Toán Lớp 9: Bạn Topone giúp mình ạ: `(2){(x^2+y^2=2x^2),((x+y)(1+xy)=4x^2y^2):}` `(3){(2x+3y=5xy),(4x^2+y^2=5xy^2):}`

Question

Toán Lớp 9:  Bạn Topone giúp mình ạ: (2){(x^2+y^2=2x^2),((x+y)(1+xy)=4x^2y^2):} (3){(2x+3y=5xy),(4x^2+y^2=5xy^2):}

ngochuong2k2 · Answer

Giải đáp+Lời giải và giải thích chi tiết:   (2) x&sup2;+y&sup2;=2x&sup2; &rArr; x&sup2;=y&sup2;   &rArr; 4x&sup2;y&sup2;=4x&sup2;.x&sup2;=4x^4   &hArr; (x+y)(1+xy)=2x(1+x&sup2;)=2x&sup3;+2x   &hArr; 2x&sup3;+2x=4x^4 &hArr; 4x^4-2x&sup3;-2x=0   &hArr; 2x(2x&sup3;-x&sup2;-x)=0   &hArr; 2x(2x&sup3;+x&sup2;-2x&sup2;-x)=0   &hArr; 2x[x&sup2;(2x+1)-x(2x+1)]=0   &hArr; 2x(x&sup2;-x)(2x+1)=0   &hArr; 2x(x-1)(x+1)(2x+1)=0   &hArr; x=0;x=1;x=-1;x=-1/2   &hArr; x&sup2;=0;1;1/4   &hArr; y=0;&plusmn;1;&plusmn;1/2   Vậy hệ phương tr&igrave;nh c&oacute; 8 nghiệm   (x;y)=(0;0);(1;-1);(1;1);(-1;1);(-1;-1);(-1/2;1/2);(-1/2;1/2)   (3) X&eacute;t phương tr&igrave;nh c&oacute; x=y=0 l&agrave; nghiệm   Với x,y kh&aacute;c 0 , nh&acirc;n phương tr&igrave;nh thứ nhất với y:   2x+3y=5xy &rArr; 2xy+3y&sup2;=5xy&sup2;   &hArr; 2xy+3y&sup2;=4x&sup2;+y&sup2;   &hArr; 2xy+2y&sup2;=4x&sup2;   &hArr; 4x&sup2;-2y&sup2;-2xy=0   &hArr; 2(2x&sup2;-y&sup2;-xy)=0   &hArr; 2x&sup2;-y&sup2;-xy=0   &hArr; 2(x-y)(2x+y)=0   TH1:x-y=0 &hArr; x=y   &hArr; 2x+3y=5x=5x&sup2;   &hArr; 5x&sup2;-5x=5x(x-1)=0   &hArr; x=0;x=1 &hArr; y=0;y=1   TH2: 2x+y=0 &hArr; y=-2x   &hArr; 2x+3y=2x-6x=-4x=5.x.-2x=-10x&sup2;   &hArr; -10x&sup2;+4x=0 &hArr; -2x(5x-2)=0   &hArr; x=0;x=2/5 &hArr; y=0;y=-4/5   Vậy hệ phương tr&igrave;nh c&oacute; 3 nghiệm   (x;y)=(0;0);(1;1);(2/5;-4/5)