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TT
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NT
26 tháng 2 2021 lúc 23:25

Ta có: \(\left(\dfrac{x+2}{3x}+\dfrac{2}{x+1}-3\right):\dfrac{2-4x}{x+1}\cdot\dfrac{3x}{x^2-3x-1}\)

\(=\left(\dfrac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\dfrac{6x}{3x\left(x+1\right)}-\dfrac{9x\left(x+1\right)}{3x\left(x+1\right)}\right):\dfrac{2-4x}{x+1}\cdot\dfrac{3x}{x^2-3x-1}\)

\(=\dfrac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\dfrac{x+1}{2-4x}\cdot\dfrac{3x}{x^2-3x-1}\)

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

\(=\dfrac{-2\left(4x^2-1\right)}{3x\cdot2\cdot\left(1-2x\right)}\cdot\dfrac{3x}{x^2-3x-1}\)

\(=\dfrac{2\left(1-2x\right)\left(2x+3\right)}{6x\left(1-2x\right)}\cdot\dfrac{3x}{x^2-3x-1}\)

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

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LN
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NT
1 tháng 1 2022 lúc 11:08

\(=\dfrac{2x+6}{x\left(3x-1\right)}+\dfrac{x+3}{3x-1}\)

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

\(=\dfrac{x^2+5x+6}{x\left(3x-1\right)}\)

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DL
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NT
23 tháng 12 2021 lúc 11:22

a: \(=x^2+2x-8-x^2-2x-1=-9\)

b: \(=\dfrac{x^2+6x+9+3x-9+2x^2-18x}{x\left(x-3\right)\left(x+3\right)}\)

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

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LN
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NT
14 tháng 8 2021 lúc 21:42

a: \(\dfrac{x^2}{3x+6}+\dfrac{4x+4}{3x+6}=\dfrac{x^2+4x+4}{3x+6}=\dfrac{x+2}{3}\)

b: \(\dfrac{x+3}{x}+\dfrac{x}{3-x}-\dfrac{9}{3x-x^2}\)

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

=0

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HH
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NT
28 tháng 7 2021 lúc 22:23

a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)

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

\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)

\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)

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

 

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NT
28 tháng 7 2021 lúc 22:31

c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)

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

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

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

\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)

 

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NL
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NM
24 tháng 12 2021 lúc 7:02

\(=\dfrac{3x+2}{\left(x-1\right)^2}-\dfrac{6}{\left(x-1\right)\left(x+1\right)}+\dfrac{2-3x}{\left(x+1\right)^2}\\ =\dfrac{\left(3x+2\right)\left(x+1\right)^2-6\left(x^2-1\right)+\left(2-3x\right)\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}\\ =\dfrac{10x^2+10}{\left(x-1\right)^2\left(x+1\right)^2}\)

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LV
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NM
11 tháng 11 2021 lúc 14:40

\(=\dfrac{-3\left(x-1\right)}{\left(1+x\right)^2}\cdot\dfrac{x+1}{6\left(x-1\right)\left(x+1\right)}\left(x\ne\pm1\right)\\ =\dfrac{-1}{2\left(x+1\right)^2}\)

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H24
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TN
23 tháng 7 2023 lúc 16:49

\(a,\dfrac{8y}{3x^2}.\dfrac{9x^2}{4y^2}=\dfrac{72x^2y}{12x^2y^2}=\dfrac{6}{y}\\b,\dfrac{3x+x^2}{x^2+x+1}.\dfrac{3x^3-3}{x+3}=\dfrac{x\left(x+3\right)3\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x+3\right)}=3x\left(x-1\right)=3x^2-3x \)

\(c,\dfrac{2x^2+4}{x-3}.\dfrac{3x+1}{x-1}.\dfrac{6-2x}{x^2+2}=\dfrac{2\left(x^2+2\right)\left(3x+1\right)2\left(3-x\right)}{\left(x-3\right)\left(x-1\right)\left(x^2+2\right)}=\dfrac{-4\left(3x+1\right)}{x-1}=\dfrac{-12x-4}{x-1}\)

\(d,\dfrac{2x^2}{3y^3}:\left(-\dfrac{4x^3}{21y^2}\right)=\dfrac{-2x^2.21y^2}{3y^3.4x^3}=\dfrac{-42x^2y^2}{12x^3y^3}=\dfrac{-7}{2xy}\)

\(e,\dfrac{2x+10}{x^3-64}:\dfrac{\left(x+5\right)^2}{2x-8}=\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}.\dfrac{2\left(x-4\right)}{\left(x+5\right)^2}=\dfrac{4}{\left(x+5\right)\left(x^2+4x+16\right)}=\dfrac{4}{x^3+9x^2+16x+80}\)

\(f,\dfrac{1}{x+y}\left(\dfrac{x+y}{xy}-x-y\right)-\dfrac{1}{x^2}:\dfrac{y}{x}=\dfrac{1}{x+y}\left(\dfrac{\left(x+y\right)\left(1-xy\right)}{xy}\right)-\dfrac{x}{x^2y}=\dfrac{1-xy}{xy}-\dfrac{x}{x^2y}=\dfrac{-x^2y}{x^2y}=-1\)

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HH
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HN
29 tháng 7 2021 lúc 9:23

\(\dfrac{4x^2-3x+5}{x^3-1}-\dfrac{1+2x}{x^2+x+1}-\dfrac{6}{x-1}\)

\(\Leftrightarrow\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1+2x}{x^2+x+1}-\dfrac{6}{x-1}\)

\(ĐKXĐ:x\ne1\)

\(\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{(1+2x)\left(x-1\right)}{(x^2+x+1)\left(x-1\right)}-\dfrac{6\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}\)

\(\Rightarrow4x^2-3x+5-\left(1+2x\right)\left(x-1\right)-6\left(x^2+x+1\right)\)

\(\Rightarrow4x^2-3x+5-\left(x-1+2x^2-2x\right)-6x^2-6x-6\)

\(\Rightarrow4x^2-3x+5-x+1-2x^2+2x-6x^2-6x-6\)

\(\Rightarrow-4x^2-8x\)

⇒-4x(x-4)

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H24
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NT
29 tháng 7 2021 lúc 23:51

a) Ta có: \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)

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

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

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

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

b) Ta có: \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)

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

\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)

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

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

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

c) Ta có: \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)

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

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

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

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

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

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