quy đồng mẫu

N

nhimcon_online

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T

trinhminh18

a/ TA có:
$\dfrac{5}{4x^2-y^2}=\dfrac{5}{(2x-y)(2x+y)}=\dfrac{20x^2-5y^2}{(2x-y)^2.(2x+y)^2}$
$\dfrac{7}{4x^2-4xy+y^2}=\dfrac{7}{(2x-y)^2}=\dfrac{7(2x+y)^2}{(2x-y)^2.(2x+y)^2}$
$\dfrac{9}{4x^2+4xy+y^2}$=$\dfrac{9}{(2x+y)^2}$
=$\dfrac{9(2x-y)^2}{(2x-y)^2.(2x+y)^2}$
 
T

trinhminh18

b/ Ta có:
$\dfrac{5x^2+1}{8x^2-18}=\dfrac{5x^3+x}{2x(2x-3)(2x+3)}$
$\dfrac{3x}{2x^2+3x}=\dfrac{3x}{2x^2+3x}=\dfrac{6x(2x-3)}{2x(2x-3)(2x+3)}$
$\dfrac{-5}{2(2x-3)}=\dfrac{-5x(2x+3)}{2x(2x-3)(2x+3)}$
 
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