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x-2-xy-4x-y-2-xy-4y-log-16-x-1-y-1-x-2-y-2-




Question Number 149040 by mathdanisur last updated on 02/Aug/21
 { ((x^2 +xy=4x)),((y^2 +xy=4y)) :}   ⇒  log_(16) ^((x_1 +y_1 +x_2 +y_2 ))  = ?
{x2+xy=4xy2+xy=4ylog16(x1+y1+x2+y2)=?
Answered by bramlexs22 last updated on 02/Aug/21
(1)+(2)⇔x^2 +2xy+y^2 =4(x+y)  ⇒(x+y)^2 =4(x+y);(x+y)(x+y−4)=0  case(1)⇒x=−y   ⇒x^2 −x^2 =4x ⇒ { ((x_1 =0)),((y_1 =0)) :}  case(2)x=4−y  ⇒x(x+y)=4x ; 4x=4x →x_2 =λ ,y_2 =4−λ  then log _(16) (x_1 +y_1 +x_2 +y_2 )=log _(16) (0+λ+4−λ)  =log _(16) (4)=(1/2)
(1)+(2)x2+2xy+y2=4(x+y)(x+y)2=4(x+y);(x+y)(x+y4)=0case(1)x=yx2x2=4x{x1=0y1=0case(2)x=4yx(x+y)=4x;4x=4xx2=λ,y2=4λthenlog16(x1+y1+x2+y2)=log16(0+λ+4λ)=log16(4)=12
Commented by mathdanisur last updated on 02/Aug/21
Thank You Ser, Cool
ThankYouSer,Cool

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