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Nguyễn Anh Quân
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Không Tên
24 tháng 11 2017 lúc 19:15

Ta có: x2 + x2y2 - 2y = 0

\(\Rightarrow\)x2 + x2y2 + y2 - 2y + 1 - y2 - 1 = 0

\(\Rightarrow\)(x- 1) + (x2y2 - y2) + (y - 1)2 = 0 

\(\Rightarrow\)(x2 - 1) + y2(x2 - 1) + (y - 1)2 = 0

\(\Rightarrow\)(x2 - 1)(1 + y2) + (y - 1)2 = 0

\(\Rightarrow\)(x2 - 1)(1 + y2) =   -(y - 1)2     \(\le\)0     V y

\(\Rightarrow\)x2 - 1 \(\le\)0  V x       ( vì 1 + y2 > 0 ,  V y )

\(\Rightarrow\)(x - 1)(x + 1) \(\le\)

\(\Rightarrow\)x - 1 và x + 1 trái dấu

Do đó  \(\hept{\begin{cases}x-1\ge0\\x+1\le0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x\ge1\\x\le-1\end{cases}}\)  ( vô lý )

Hoặc \(\hept{\begin{cases}x-1\le0\\x+1\ge0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x\le1\\x\ge-1\end{cases}}\)  \(\Leftrightarrow\)-1\(\le\)\(\le\)1     (*)

Lại có:  x3 + 2y2 - 4y + 3 = 0

\(\Rightarrow\)(x3 + 1) + 2(y2 - 2y + 1) = 0

\(\Rightarrow\)(x3 + 1) + 2(y - 1)2 = 0

\(\Rightarrow\)x3 + 1 =   -2(y - 1)2  \(\le\)0,    V  y 

\(\Rightarrow\)x3 + 1 \(\le\)0,   V  x

\(\Rightarrow\)(x + 1)(x2 - x + 1) \(\le\)

\(\Rightarrow\)x + 1 \(\le\)0   ( vì x2 - x + 1 = (x - 1/2 )2 + 3/4  > 0, V x   )

\(\Rightarrow\)\(\le\)-1  (**)

Từ (*) và (**) suy ra   x = -1 \(\Rightarrow\)(-1)2 + (-1)2 . y2 - 2y = 0

                                            \(\Rightarrow\)1 + y2 - 2y = 0

                                            \(\Rightarrow\)( y - 1 )2 = 0  \(\Rightarrow\)y = 1

\(\Rightarrow\)x2 + y2 = (-1)2 + 12 = 2

phan tuấn anh
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Vũ Thị Thùy Trang
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Pé Ken
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Đinh Thùy Linh
26 tháng 6 2016 lúc 18:22

1) a thỏa mãn: a2 + a + 1 = 0, rõ ràng a khác 0. Chia cả 2 vế cho a ta được: \(a+\frac{1}{a}=-1\)

Mặt khác ta có: \(\left(a+\frac{1}{a}\right)^3=-1\Rightarrow a^3+3\cdot\left(a+\frac{1}{a}\right)+\frac{1}{a^3}=-1\Rightarrow a^3+\frac{1}{a^3}=2\)\(\Rightarrow\left(a^3+\frac{1}{a^3}\right)^2=4\Rightarrow a^6+\frac{1}{a^6}=2\)\(\Rightarrow\left(a^6+\frac{1}{a^6}\right)\left(a^3+\frac{1}{a^3}\right)=4\Rightarrow a^9+\frac{1}{a^9}+a^3+\frac{1}{a^3}=4\Rightarrow a^9+\frac{1}{a^9}=2\)... \(\Rightarrow a^{3k}+\frac{1}{a^{3k}}=2\)\(\Rightarrow a^{2013}+\frac{1}{a^{2013}}=2\)

2) Từ: \(x^2+x^2y^2-2y=0\Rightarrow x^2\left(y^2+1\right)=2y\Rightarrow x^2=\frac{2y}{y^2+1}\)

Với mọi y thì: \(\left(y-1\right)^2\ge0\Leftrightarrow2y\le y^2+1\Leftrightarrow\frac{2y}{y^2+1}\le1\)Do đó \(x^2=\frac{2y}{y^2+1}\le1\Rightarrow-1\le x\le1\)(1)

Mặt khác: \(x^3+2y^2-4y+3=0\Leftrightarrow x^3+1+2\left(y-1\right)^2=0\)(2)

Từ (1) => \(x^3+1\ge0\forall x\Rightarrow VT\left(2\right)\ge VP\left(2\right)\forall x;y\)

Để TM (2) thì dấu "=" xảy ra, khi đó x = -1; y = 1

và suy ra \(Q=x^2+y^2=2\)

Nguyễn Bá Minh
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alibaba nguyễn
12 tháng 8 2017 lúc 10:25

Ta có:

\(x^2+x^2y^2-2y=0\)

\(\Leftrightarrow x^2=\frac{2y}{y^2+1}\le1\)(cái này chứng minh đơn giản b tự làm lấy nhé)

\(\Leftrightarrow-1\le x\le1\left(1\right)\)

Ta lại có:

\(x^3+2y^2-4y+3=0\)

\(\Leftrightarrow x^3=-1-2\left(y-1\right)^2\le-1\left(2\right)\)

Từ (1) và (2) \(\Rightarrow x=-1\)

\(\Rightarrow y=1\)

\(\Rightarrow x^2+y^2=1+1=2\)

Quách Thanh Bình
1 tháng 5 2020 lúc 16:57

kdfjeuy;r;

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Nguyễn Hải Anh
1 tháng 5 2020 lúc 17:50

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Ngô Phương Quý
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loan cao thị
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Chung Nguyễn Thành
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Sherry
28 tháng 12 2017 lúc 20:56

Có x^2 + 2xy + 4x + 4y + 2y^2 + 3 = 0

--> (x+y)^2 + 4(x+y) + 4+ y^2 - 1 = 0

--> (x+y+2)^2 + y^2 = 1

-->(x+y+2)^2 <= 1 ( vì y^2 >=1)

--> -1 <= x+y+2 <=1

--> 2015 <= x+y+2018 <= 2017

hay 2015 <= Q , dau bang xay ra khi x+y+2=-1 --> x+y=-3

Q<=2017, dau bang xay ra khi  x+y+2=1 --> x+y=-1

Vậy giá trị nhỏ nhất của Q là 2015 khi x+y =-3

 giá trị lớn nhất của Q là 2017 khi x+y=-1

Le Thi Phuong Anh
14 tháng 5 2020 lúc 14:20

giá trị lớn nhất là 2017

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