a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)

\(=\left(\dfrac{\left(2x+1\right)\left(2x+1\right)}{2x^2-1}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)

\(=\left(\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{2x^2-1}\right):\dfrac{4x}{10x-5}\)

\(=\left(\dfrac{\left(2x+1-2x-1\right)\left(2x+1+2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)

\(=\dfrac{4x}{2x^2-1}.\dfrac{5\left(2x-1\right)}{4x}\)

\(=\dfrac{5}{2x+1}\)

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

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

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

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

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

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

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

c) d) Tự làm đi mình làm biếng quass >.< ^^

\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)

a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)

\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)

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

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

a: \(=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}\)

b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)

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

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

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

=1/3

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

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

ĐKXĐ: \(x\ne\pm y\)

\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)}-\dfrac{2xy^2}{x^4-2x^2y^2+y^4}+\dfrac{7^2}{\left(x^2-y^2\right)\left(x+y\right)}\)

\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)}-\dfrac{2xy^2}{\left(\left(x+y\right).\left(x-y\right)\right)^2}+\dfrac{49}{\left(x+y\right)^2.\left(x-y\right)}\)

\(A=\dfrac{x^2}{\left(x-y\right)^2.\left(x+y\right)^{ }}-\dfrac{2xy^2}{\left(x-y\right)^2.\left(x+y\right)^2}+\dfrac{49}{\left(x+y\right)^2.\left(x-y\right)}\)

\(A=\dfrac{x^2.\left(x+y\right)-2xy^2+49.\left(x-y\right)}{\left(x-y\right)^2.\left(x+y\right)^2}\)

\(A=\dfrac{x^3+x^2y-2xy^2+49x-49y}{\left(x-y\right)^2.\left(x+y\right)^2}\)

ĐKXĐ: \(x\ne\pm1\)

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

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

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

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

a, \(\dfrac{x^2+y^2}{4\left(x+y\right)}+\dfrac{2xy}{4\left(x+y\right)}\)=\(\dfrac{x^2+2xy+y^2}{4\left(x+y\right)}\)   = \(\dfrac{\left(x+y\right)^2}{4\left(x+y\right)}\)  =\(\dfrac{x+y}{4}\) 

a. \(\dfrac{x^2+y^2}{4\left(x+y\right)}+\dfrac{2xy}{4\left(x+y\right)}\)

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

\(=\dfrac{\left(x+y\right)^2}{4\left(x+y\right)}\)

\(=\dfrac{x+y}{4}\)

b. \(\dfrac{x+5}{2x-2}-\dfrac{4}{x^2-1}:\dfrac{2}{x+1}\)

\(=\dfrac{x+5}{2\left(x-1\right)}-\dfrac{4}{\left(x+1\right)\left(x-1\right)}:\dfrac{2}{x+1}\)

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

\(=\dfrac{x+5}{2\left(x-1\right)}-\dfrac{4}{2\left(x-1\right)}\)

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