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30 tháng 1 2017

\(P=\frac{\left(\frac{2x-3}{4x^2-12x+5}+\frac{2x-8}{13x-2x^2-20}-\frac{3}{2x-1}\right)}{\left(\frac{21+2x-8x^2}{4x^2+4x-3}\right)}+1\)

\(=\frac{\left(\frac{2x-3}{4x^2-2x-10x+5}+\frac{-2\left(4-x\right)}{8x-2x^2+5x-20}-\frac{3}{2x-1}\right)}{\left(\frac{21-12x+14x-8x^2}{4x^2+6x-2x-3}\right)}+1\)

\(=\frac{\left(\frac{2x-3}{2x\left(2x-1\right)-5\left(2x-1\right)}+\frac{-2\left(4-x\right)}{2x\left(4-x\right)-5\left(4-x\right)}-\frac{3}{2x-1}\right)}{\left(\frac{3\left(7-4x\right)+2x\left(7-4x\right)}{2x\left(2x+3\right)-\left(2x+3\right)}\right)}+1\)

\(=\frac{\left(\frac{2x-3}{\left(2x-1\right)\left(2x-5\right)}+\frac{-2\left(4-x\right)}{\left(2x-5\right)\left(4-x\right)}-\frac{3}{2x-1}\right)}{\frac{\left(7-4x\right)\left(3+2x\right)}{\left(2x+3\right)\left(2x-1\right)}}+1\)

\(=\frac{\left(\frac{2x-3}{\left(2x-1\right)\left(2x-5\right)}+\frac{-2\left(2x-1\right)}{\left(2x-1\right)\left(2x-5\right)}-\frac{3\left(2x-5\right)}{\left(2x-1\right)\left(2x-5\right)}\right)}{\frac{7-4x}{2x-1}}+1\)

\(=\frac{2x-3-4x+2-6x+15}{\left(2x-1\right)\left(2x-5\right)}\times\frac{2x-1}{7-4x}+1\)

\(=\frac{14-8x}{2x-5}\times\frac{1}{7-4x}+1\)

\(=\frac{2\left(7-4x\right)}{2x-5}\times\frac{1}{7-4x}+\frac{2x-5}{2x-5}\)

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

\(=\frac{2x-3}{2x-5}\)

26 tháng 2 2022

hic, mk chx học

23 tháng 3 2019

\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\) \(ĐK:x\ne-1;x\ne-3\)

\(\Leftrightarrow\frac{4x}{x^2+4x+3}-\frac{x^2+4x+3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)}{2\left(x+3\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)\left(x+3\right)}\right]\)

\(\Leftrightarrow\frac{4x-x^2-4x-3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)-x-3}{2\left(x+3\right)\left(x+1\right)}\right]\)

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

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

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

\(\Leftrightarrow-x^2-3=3x-3\)

\(\Leftrightarrow-x^2-3x=0\)

\(\Leftrightarrow-x\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\left(loại\right)\end{cases}}\) 

Vậy x = 0 

23 tháng 3 2019

\(ĐK:x\ne\frac{-1}{2};x\ne\frac{-3}{2}\)

\(\frac{3}{2x+1}=\frac{6}{2x+3}+\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow\frac{3}{2x+1}-\frac{6}{2x+3}=\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow\frac{3\left(2x+3\right)-6\left(2x+1\right)}{\left(2x+1\right)\left(2x+3\right)}=\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow\frac{6x+9-12x-6}{4x^2+8x+3}=\frac{8}{4x^2+8x+3}\)

\(\Leftrightarrow-6x+3=8\)

\(\Leftrightarrow x=-\frac{5}{6}\)

Vậy ...