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

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

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

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

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

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

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

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

\(=\frac{3}{3-x}\left(đpcm\right)\)

a: \(VT=\dfrac{a^2\left(a-4\right)-\left(a-4\right)}{\left(a-2\right)\left(a^2+2a+4\right)-7a\left(a-2\right)}\)

\(=\dfrac{\left(a-4\right)\left(a-1\right)\left(a+1\right)}{\left(a-1\right)\left(a^2-5a+4\right)}\)

\(=\dfrac{\left(a-4\right)\left(a+1\right)}{\left(a-4\right)\left(a-1\right)}=\dfrac{a+1}{a-1}=VP\)

b: \(VT=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)

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

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

14 tháng 2 2020

Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\) 

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

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

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

\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\) 

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

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

\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)

16 tháng 12 2020

Bài 1.

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

b) \(\frac{2x+1}{2x^2-x}+\frac{32x^2}{1-4x^2}+\frac{1-2x}{2x^2+x}=\frac{2x+1}{x\left(2x-1\right)}+\frac{-32x^2}{4x^2-1}+\frac{1-2x}{x\left(2x+1\right)}\)

\(=\frac{\left(2x+1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{\left(1-2x\right)\left(2x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\frac{4x^2+4x+1}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\frac{4x^2+4x+1-32x^3-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-32x^3+8x}{x\left(2x-1\right)\left(2x+1\right)}\)

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

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

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

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

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

\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}\times\frac{x-1}{4x}=\frac{1}{x+1}\)

Bài 3.

N = ( 4x + 3 )2 - 2x( x + 6 ) - 5( x - 2 )( x + 2 )

= 16x2 + 24x + 9 - 2x2 - 12x - 5( x2 - 4 )

= 14x2 + 12x + 9 - 5x2 + 20

= 9x2 + 12x + 29

= 9( x2 + 4/3x + 4/9 ) + 25

= 9( x + 2/3 )2 + 25 ≥ 25 > 0 ∀ x 

=> đpcm