`1/(4x2 -12x9)-3/(9-4x2)=(-2)/(4x2 12x9)`

Question

`1/(4x^2 -12x+9)-3/(9-4x^2)=(-2)/(4x^2 +12x+9)`

lantrungbach · Answer

1/(4x^2-12x+9) - 3/(9-4x^2) = -2/(4x^2+12x+9) ĐKXĐ: x≠±3/2 <=> 1/[(2x-3)^2] + 3/[(2x-3)(2x+3)] = -2/(2x+3)^2 MTC: (2x+3)^2(2x-3)^2 <=> (2x+3)^2 + 3(2x-3)(2x+3) = -2(2x-3)^2 <=> 4x^2 + 12x + 9 + 3(4x^2 + 6x - 6x - 9) = -2(4x^2 - 12x + 9) <=> 4x^2 + 12x + 9 + 12x^2 - 27 = -8x^2 + 24x - 18 <=> 4x^2 + 12x + 12x^2 + 8x^2 - 24x = -18 - 9 + 27 <=> 24x^2 - 12x = 0 <=> 12x(2x-1)=0 <=> ( left[ begin{array}{l}12x=0 2x-1=0 end{array} right. ) <=> ( left[ begin{array}{l}x=0 x= frac{1}{2} end{array} right. ) (t/mãn) Vậy S={0;1/2} $ color{lightblue}{eriet}$

kimlien · Answer

Giải đáp: frac{1}{4x^{2}-12x+9}- frac{3}{9-4x^{2}}= frac{-2}{4x^{2}+12x+9} (đk: x ne pm3/2) <=> frac{1}{(2x)^{2}-2.2x.3+3^{2}}+ frac{3}{(2x)^{2}-3^{2}}= frac{-2}{(2x)^{2}+2.2x.3+3^{2}} <=> frac{1}{(2x-3)^{2}}+ frac{3}{(2x-3).(2x+3)}= frac{-2}{(2x+3)^{2}} <=> frac{1.(2x+3)^{2}}{(2x-3)^{2}.(2x+3)^{2}}+ frac{3.(2x-3).(2x+3)}{(2x-3)^{2}.(2x+3)^{2}}= frac{-2.(2x-3)^{2}}{(2x-3)^{2}.(2x+3)^{2}} =>1.[(2x)^{2}+2.2x.3+3^{2}]+3.[(2x)^{2}-3^{2}]=-2.[(2x)^{2}-2.2x.3+3^{2}] <=>4x^{2}+12x+9+3.(4x^{2}-9)=-2.(4x^{2}-12x+9) <=>4x^{2}+12x+9+12x^{2}-27=-8x^{2}+24x-18 <=>16x^{2}+12x-18+8x^{2}-24x+18=0 <=>24x^{2}-12x=0 <=>12x.(2x-1)=0 <=>12x=0 hoặc 2x-1=0 <=>x=0(tmđk) hoặc x=1/2(tmđk) Vậy S={0;1/2}

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