Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 9:\\ g.\ x^{2} -x+1\\ h.\ \frac{xy}{x+y}\\ i.\ \frac{x^{2} +1}{x+2}\\ k.\ \frac{x+4}{x+2}\\ Bài\ 11:\\ a.\ P=\frac{x^{8} +x^{4} +1}{x^{2} +1}\\ b.\ Q=\frac{1}{x^{5} +1}\\ Bài\ 12:\\ a.\ x=3a^{2} -12\\ b.\ x=\frac{a-5}{a} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 9:\\ g.\ \frac{x^{5} +x^{3} +x^{2} +1}{x^{3} +x^{2} +x+1} =\frac{x^{3}\left( x^{2} +1\right) +\left( x^{2} +1\right)}{x^{2}( x+1) +( x+1)}\\ =\frac{\left( x^{3} +1\right)\left( x^{2} +1\right)}{( x+1)\left( x^{2} +1\right)} =\frac{( x+1)\left( x^{2} -x+1\right)}{x+1}\\ =x^{2} -x+1\\ h.\ \frac{x^{3} y+xy^{3} +xy}{x^{3} +y^{3} +x^{2} y+xy^{2} +x+y}\\ =\frac{xy\left( x^{2} +y^{2}\right) +xy}{( x+y)\left( x^{2} -xy+y^{2}\right) +xy( x+y) +( x+y)}\\ =\frac{xy\left( x^{2} +y^{2} +1\right)}{( x+y)\left( x^{2} -xy+y^{2} +xy+1\right)}\\ =\frac{xy}{x+y}\\ i.\ \frac{x^{4} -1}{x^{3} +2x^{2} -x-2} =\frac{\left( x^{2} -1\right)\left( x^{2} +1\right)}{x^{2}( x+2) -( x+2)}\\ =\frac{\left( x^{2} -1\right)\left( x^{2} +1\right)}{\left( x^{2} -1\right)( x+2)} =\frac{x^{2} +1}{x+2}\\ k.\ \frac{x^{2} +7x+12}{x^{2} +5x+6} =\frac{x^{2} +3x+4x+12}{x^{2} +2x+3x+6}\\ =\frac{x( x+3) +4( x+3)}{x( x+2) +3( x+2)} =\frac{( x+4)( x+3)}{( x+2)( x+3)} =\frac{x+4}{x+2}\\ Bài\ 10:\\ a.\ VT=\frac{( x-2)\left( 2x+2x^{2}\right)}{( x+1)\left( 4x-x^{3}\right)} =\frac{( x-2) 2x( x+1)}{( x+1) x\left( 4-x^{2}\right)}\\ =\frac{2( x-2)}{( 2-x)( 2+x)} =\frac{-2}{x+2} =VP\Rightarrow đpcm\\ b.\ VT=\frac{x^{2} +y^{2} +2xy-1}{x^{2} -y^{2} +1+2x} =\frac{( x+y)^{2} -1}{( x+1)^{2} -y^{2}}\\ =\frac{( x+y-1)( x+y+1)}{( x+1-y)( x+1+y)} =\frac{x+y+1}{x+1-y} =VP\Rightarrow đpcm\\ c.\ VT=\frac{\left( x^{2} +2\right)^{2} -4x^{2}}{y\left( x^{2} +2\right) -2xy-( x-1)^{2} -1}\\ =\frac{\left( x^{2} +2-2x\right)\left( x^{2} +2+2x\right)}{y\left( x^{2} +2-2x\right) -\left( x^{2} -2x+1+1\right)}\\ =\frac{\left( x^{2} +2-2x\right)\left( x^{2} +2+2x\right)}{( y-1)\left( x^{2} -2x+2\right)} =\frac{x^{2} +2x+2}{y-1} =VP\Rightarrow đpcm\\ d.\ VT=\frac{3y-2-3xy+2x}{1-3x-x^{3} +3x^{2}} =\frac{( 3y-2) -x( 3y-2)}{1-x^{3} -3x( 1-x)}\\ =\frac{( 3y-2)( 1-x)}{( 1-x)\left( 1+x+x^{2} -3x\right)} =\frac{3y-2}{x^{2} -2x+1} =\frac{3y-2}{( x-1)^{2}} =VP\Rightarrow đpcm\\ Bài\ 11:\\ a.\ P=\frac{x^{10} -x^{8} +x^{6} -x^{4} +x^{2} -1}{x^{4} -1} ;\ ĐKXĐ:\ x\neq \pm 1\\ =\frac{x^{8}\left( x^{2} -1\right) +x^{4}\left( x^{2} -1\right) +\left( x^{2} -1\right)}{\left( x^{2} -1\right)\left( x^{2} +1\right)} =\frac{\left( x^{2} -1\right)\left( x^{8} +x^{4} +1\right)}{\left( x^{2} -1\right)\left( x^{2} +1\right)}\\ =\frac{x^{8} +x^{4} +1}{x^{2} +1}\\ b.\ Q=\frac{x^{40} +x^{30} +x^{20} +x^{10} +1}{x^{45} +x^{40} +x^{35} +...+x^{10} +x^{5} +1}\\ \Rightarrow Q.x^{5} =\frac{x^{45} +x^{35} +x^{25} +x^{15} +x{^{5}}}{x^{45} +x^{40} +x^{35} +...+x^{10} +x^{5} +1}\\ \Rightarrow Q.x^{5} +Q=\frac{x^{45} +x^{40} +x^{35} +...+x^{10} +x^{5} +1}{x^{45} +x^{40} +x^{35} +...+x^{10} +x^{5} +1} =1\\ \Rightarrow Q\left( x^{5} +1\right) =1\Leftrightarrow Q=\frac{1}{x^{5} +1}\\ Bài\ 12:\\ a.\ a^{2} x+4x=3a^{4} -48\\ \Leftrightarrow x\left( a^{2} +4\right) =3\left( a^{4} -4^{2}\right)\\ \Leftrightarrow x=\frac{3\left( a^{2} -4\right)\left( a^{2} +4\right)}{a^{2} +4} =3a^{2} -12\\ b.\ a^{2} x+5ax+25=a^{2}\\ \Leftrightarrow x\left( a^{2} +5a\right) =a^{2} -25\\ \Leftrightarrow x=\frac{( a-5)( a+5)}{a( a+5)} =\frac{a-5}{a}( ĐK:\ a\neq 0;\ a\neq -5) \end{array}$
