Nhờ các bạn giải giùm bài hệ vòng quanh (Khẩn cấp ) kh

[hellboyace159]

[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
Chắc suất Đại học top - Giữ chỗ ngay!!

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( 2x mu 2 ) chia ( X binh + 1 ) =y
( 2y mu 2 ) chia ( y binh + 1 ) =z
( 2z mu 2 ) chia ( z binh + 1 ) =x


Các bạn giúp mình khẩn cấp nhé, Thằng cu gần nhà nó thi vào 10 lấy trong đề thi tuyển Sinh nhờ anh giải giùm
Mình thấy quen quen mà giải ko đc
Các bạn giải đừng dùng Đạo hàm nhé, Cái dạng này mình search trên google thấy khó khó,Mà ko biết làm tương tự....


(^_^)
thank nhiu

 

[hellboyace159]

Học mãi toán học ui giúp em giải bài này với ạ

 

[congiomuahe]

Nhìn vào hệ phương trình ta thấy phương trình đối xứng, thay x=y=z.
Nên ta giải ra được: (x;y;z)=(0;0;0) hoặc (1;1;1)

 

[Friefttar]

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[hellboyace159]

Bạn có thể giải thích giùm mình cái chỗ ta thấy hệ đối xưng ùi thay x=y=z giùm mình nhé,thank nhìu

 

[Jenniferito]

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[conga222222]

( 2x mu 2 ) chia ( X binh + 1 ) =y
( 2y mu 2 ) chia ( y binh + 1 ) =z
( 2z mu 2 ) chia ( z binh + 1 ) =x


Các bạn giúp mình khẩn cấp nhé, Thằng cu gần nhà nó thi vào 10 lấy trong đề thi tuyển Sinh nhờ anh giải giùm
Mình thấy quen quen mà giải ko đc
Các bạn giải đừng dùng Đạo hàm nhé, Cái dạng này mình search trên google thấy khó khó,Mà ko biết làm tương tự....


(^_^)
thank nhiu

$\eqalign{
& \left\{ \matrix{
{{2{x^2}} \over {{x^2} + 1}} = y \cr
{{2{y^2}} \over {{y^2} + 1}} = z \cr
{{2{z^2}} \over {{z^2} + 1}} = x \cr} \right. \cr
& \to {{2{x^2}} \over {{x^2} + 1}} + {{2{y^2}} \over {{y^2} + 1}} + {{2{z^2}} \over {{z^2} + 1}} = x + y + z\;\;(1) \cr
& x = {{2{z^2}} \over {{z^2} + 1}} = 2 - {2 \over {{z^2} + 1}} \to 0 \le x < 2 \cr
& tuong\;tu \cr
& 0 \le y < 2 \cr
& 0 \le z < 2 \cr
& TH1:\;x = y = z = 0 \to t/m \cr
& TH2:x,y,z > 0 \cr
& \cos i: \cr
& {x^2} + 1 \ge 2x \to {{2{x^2}} \over {{x^2} + 1}} \le {{2{x^2}} \over {2x}} = x \cr
& tuong\;tu: \cr
& {{2{y^2}} \over {{y^2} + 1}} \le y \cr
& {{2{z^2}} \over {{z^2} + 1}} \le z \cr
& \to {{2{x^2}} \over {{x^2} + 1}} + {{2{y^2}} \over {{y^2} + 1}} + {{2{z^2}} \over {{z^2} + 1}} \le x + y + z \leftrightarrow VT\;cua\;(1) \le VP\;cua\;(1) \cr
& dau = \leftrightarrow x = y = z = 1\;tm \cr
& vay\;he\;co\;2nghiem\;(0,0,0)\;va\;(1,1,1) \cr} $

 

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