1.\(\left\{{}\begin{matrix}x^2+2xy-2x-y=0\\x^4-4\left(x+y-1\right)x^2+y^2+2xy=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+y\right)\left(x-1\right)=0\\x^4-4\left(x+y-1\right)x^2+y^2+2xy=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\1^4-4\left(1+y-1\right)1^2+y^2+2.1.y=0\end{matrix}\right.\)(1)

hoặc \(\Leftrightarrow\left\{{}\begin{matrix}y=-2x\\x^4-4\left(x-2x-1\right)x^2+\left(-2x\right)^2+2x.\left(-2x\right)=0\end{matrix}\right.\)(2)

(1)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\1-4y+y^2+2y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y^2-2y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

(2)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2x\\x^4-4\left(-x-1\right)x^2+4x^2-4x^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2x\\x^2\left(x^2+4x+4\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2x\\x^2\left(x+2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=0\end{matrix}\right.\)hoặc\(\left\{{}\begin{matrix}y=4\\x=-2\end{matrix}\right.\)

Vậy nghiệm của hệ pt là (1;1),(0;0),(-2;4)

2. \(x^4-x^3+1-y^2=0\)

\(\Leftrightarrow x^3\left(x-1\right)+\left(1-y\right)\left(1+y\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^3\left(x-1\right)=0\\\left(1-y\right)\left(1+y\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\pm1\end{matrix}\right.\)(tm)hoặc\(\left\{{}\begin{matrix}x=1\\y=\pm1\end{matrix}\right.\)(tm)

Vậy nghiệm nguyên cuar pt là (0;1),(0;-1),(1;1),(1;-1)

Câu 1:

\(\left\{\begin{matrix} x^2+2xy-2x-y=0(1)\\ x^4-4(x+y-1)x^2+y^2+2xy=0(2)\end{matrix}\right.\)

Bình phương (1)

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

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

Lấy \((3)-(2)\) thu được:

\(4x^3y+4x^2y^2-6x^2y-4xy^2+2xy=0\)

\(\Leftrightarrow 2xy[2x^2+2xy-3x-2y+1]=0\)

\(\Leftrightarrow 2xy[2x(x-1)+2y(x-1)-(x-1)]=0\)

\(\Leftrightarrow 2xy(2x+2y-1)(x-1)=0\)

Do đó xét các TH sau:

TH1: \(x=0\) thay vào (1) suy ra \(y=0\)

TH2: \(y=0\Rightarrow x^2-2x=0\Leftrightarrow x=0;2\)

TH3: \(x=1\). Thay vào (1) suy ra \(y=1\). Thử lại thấy đúng.

TH4: \(2x+2y-1=0\)

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

\(\Leftrightarrow y^2-y+1=(\frac{1}{2}-1)^2=\frac{1}{4}\)

\(\Leftrightarrow y^2-y+\frac{3}{4}=0\)

\(\Leftrightarrow (y-\frac{1}{2})^2+\frac{1}{2}=0\) (vô lý)

Vậy \((x,y)=(0,0); (2,0); (1,1)\)

Ap dụng bất đẳng thức BDT Caucchy Schwarz ta có :

\(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2zx}+\frac{z^2}{z^2+2xy}\)

\(=\frac{\left(x+y+z\right)^2}{x^2+2yz+y^2+2zx+z^2+2xy}\)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)

Cái này mình mới xét TH a#0, bạn xét thêm TH a=0 nữa nhé

\(x^2+2xy+2y^2+y+\frac{1}{2}\)

\(=x^2+2xy+y^2+y^2+y+\frac{1}{2}\)

\(=\left(x+y\right)^2+y^2+y+\frac{1}{2}\)

Có: \(\left(x+y\right)^2\ge0\)

\(y^2\ge y\ge0\Rightarrow y^2+y\ge0\)

\(\frac{1}{2}>0\)

\(\Rightarrow x^2+2xy+2y^2+y+\frac{1}{2}>0\) với mọi x

xét vế trái:     \(x^2+2xy+2y^2+y+\frac{1}{2}\)    =\(x^2+2xy+y^2+y^2+y+\frac{1}{2}\)

                                                                        = \(\left(x^2+2xy+y^2\right)+\left(y^2+y+\frac{1}{2}\right)\)

                                                                        = \(\left(x+y\right)^2+\left(y^2+2.\frac{1}{2}.y+\frac{1}{4}-\frac{1}{4}+\frac{1}{2}\right)\)

                                                                        =  \(\left(x+y\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{1}{4}\)

  vì \(\left(x+y\right)^2>=0\) và \(\left(y+\frac{1}{2}\right)^2>=0\)  =>   \(\left(x+y\right)^2+\left(y+\frac{1}{2}\right)^2>=0\)

   mà    1/4 >0    =>     \(\left(x+y\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{1}{4}>0\)

bn ... ơi...mik ...bỏ...cuộc ...hu...hu

. Huhu T^T mong sẽ có ai đó giúp mình "((

Vì |x + 1| ≥ 0 ; |y + 2| ≥ 0

=> |x + 1| + |y + 2| ≥ 0

Dấu "=" xảy ra khi |x + 1| = 0 ;|y + 2| = 0

=> x = - 1; y = - 2

Vậy x = - 1; y = - 2

|x+1|+|y+2|=0

=>x+1=0 và y+2=0

=>x=-1 và y=-2

vậy...

mk ko chắc chắn đâu.ai thấy sai nhắn cho mk

a. Ta có : \(4x^2-6x+9=4x^2-6x+\dfrac{9}{4}+\dfrac{27}{4}\)

\(=\left[\left(2x\right)^2-6x+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{27}{4}\)

\(=\left(2x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\)

Vì \(\left(2x-\dfrac{3}{2}\right)^2\ge0\forall x\)

nên \(\left(2x-\dfrac{3}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\forall x\)

b.Ta có : \(x^2+2y^2-2xy+y+1=\left(x^2+y^2-2xy\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\)

\(=\left(x-y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Vì \(\left(x-y\right)^2\ge0\forall x;y\)

\(\left(y+\dfrac{1}{2}\right)^2\ge0\forall y\)

nên \(\left(x-y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\forall x;y\)

a)\(x^2+2xy+1+y^2=\left(x+y\right)^2+1\)

Vì \(\left(x+y\right)^2\ge0\)với mọi \(x,y\in\)

nên \(\left(x+y\right)^2+1>0\)với mọi \(x,y\in R\)

Vậy biểu thức \(x^2+2xy+y^2+1>0\left(x;y\in R\right)\)

b) \(-x^2+x-1=-\left(x^2-2x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\right)=-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]=-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}\)

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\left(x\in R\right)\)

nên \(-\left(x-\frac{1}{2}\right)^2\le0\left(x\in R\right)\)

do đó \(-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}< 0\left(x\in R\right)\)

Vậy biểu thức \(x-x^2-1< 0\left(x\in R\right)\)

a) x² + 2xy + 1 +y² = (x²+2xy+y²)+1=(x+y)²+1 mà (x+y)² luôn lớn hơn hoặc bằng 0 với mọi x,y

=>x²+2xy+1+y²>1>0

b)x-x²-1=-(x²-x+1)=-((x²-2.x.0,5+0,25)+0,75)=-((x-0,5)²+0,75) mà (x-0,5)² luôn lớn hơn hoặc bằng 0 vớ mọi x

=>x-x²-1<0

TƯỞNG KHÔNG DỄ NHƯNG DỄ KHÔNG TƯỞNG!

\(x^2+2xy+2y^2+y+\frac{1}{2}\)

\(=x^2+2xy+y^2+y^2+y+\frac{1}{2}\)

\(=\left(x+y\right)^2+y^2+y+\frac{1}{2}\)

Có : \(\left(x+y\right)^2\ge0\)

\(y^2\ge y\ge0\Rightarrow y^2+y\ge0\)

\(\frac{1}{2}>0\)

\(\Rightarrow x^2+2xy+2y^2+y+\frac{1}{2}>0\) với mọi x y 

Ta có

\(x^2+2xy+2y^2+y+\frac{1}{2}\)

\(=\left(x^2+2xy+y^2\right)+\left(y^2+2.y.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}\)

\(=\left(x+y\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{1}{2}\)

Mà \(\begin{cases}\left(x^2+2xy+y^2\right)\ge0\\\left(y^2+2.y.\frac{1}{2}+\frac{1}{4}\right)\ge0\\\frac{1}{4}>0\end{cases}\)

\(\Rightarrow\left(x^2+2xy+y^2\right)+\left(y^2+2.y.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}>0\)