}{3x}-x-1\right)\right]:\frac{x-1}{x}\)

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

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

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

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

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

24 tháng 11 2019

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

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

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

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

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

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

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

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

\(=\frac{3}{3-x}\left(đpcm\right)\)