Rút gọn: \(\frac{x+3}{2x-2}-\frac{4}{x^2-1}\times\frac{x+1}{2}\)
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2 tháng 2 2022
\(=\dfrac{4x\left(x+1\right)+1}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{\left(2x-1\right)^2}{2x+1}\right)-\dfrac{1}{2x}\)
\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\left(\dfrac{-\left(2x-1\right)}{2x+1}+\dfrac{2x-1}{\left(2x+1\right)^2}\right)-\dfrac{1}{2x}\)
\(=\dfrac{\left(2x+1\right)^2}{4x^2}\cdot\dfrac{-\left(2x-1\right)\left(2x+1\right)+2x-1}{\left(2x+1\right)^2}-\dfrac{1}{2x}\)
\(=\dfrac{-4x^2+1+2x-1}{4x^2}-\dfrac{1}{2x}\)
\(=\dfrac{-4x^2+2x}{4x^2}-\dfrac{1}{2x}\)
\(=\dfrac{-2x\left(2x-1\right)}{2x\cdot2x}-\dfrac{1}{2x}\)
\(=\dfrac{-2x+1-1}{2x}=\dfrac{-2x}{2x}=-1\)
27 tháng 5 2016
\(A=\left(\frac{1+2x}{2.\left(2+x\right)}-\frac{x}{3.\left(x-2\right)}+\frac{2x^2}{3.\left(4-x^2\right)}\right).\frac{24-12x}{6+13x}\)
\(=\left[\frac{3.\left(1+2x\right)\left(2-x\right)-2x\left(x+2\right)+4x^2}{2.3.\left(x+2\right)\left(2-x\right)}\right].\frac{24-12x}{6+13x}\)
\(=\frac{6+9x-6x^2-2x^2-4x+4x^2}{6.\left(4-x^2\right)}.\frac{24-12x}{6+13x}\)
\(=\frac{6+5x-4x^2}{6.\left(4-x^2\right)}.\frac{12.\left(2-x\right)}{6+13x}\) \(=\frac{\left(6+5x-4x^2\right).2}{\left(x+2\right)\left(6+13x\right)}=\frac{12+10x-8x^2}{13x^2+32x+12}\)
18 tháng 6 2016
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right):\frac{4}{4x^2-4}\)
\(=\left(\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+2\right)}+\frac{6}{2.\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\frac{4}{4\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}.\left(x-1\right)\left(x+1\right)=\frac{4}{2}=2\)
31 tháng 1 2022
\(A=\left(\dfrac{x^2-2x+1}{x^2+x+1}-\dfrac{-2x^2+4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right):\dfrac{2x}{x^3+x}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}=\dfrac{x^2+1}{2}\)
PA
21 tháng 10 2016
\(\frac{2+x}{2-x}\div\frac{4x^2}{4-4x+x^2}\times\left(\frac{2}{2-x}-\frac{8}{8+x^3}\times\frac{4-2x+x^2}{2-x}\right)\)
\(=\frac{2+x}{2-x}\times\frac{4-4x+x^2}{4x^2}\times\left(\frac{2}{2-x}-\frac{8}{\left(2+x\right)\left(4-2x+x^2\right)}\times\frac{4-2x+x^2}{2-x}\right)\)
\(=\frac{2+x}{2-x}\times\frac{\left(2-x\right)^2}{4x^2}\times\left(\frac{2\left(2+x\right)}{\left(2+x\right)\left(2+x\right)}-\frac{8}{\left(2+x\right)\left(2-x\right)}\right)\)
\(=\frac{\left(2+x\right)\left(2-x\right)}{4x^2}\times\frac{4+2x-8}{\left(2+x\right)\left(2-x\right)}\)
\(=\frac{2\left(2+x-4\right)}{4x^2}\)
\(=\frac{x-2}{2x^2}\)
21 tháng 5 2016
- Ta chứng minh bất đẳng thức phụ dưới đây: \(\frac{1}{\sqrt{x}\left(x+1\right)}=\frac{\sqrt{x}}{x\left(x+1\right)}=\sqrt{x}\left(\frac{1}{x}-\frac{1}{x+1}\right)=\sqrt{x}\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x+1}}\right)\)\(=\left(1+\frac{\sqrt{x}}{\sqrt{x+1}}\right)\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)< 2\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\right)\)
Áp dụng : \(\frac{1}{\sqrt{1}.2}< 2.\left(1-\frac{1}{\sqrt{2}}\right)\)
\(\frac{1}{\sqrt{2}.3}< 2.\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)\)
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\(\frac{1}{\sqrt{2015}.2016}< 2.\left(\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\right)\)
Cộng các BĐT trên với nhau được : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}}< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\right)=2\left(1-\frac{1}{\sqrt{2016}}\right)< 2\left(1-\frac{1}{\sqrt{2025}}\right)=\frac{88}{45}\)
Từ đó suy ra đpcm
21 tháng 5 2016
Cái ............... là gì vậy bn
18 tháng 6 2016
ĐK: x khác 1 ; -1
\(B=\frac{1}{x-1}-\frac{x^3-x}{x^2+1}.\left(\frac{1}{1-2x+x^2}+\frac{1}{1-x^2}\right)\)
\(=\frac{1}{x-1}-\frac{x^3-x}{x^2+1}.\left(\frac{1+x}{\left(1-x\right)^2\left(1+x\right)}+\frac{1-x}{\left(1-x\right)^2\left(1+x\right)}\right)\)
=\(\frac{1}{x-1}-\frac{x\left(x-1\right)\left(x+1\right)}{x^2+1}.\frac{2}{\left(1-x\right)^2\left(1+x\right)}=\frac{1}{x-1}-\frac{2x}{\left(x^2+1\right)\left(1-x\right)}\)
\(=\frac{x^2+1}{\left(x^2+1\right)\left(x-1\right)}+\frac{2x}{\left(x^2+1\right)\left(x-1\right)}=\frac{x^2+2x+1}{\left(x^2+1\right)\left(x-1\right)}=\)
19 tháng 7 2019
\(A = \left( {\dfrac{3}{{2x + 4}} + \dfrac{x}{{2 - x}} - \dfrac{{2{x^2} + 3}}{{{x^2} - 4}}} \right):\dfrac{{2x - 1}}{{4x - 8}}\\ A = \left[ {\dfrac{3}{{2\left( {x + 2} \right)}} - \dfrac{x}{{x - 2}} - \dfrac{{2{x^2} + 3}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right].\dfrac{{4x - 8}}{{2x - 1}}\\ A = \dfrac{{3\left( {x - 2} \right) - 2x\left( {x + 2} \right) - 2\left( {2{x^2} + 3} \right)}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{4\left( {x - 2} \right)}}{{2x - 1}}\\ A = \dfrac{{3x - 6 - 2{x^2} - 4x - 4{x^2} - 6}}{{x + 2}}.\dfrac{2}{{2x - 1}}\\ A = \dfrac{{ - x - 12 - 6{x^2}}}{{x + 2}}.\dfrac{2}{{2x - 1}}\\ A = \dfrac{{ - 2x - 24 - 12{x^2}}}{{2{x^2} - x + 4x - 2}}\\ A = \dfrac{{ - 12{x^2} - 2x - 24}}{{2{x^2} + 3x - 2}}\\ \)
\(\frac{x+3}{2x-2}-\frac{4}{x^2-1}.\frac{x+1}{2}\)
\(=\frac{x+3}{2x-2}-\left(\frac{4}{x^2-1}.\frac{x+1}{2}\right)\)
\(=\frac{x+3}{2\left(x-1\right)}-\frac{4\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+3}{2\left(x-1\right)}-\frac{4}{2\left(x-1\right)}\)
\(=\frac{x+3-4}{2\left(x-1\right)}\)
\(=\frac{x-1}{2\left(x-1\right)}\)
\(=\frac{1}{2}\)