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1.Simplificati raportul algebric pana la un raport ireductibil:
a)3x+9 supra lui x la 2+6x+9;
b)a la 2-2ab+b la 2 supra la a la 2-b la 2;
c)a la 2-ax supra la x la 2-a la 2;
d)x la 2-4 supra la 2x-x la 2.
2.Efectuati:
a)1 supra lui x la 2 y + 2 supra lui xy la 2;
b)5 supra lui x-2y - 3 supra lui x+2y;
c)x-1 supra lui 2x-6 + 1 supra lui 3-x;
d)5 supra lui x la 2-16 - 7 supra lui x-4.
3.Efectuati:
a)(x+1)la 3 supra lui (x-2)la 4 inmultit (x-2)la 3 supra lui x+1;
b)xy2 supra lui x+y inmultit (x+y)la 2 supra lui xy;
c)3ab supra lui 25 yx la 4 inmultit 15x la 2 y la 2 supra lui 18a la 2 b la 3;



Răspuns :

1)a)[tex] \frac{3x+9}{ x^{2} +6x+9}= \frac{3(x+3)}{ (x+3)^{2} }= \frac{3}{x+3} [/tex]
b)[tex] \frac{ a^{2}-2ab+b ^{2} }{a ^{2} -b^{2} }= \frac{(a-b) ^{2} }{(a-b)(a+b)}= \frac{a-b}{a+b} [/tex]
c)[tex] \frac{ a^{2}-ax }{ x^{2} - a^{2} } = \frac{-a(x-a)}{(x-a)(x+a)}= \frac{-a}{x+a} [/tex]
d)[tex] \frac{ x^{2} -4}{2x- x^{2} }= \frac{(x-2)(x+2)}{-x(x-2)}= \frac{x+2}{-x}= \frac{-x-2}{x} [/tex]
2)a)[tex] \frac{1}{ x^{2} y}+ \frac{2}{x y^{2} }= \frac{y}{ x^{2} y^{2} }+ \frac{2x}{ x^{2} y^{2} }= \frac{y+2x}{ x^{2} y^{2} } [/tex]
b)[tex] \frac{5}{ x-2y} - \frac{3}{x+2y}= \frac{5(x+2y)}{(x-2y)(x+2y)}- \frac{3(x-2y)}{(x-2y)(x+2y)} = \frac{5x+10y-3x+6y}{(x-2y)(x+2y)}= \\ \frac{2x+16y}{(x-2y)(x+2y)} [/tex]
c)[tex] \frac{x-1}{2x-6}+ \frac{1}{3-x}= \frac{x-1}{2(x-3)}- \frac{1}{x-3}= \frac{x-1-2}{2(x-3)}= \frac{x-3}{2(x-3)}= \frac{1}{2} [/tex]
d)[tex] \frac{5}{ x^{2} -16} - \frac{7}{x-4}= \frac{5}{(x-4)(x+4)}- \frac{7}{(x-4)}= \frac{5-7(x+4) }{(x-4)(x+4)}= \frac{5-7x-28}{(x-4)(x+4)}= \\ \frac{-7x-23}{(x-4)(x+4)} [/tex]
3)a)[tex] \frac{(x+1) ^{3} }{(x-2) ^{4} }* \frac{(x-2) ^{3} }{(x+1)}= \frac{(x+1) ^{2} }{(x-2)}= \frac{ x^{2} +2x+1}{x-2} [/tex]
b)[tex] \frac{xy2}{x+y}* \frac{(x+y) ^{2} }{xy}= 2(x+y)=2x+2y [/tex]
c)[tex] \frac{3ab}{25y x^{4} }* \frac{15 x^{2} y^{2} }{18 a^{2} b^{3} }= \frac{3y}{5 x^{2}*6 a b^{2} }= \frac{y}{10 x^{2} a b^{2} } [/tex]