toán khó

T

thinhrost1

$(3x+5)^2 . (3x+4)(x+2)-4 =(x+1)(3x+7)(9x^2+30x+28)$

$2x^2+3xy+x+12y-9y^2-3=-(3y-2x-3)(3y+x-1)$

$2x^2+xy-4x+13y-6y^2-6=-(2y+x-3)*(3y-2x-2)$

2)$2x^2+2y^2=5xy\\\Rightarrow (x-y)^2=\dfrac{xy}{2}\\\Rightarrow (x+y)^2=\dfrac{9xy}{2}\\\dfrac{(x+y)^2}{(x-y)^2}=\dfrac{xy:2}{9xy:2}=\dfrac{1}{9}\\\Rightarrow \dfrac{x+y}{x-y}=\dfrac{1}{3}$
 
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