toan cua truong Andersen

H

huongmot

[TEX]x^2[/TEX]+[TEX]y^2[/TEX]+[TEX]z^2[/TEX]= xy+3y+2z-4
<=>[TEX]x^2[/TEX] + [TEX]y^2[/TEX] + [TEX]z^2[/TEX] - xy - 3y - 2z + 4 = 0
<=>[TEX]x^2[/TEX] + 1/4.[TEX]y^2[/TEX] + 3/4.[TEX]y^2[/TEX] + [TEX]z^2[/TEX] - xy - 3y - 2z + 4=0
<=>([TEX]x^2[/TEX]- xy + 1/4.[TEX]y^2[/TEX]) + (3/4.[TEX]y^2[/TEX] - 3y + 3)+([TEX]z^2[/TEX] - 2z + 1)=0
<=>[TEX](x - 1/2.y)^2[/TEX] + 3/4([TEX]y^2[/TEX] - 4y + 4) + [TEX](z-1)^2[/TEX] = 0
<=>[TEX](x - 1/2.y)^2 [/TEX]+ 3/4[TEX](y-2)^2[/TEX] + [TEX](z-1)^2[/TEX] = 0
Vì [TEX](x - 1/2.y)^2[/TEX] >= 0
[TEX] 3/4(y-2)^2[/TEX] >= 0
[TEX] (z-1)^2[/TEX] >= 0
Mà [TEX](x - 1/2.y)^2[/TEX] + [TEX]3/4(y-2)^2[/TEX] +[TEX] (z^2-1)^2[/TEX] = 0
-> [TEX](x - 1/2.y)^2[/TEX] = 0
<->[TEX] x - 1/2y = 0[/TEX]
<-> x = 1/2y
[TEX]3/4[/TEX][TEX](y-2)^2[/TEX] = 0
-> y-2 = 0
<-> y= 2
-> x= 1
[TEX](z-1)^2[/TEX] = 0
-> z - 1 =0
<-> z=1
Vậy(x,y,z) là (1;2;1)
 
Last edited by a moderator:
C

congnhatso1

huongmot said:
[TEX]x^2[/TEX]+[TEX]y^2[/TEX]+[TEX]z^2[/TEX]= xy+3y+2z-4
=>[TEX]x^2[/TEX] + [TEX]y^2[/TEX] + [TEX]z^2[/TEX] - xy - 3y - 2z + 4 = 0
=>[TEX]x^2[/TEX] + 1/4.[TEX]y^2[/TEX] + 3/4.[TEX]y^2[/TEX] + [TEX]z^2[/TEX] - xy - 3y - 2z + 4=0
=>([TEX]x^2[/TEX]- xy + 1/4.[TEX]y^2[/TEX]) + (3/4.[TEX]y^2[/TEX] - 3y + 3)+([TEX]z^2[/TEX] - 2z + 1)=0
=>[TEX](x - 1/2.y)^2[/TEX] + 3/4([TEX]y^2[/TEX] - 4y + 4) + [TEX](z-1)^2[/TEX] = 0
=>[TEX](x - 1/2.y)^2 [/TEX]+ 3/4[TEX](y-2)^2[/TEX] + [TEX](z-1)^2[/TEX] = 0
Vì [TEX](x - 1/2.y)^2[/TEX] >= 0
[TEX] 3/4(y-2)^2[/TEX] >= 0
[TEX] (z-1)^2[/TEX] >= 0
Mà [TEX](x - 1/2.y)^2[/TEX] + [TEX]3/4(y-2)^2[/TEX] +[TEX] (z^2-1)^2[/TEX] = 0
-> [TEX](x - 1/2.y)^2[/TEX] = 0
->[TEX] x - 1/2y = 0[/TEX]
-> x = 1/2y
[TEX]3/4[/TEX][TEX](y-2)^2[/TEX] = 0
-> y-2 = 0
-> y= 2
-> x= 1
[TEX](z-1)^2[/TEX] = 0
-> z - 1 =0
-> z=1
Vậy: x=1
y=2
z=1
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đây là bài toán giải phương trình nên các chỗ đầu phải dùng dấu "\Leftrightarrow" mới đúng, còn kết luận thì phải : vậy,(x,y,z) là (1;2;1)
 
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