Bài 2:
a)
\(C=\dfrac{2x+1}{4x-2}+\dfrac{1-2x}{4x+2}-\dfrac{2}{1-4x^2}\left(x\ne\pm\dfrac{1}{2}\right)\\ =\dfrac{2x+1}{2\left(2x-1\right)}+\dfrac{1-2x}{2\left(2x+1\right)}+\dfrac{2}{4x^2-1}\\ =\dfrac{\left(2x+1\right)^2}{2\left(2x-1\right)\left(2x+1\right)}+\dfrac{\left(1-2x\right)\left(2x-1\right)}{2\left(2x-1\right)\left(2x+1\right)}+\dfrac{4}{2\left(2x+1\right)\left(2x+1\right)}\\ =\dfrac{\left(2x+1\right)^2}{2\left(2x+1\right)\left(2x-1\right)}-\dfrac{\left(2x-1\right)^2}{2\left(2x+1\right)\left(2x-1\right)}+\dfrac{4}{2\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2+4}{2\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{4x^2+4x+1-4x^2+4x-1+4}{2\left(2x+1\right)\left(2x-1\right)}=\dfrac{8x+4}{2\left(2x+1\right)\left(2x-1\right)}=\dfrac{4\left(2x+1\right)}{2\left(2x+1\right)\left(2x-1\right)}=\dfrac{2}{2x-1}\)
b)
\(D=\dfrac{3x+1}{x^2-2x+1}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\left(x\ne\pm1\right)\\ =\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}-\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(3x+1\right)\left(x+1\right)-\left(x-1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{3x^2+3x+x+1-\left(x^2-2x+1\right)-\left(x^2-x+3x-3\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+3x+x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{\left(x+3\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x+3}{x^2-2x+1}\)
Bài 1:
a. \(A=\dfrac{1}{x+1}-\dfrac{1}{x-1}-\dfrac{2x^2}{1-x^2}\left(dk:x\ne\pm1\right)\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x^2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x-1-x-1+2x^2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2x^2-2}{x^2-1}=\dfrac{2\left(x^2-1\right)}{x^2-1}=2\)
b. \(B=\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{2x-1}{x^2+x+1}-\dfrac{6}{x-1}\left(dk:x\ne1\right)\)
\(=\dfrac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(2x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{4x^2-3x+17+2x^2-3x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-12}{x^2+x+1}\)
c. \(C=\dfrac{3x+2}{x^2-2x+1}-\dfrac{6}{x^2-1}-\dfrac{3x-2}{x^2+2x+1}\left(dk:x\ne\pm1\right)\)
\(=\dfrac{\left(3x+2\right)\left(x+1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}-\dfrac{6\left(x^2-1\right)}{\left(x-1\right)^2\left(x+1\right)^2}-\dfrac{\left(3x-2\right)\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}\)
\(=\dfrac{\left(3x+2\right)\left(x^2+2x+1\right)-6x^2+6-\left(3x-2\right)\left(x^2-2x+1\right)}{\left(x-1\right)^2\left(x+1\right)^2}\)
\(=\dfrac{3x^3+6x^2+3x+2x^2+4x+2-6x^2+6-\left(3x^3-6x^2+3x-2x^2+4x-2\right)}{\left(x^2-1\right)^2}\\ =\dfrac{10x^2+10}{x^4-2x^2+1}\)
d. \(D=\dfrac{18}{\left(x-3\right)\left(x^2-9\right)}-\dfrac{3}{x^2-6x+9}-\dfrac{x}{x^2-9}\left(dk:x\ne\pm3\right)\)
\(=\dfrac{18}{\left(x-3\right)^2\left(x+3\right)}-\dfrac{3\left(x+3\right)}{\left(x-3\right)^2\left(x+3\right)}-\dfrac{x\left(x-3\right)}{\left(x-3\right)^2\left(x+3\right)}\)
\(=\dfrac{18-3x-9-x^2+3x}{\left(x-3\right)^2\left(x+3\right)}\)
\(=\dfrac{9-x^2}{\left(x^2-9\right)\left(x-3\right)}=\dfrac{1}{3-x}\)
Bài 3:
a) \(\dfrac{x+6}{x^2-4}-\dfrac{2}{x^2+2x}\left(x\ne0;x\ne\pm2\right)\)
\(=\dfrac{x+6}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\)
\(=\dfrac{2\left(x+6\right)}{2\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x+12-2x+4}{2\left(x-2\right)\left(x+2\right)}=\dfrac{16}{2\left(x^2-4\right)}=\dfrac{8}{x^2-4}\)
b) \(\dfrac{1}{3y-1}-\dfrac{1}{3y+1}-\dfrac{2y-3}{1-9y^2}\left(y\ne\pm\dfrac{1}{3}\right)\)
\(=\dfrac{3y+1}{\left(3y-1\right)\left(3y+1\right)}-\dfrac{3y-1}{\left(3y-1\right)\left(3y+1\right)}+\dfrac{2y-3}{\left(3y-1\right)\left(3y+1\right)}\)
\(=\dfrac{3y+1-3y+1+2y-3}{\left(3y-1\right)\left(3y+1\right)}\) \(=\dfrac{2y-1}{9y^2-1}\)
c) \(\dfrac{3a+5}{6a^3b}-\dfrac{5-11a}{6a^3b}=\dfrac{3a+5-5+11a}{6a^3b}=\dfrac{14a}{6a^3b}=\dfrac{7}{3a^2b}\left(a,b\ne0\right)\)
