Đáp án+Giải thích các bước giải:

$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\\ a.\ \frac{14xy^{5}( 2x-3y)}{21x^{2} y( 2x-3y)^{2}} ;\ ĐKXĐ:\ x,y\neq 0;2x\neq 3y\\ =\frac{14xy^{4}}{21x( 2x-3y)}\\ b.\ \frac{8xy( 3x-1)^{3}}{12x^{3}( 1-3x)} ;\ ĐKXĐ\ x\neq 0;\ x\neq \frac{1}{3}\\ =\frac{2y( 3x-1)^{3}}{-3x^{2}( 3x-1)} =\frac{2y( 3x-1)^{2}}{-3x^{2}}\\ c.\ \frac{20x^{2} -45}{( 2x+3)^{2}} ;\ ĐKXĐ:\ x\neq -\frac{3}{2}\\ =\frac{5\left( 4x^{2} -9\right)}{( 2x+3)^{2}} =\frac{5( 2x-3)( 2x+3)}{( 2x+3)^{2}} =\frac{5( 2x-3)}{2x+3}\\ d.\ \frac{5x^{2} -10xy}{2( 2y-3)^{2}} ;\ ĐKXĐ:\ y\neq \frac{3}{2}\\ =\frac{5x( x-2y)}{2( 2y-3)^{2}} =\frac{-5x( 2y-3)}{2( 2y-3)^{2}} =\frac{-5x}{2( 2y-3)}\\ e.\ \frac{80x^{3} -125x}{3( x-3) -( x-3)( 8-4x)} ;\ ĐKXĐ:x\neq 3;x\neq \frac{5}{4} \ \\ =\frac{5x\left( 16x^{2} -25\right)}{( x-3)( 3-8+4x)} =\frac{5x( 4x-5)( 4x+5)}{( x-3)( 4x-5)}\\ =\frac{5x( 4x+5)}{x-3}\\ f.\ \frac{9-( x+5)^{2}}{x^{2} +4x+4} ;\ ĐKXĐ:x\neq -2\\ =\frac{( 3-x-5)( 3+x+5)}{( x+2)^{2}} =\frac{( -x-2)( x+2)}{( x+2)^{2}} =\frac{-( x+2)^{2}}{( x+2)^{2}} =-1\\ g.\ \frac{32x-8x^{2} +2x^{3}}{x^{3} +64} \ ;\ ĐKXĐ:x\neq -4\\ =\frac{2x\left( x^{2} -4x+16\right)}{( x+4)\left( x^{2} -4x+16\right)} =\frac{2x}{x+4}\\ h.\ \frac{5x^{3} +5x}{x^{4} -1} ;\ ĐKXĐ:x\neq \pm 1\\ =\frac{5x\left( x^{2} +1\right)}{\left( x^{2} -1\right)\left( x^{2} +1\right)} =\frac{5x}{x^{2} -1}\\ i.\ \frac{x^{2} +5x+6}{x^{2} +4x+4} ;\ ĐKXĐ:x\neq -2\\ =\frac{x^{2} +2x+3x+6}{( x+2)^{2}} =\frac{x( x+2) +3( x+2)}{( x+2)^{2}}\\ =\frac{( x+3)( x+2)}{( x+2)^{2}} =\frac{x+3}{x+2}\\ j.\ \frac{10xy^{2}( x+y)}{15xy( x+y)^{3}} ;\ ĐKXĐ:\ x,y\neq 0;\ x\neq -y\\ =\frac{2y}{3( x+y)^{2}}\\ k.\ \frac{x^{2} -xy-x+y}{x^{2} +xy-x-y} ;\ ĐKXĐ:x\neq 1;\ x\neq -y\\ =\frac{x( x-y) -( x-y)}{x( x+y) -( x+y)} =\frac{( x-1)( x-y)}{( x-1)( x+y)} =\frac{x-y}{x+y}\\ l.\ \frac{3x^{2} -12x+12}{x^{4} -8x} ;\ ĐKXĐ:x\neq 0;\ x\neq 2\\ =\frac{3\left( x^{2} -4x+4\right)}{x\left( x^{3} -2^{3}\right)} =\frac{3( x-2)^{2}}{x( x-2)\left( x^{2} +2x+4\right)} =\frac{3( x-2)}{x\left( x^{2} +2x+4\right)}\\ n.\ \frac{7x^{2} +14x+7}{3x^{2} +3x} ;\ ĐKXĐ:x\neq 0;\ x\neq -1\\ =\frac{7\left( x^{2} +2x+1\right)}{3x( x+1)} =\frac{7( x+1)^{2}}{3x( x+1)} =\frac{7( x+1)}{3x}\\ m.\ \frac{2a^{2} -2ab}{ac+ad-bc-bd} ;\ ĐKXĐ:a\neq b;\ c\neq -d\\ =\frac{2a( a-b)}{a( c+d) -b( c+d)} =\frac{2a( a-b)}{( a-b)( c+d)} =\frac{2a}{c+d}\\ o.\ \frac{x^{2} -xy}{y^{2} -x^{2}} ;\ ĐKXĐ:\ x\neq \pm y\\ =\frac{x( x-y)}{( y-x)( y+x)} =\frac{-x( y-x)}{( y-x)( y+x)} =\frac{-x}{y+x}\\ ơ.\ \frac{2x-2y}{x^{2} -2xy+y^{2}} ;\ ĐKXĐ:x\neq y\\ =\frac{2( x-y)}{( x-y)^{2}} =\frac{2}{x-y}\\ p.\ \frac{2-2a}{a^{3} -1} ;\ ĐKXĐ:a\neq 1\\ =\frac{2( 1-a)}{( a-1)\left( a^{2} +a+1\right)} =\frac{-2( a-1)}{( a-1)\left( a^{2} +a+1\right)} =\frac{-2}{a^{2} +a+1}\\ q.\ \frac{x^{3} -6x+9}{x^{2} -8x+15} ;\ ĐKXĐ:\\ =\frac{( x-3)^{2}}{x^{2} -5x-3x+15} =\frac{( x-3)^{2}}{x( x-5) -3( x-5)}\\ =\frac{( x-3)^{2}}{( x-3)( x-5)} =\frac{x-3}{x-5}\\ v.\ \frac{x^{4} -2x^{3}}{2x^{4} -x^{3}} ;\ ĐKXĐ:x\neq 0;\ x\neq \frac{1}{2}\\ =\frac{x^{3}( x-2)}{x^{3}( 2x-1)} =\frac{x-2}{2x-1}\\ u.\ \frac{x^{7} -x^{4}}{x^{6} -1} ;\ ĐKXĐ:x\neq \pm 1\\ =\frac{x^{4}\left( x^{3} -1\right)}{\left( x^{3} -1\right)\left( x^{3} +1\right)} =\frac{x^{4}}{x^{3} +1}\\ ư.\ \frac{( x+2)^{2} -( x-2)^{2}}{16x} ;\ ĐKXĐ:x\neq 0\\ =\frac{( x+2-x+2)( x+2+x-2)}{16x} =\frac{4.2x}{16x} =\frac{1}{2}\\ x.\ \frac{24,5x^{2} -0,5y^{2}}{3,5x^{2} -0,5xy} ;\ ĐKXĐ:\ y\neq 7x\\ =\frac{0,5.\left( 49x^{2} -y^{2}\right)}{0,5x( 7x-y)} =\frac{( 7x-y)( 7x+y)}{( 7x-y)} =7x+y\\ y.\ \frac{a^{3} -3a^{2} +2a-6}{a^{2} +2} ;\ ĐKXĐ:\forall a\\ =\frac{a^{2}( a-3) +2( a-3)}{a^{2} +2} =\frac{\left( a^{2} +2\right)( a-3)}{a^{2} +2} =a-3\\ z.\ \frac{( a-b) .( c-d)}{\left( b^{2} -a^{2}\right)\left( d^{2} -c^{2}\right)} ;\ ĐKXĐ:a\neq \pm b;\ c\neq \pm d\\ =\frac{( b-a)( d-c)}{( b-a)( b+a)( d-c)( d+c)} =\frac{1}{( b+a)( d+c)}\\ Bài\ 2\\ a.\ \frac{x^{2} y+2xy^{2} +y^{3}}{2x^{2} +xy-y^{2}} =\frac{y\left( x^{2} +2xy+y^{2}\right)}{2x^{2} -xy+2xy-y^{2}}\\ =\frac{y( x+y)^{2}}{x( 2x-y) +y( 2x-y)} =\frac{y( x+y)^{2}}{( x+y)( 2x-y)} =\frac{xy+y^{2}}{2x-y} \Rightarrow đpcm\\ b.\ \frac{x^{2} +3xy+2y^{2}}{x^{3} +2x^{2} y-xy^{2} -2y^{3}} =\frac{x^{2} +2xy+xy+2y^{2}}{x^{2}( x+2y) -y^{2}( x+2y)}\\ =\frac{x( x+2y) +y( x+2y)}{\left( x^{2} -y^{2}\right)( x+2y)} =\frac{( x+y)( x+2y)}{( x-y)( x+y)( x+2y)} =\frac{1}{x-y} \Rightarrow đpcm \end{array}$