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

14.C

15.D

Câu 14: C

Câu 15: D

\(=\dfrac{1}{y-x}\cdot x^3\cdot\left(x-y\right)=-x^3\)

\(a,\dfrac{21x^2y^3}{24x^3y^2}=\dfrac{7y}{8x}\)

\(b,\dfrac{15xy^3\left(x^2-y^2\right)}{20x^2y\left(x+y\right)^2}=\dfrac{15xy^3\left(x-y\right)\left(x+y\right)}{20x^2y\left(x+y\right)^2}=\dfrac{3y^2\left(x-y\right)}{4x\left(x+y\right)}=\dfrac{3xy^2-3y^3}{4x^2+4xy}\)

a) Ta có: \(\dfrac{21x^2y^3}{24x^3y^2}\)

\(=\dfrac{21x^2y^3:3x^2y^2}{24x^3y^2:3x^2y^2}\)

\(=\dfrac{7y}{8x}\)

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

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

\(=2x^2-4xy+\dfrac{15}{4}y^2\)

b: \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)

\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)

\(=2x^2+2x+13-2x^2+2\)

=2x+15

a) \(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2=2x^2-4xy+\dfrac{15}{4}y^2\)

b) \(=x^2-4x+4+x^2+6x+9-2x^2+2\)

\(=2x+15\)

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

\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)

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

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

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

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

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

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

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

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

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

=>A+4=\(\frac{x^2+2xy+y^2}{xy}=\frac{\left(x+y\right)^2}{xy}>0\forall x,y\) thỏa mãn ĐKXĐ

=>A>-4∀x thỏa mãn ĐKXĐ