1: \(MTC=2\left(x-y\right)\left(x+y\right)\)

\(\dfrac{x-y}{2x^2-4xy+2y^2}=\dfrac{x-y}{2\left(x-y\right)^2}=\dfrac{1}{2\left(x-y\right)}=\dfrac{1\cdot\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{2\left(x-y\right)\left(x+y\right)}\)

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

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

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

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

2: \(\dfrac{1}{x^2+8x+15}=\dfrac{1}{\left(x+3\right)\left(x+5\right)}=\dfrac{x+3}{\left(x+3\right)^2\cdot\left(x+5\right)}\)

\(\dfrac{1}{x^2+6x+9}=\dfrac{1}{\left(x+3\right)^2}=\dfrac{x+5}{\left(x+3\right)^2\cdot\left(x+5\right)}\)

3: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}=\dfrac{1\cdot\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\dfrac{1}{\left(c-b\right)\left(c-a\right)}=\dfrac{1}{\left(b-c\right)\left(a-c\right)}=\dfrac{a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\dfrac{1}{\left(b-a\right)\left(a-c\right)}=\dfrac{-1}{\left(a-b\right)\left(a-c\right)}=\dfrac{-\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

a) \(\dfrac{3x^2y}{2xy^5}=\dfrac{3x}{2y^4}\)

b) \(\dfrac{3x^2-3x}{x-1}=\dfrac{3x\left(x-1\right)}{x-1}=3x\)

c) \(\dfrac{ab^2-a^2b}{2a^2+a}=\dfrac{ab\left(b-a\right)}{a\left(2a+1\right)}=\dfrac{b\left(b-a\right)}{2a+1}=\dfrac{b^2-ab}{2a+1}\)

d) \(\dfrac{12\left(x^4-1\right)}{18\left(x^2-1\right)}=\dfrac{2\left(x^2-1\right)\left(x^2+1\right)}{3\left(x^2-1\right)}=\dfrac{2\left(x^2+1\right)}{3}\)

`a, (3x^2y)/(2xy^5)`

`= (3x)/(2y^4)`

`b, (3x^2-3x)/(x-1)`

`= (3x(x-1))/(x-1)`

`= 3x`

`c, (ab^2-a^2b)/(2a^2+a)`

`= (b(a-b))/((2a+1))`

`d, (12(x^4-1))/(18(x^2-1)) = (2(x^2+1))/3`.

a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)

b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)

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

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

e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)

\(a,=\dfrac{5x}{4y^3}\times\left(\dfrac{-20y}{x^4}\right)=\dfrac{-100xy}{4x^4y^3}=\dfrac{-25}{x^3y^2}\\ b,=\dfrac{\left(x-4\right)\left(x+4\right)}{\left(x+4\right)}\times\dfrac{x}{2\left(x-4\right)}=\dfrac{x}{2}\)

\(c,=\dfrac{2\left(x+3\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\times\dfrac{2\left(x-2\right)}{\left(x+3\right)^3}=\dfrac{4}{\left(x+3\right)^2.\left(x^2+2x+4\right)}\)

a) \(\dfrac{5x}{4y^3}:\left(-\dfrac{x^4}{20y}\right)=\dfrac{5x}{4y^3}\cdot\left(-\dfrac{20y}{x^4}\right)=\dfrac{5\cdot-5}{y^2\cdot x^3}=\dfrac{-25}{x^3y^2}\)

b) \(\dfrac{x^2-16}{x+4}:\dfrac{2x-8}{x}=\left(x-4\right)\cdot\dfrac{x}{2\left(x-4\right)}=\dfrac{x}{2}\)

c) \(\dfrac{2x+6}{x^3-8}:\dfrac{\left(x+3\right)^3}{2x-4}=\dfrac{2\left(x+3\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\cdot\dfrac{2\left(x-2\right)}{\left(x+3\right)^3}=\dfrac{4}{\left(x^2+2x+4\right)\left(x+3\right)^2}\)

\(\dfrac{3}{x-5}-\dfrac{x+1}{x\left(x-5\right)}\left(dkxd:x\ne0,x\ne5\right)\\ =\dfrac{3x-x-1}{x\left(x-5\right)}=\dfrac{2x-1}{x^2-5x}\)

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\(\dfrac{8\left(y+2\right)}{3x^2}.\dfrac{15x^5}{4\left(y+2\right)^2}\left(dkxd:x\ne0,y\ne-2\right)\\ =\dfrac{8}{4}.\dfrac{15x^2.x^3}{3x^2}=10x^3\)

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\(\dfrac{8\left(y-1\right)}{3x^2-3}:\dfrac{4\left(y-1\right)^3}{x^2-2x+1}\left(dkxd:x\ne1,x\ne-1\right)\\ =\dfrac{8\left(y-1\right)}{3\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)^2}{4\left(y-1\right)^3}\\ =\dfrac{2\left(x-1\right)}{3\left(x+1\right)\left(y-1\right)^2}\)

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: \(x\ne\pm1\)

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

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

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

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