a, \(\frac{4x+1}{2}-\frac{3x+2}{3}=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{12x+3-6x-4}{6}=\frac{6x-1}{6}\)

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

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

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

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

\(\frac{4x+1}{2}-\frac{3x+2}{3}\)

\(=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{6x-1}{6}\)

tương tự đến hết nha a hay cj gì đps ! 

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

                                                \(=\frac{12.x+3-6.x-4}{6}\)

                                                   \(=\frac{6.x-1}{6}\)

b)\(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}\)

\(=\frac{x+3}{\left(x-1\right).\left(x+1\right)}-\frac{1}{x.\left(x+1\right)}\)

\(=\frac{x.\left(x+3\right)-\left(x-1\right)}{x.\left(x-1\right).\left(x+1\right)}\)

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

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

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

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

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

c)\(\frac{3}{2.x^2+2.x}+\frac{2.x-1}{x^2-1}-\frac{1}{2}\)

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

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

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

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

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

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

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

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

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

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

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

Bài 1:

a) Ta có: \(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)

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

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

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

\(=\frac{2x-4y+xy+2y^2+4}{\left(x-2y\right)\cdot\left(x+2y\right)}\)

b) Ta có: \(\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}\)

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

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

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

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

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

\(=\frac{2x-2y}{x^2+xy+y^2}\)

c) Ta có: \(\frac{xy}{2x-y}-\frac{x^2-1}{y-2x}\)

\(=\frac{xy}{2x-y}+\frac{x^2-1}{2x-y}\)

\(=\frac{x^2+xy-1}{2x-y}\)

d) Ta có: \(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}\)

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

\(=\frac{2x^2-2y^2+2y^2}{x}\)

\(=\frac{2x^2}{x}=2x\)

Bài 2:

a) Ta có: \(\frac{4x+1}{2}-\frac{3x+2}{3}\)

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

\(=\frac{12x+3-6x-4}{6}\)

\(=\frac{6x-1}{6}\)

b) Ta có: \(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\)

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

\(=\frac{x^2-9-x^2+9}{x\left(x-3\right)}=\frac{0}{x\left(x-3\right)}=0\)

c) Ta có: \(\frac{x+3}{x^2+1}-\frac{1}{x^2+2}\)

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

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

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

e) Ta có: \(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}\)

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

\(=\frac{3\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}+\frac{2x\left(2x-1\right)}{2x\left(x+1\right)\left(x-1\right)}-\frac{2\cdot2\cdot\left(x+1\right)\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}\)

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

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

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

d) Ta có: \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)

\(=\frac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\frac{3x+2-12x+8+10x-8}{\left(3x-2\right)\left(3x+2\right)}\)

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

f) Ta có: \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)

\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)

\(=\frac{3x\cdot2\cdot\left(x-y\right)}{10\left(x+y\right)\left(x-y\right)}-\frac{x\cdot\left(x+y\right)}{10\left(x-y\right)\left(x+y\right)}\)

\(=\frac{6x^2-6xy-x^2-xy}{10\left(x-y\right)\left(x+y\right)}\)

\(=\frac{5x^2-7xy}{10\left(x-y\right)\left(x+y\right)}\)

a)\(ĐKXĐ:x\ne0;-1\)

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

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

1,\(\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)

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

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

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

3,1/x(y-x)-1/y(y-x)=y/xy(y-x)-x/xy(y-x)=(y-x)/xy(x-y)=1/xy

a) \(\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}:\frac{10x-10y}{x^3+y^3}\)

\(=\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}.\frac{x^3+y^3}{10x-10y}\)

\(=\frac{3\left(x^2-2xy+y^2\right)}{5\left(x^2-xy+y^2\right)}.\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{10\left(x-y\right)}\)

\(=\frac{3\left(x^2-2xy+y^2\right)}{5}.\frac{x+y}{10\left(x-y\right)}\)

\(=\frac{3\left(x-y\right)^2}{5}.\frac{x+y}{10\left(x-y\right)}\)

\(=\frac{3\left(x-y\right)}{5}.\frac{x+y}{10}\)

\(=\frac{3x^2-3y^2}{50}\)

c) \(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)-\frac{x^2-y^2}{\left(x-y\right)^2}\)

\(=\frac{2}{xy}:\frac{y-x}{xy}-\frac{\left(x+y\right)\left(x-y\right)}{\left(x-y\right)^2}\)

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

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

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

d) \(\frac{\frac{x-y}{x+y}-\frac{x+y}{x-y}}{1-\frac{x^2}{x^2+y^2}}\)

\(=\frac{\frac{x^2-2xy+y^2}{x^2-y^2}-\frac{x^2+2xy+y^2}{x^2-y^2}}{\frac{y^2}{x^2+y^2}}\)

\(=\frac{\frac{2x^2+2y^2}{x^2-y^2}}{\frac{y^2}{x^2+y^2}}\)

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

\(=\frac{2x^4+4x^2y^2+2y^4}{x^2y^2-y^4}\)