a)2x^2+xy-y^2-x+2y-1

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

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

=2a^2+ab-b^2         với a=x,b=y-1

=2a^2+2ab-ab-b^2

=(2a-b)(a+b)

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

Bạn xem ở http://olm.vn/hoi-dap/question/106067.html

hoi-dap/question/106067.html

Vì Dũng bắt chước mình nên thấy lạ lạ chứ gì !

Lời giải:
$2x^2+y^2+2xy-6x-2y=8$

$\Leftrightarrow (x^2+y^2+2xy)+x^2-6x-2y=8$
$\Leftrightarrow (x+y)^2-2(x+y)+x^2-4x=8$

$\Leftrightarrow (x+y)^2-2(x+y)+1+(x^2-4x+4)=13$

$\Leftrightarrow (x+y-1)^2+(x-2)^2=13$
$\Rightarrow (x-2)^2=13-(x+y-1)^2\leq 13$
Mà $(x-2)^2$ là scp với mọi $x$ nguyên nên $(x-2)^2\in\left\{0; 1; 4; 9\right\}$

Nếu $(x-2)^2=0\Rightarrow (x+y-1)^2=13-(x-2)^2=13$ (không là scp - loại) 

Nếu $(x-2)^2=1\Rightarrow (x+y-1)^2=12$ (không là scp - loại)

Nếu $(x-2)^2=4\Rightarrow (x+y-1)^2=9$

$\Rightarrow x-2=\pm 2$ và $x+y-1=\pm 3$
TH1: $x-2=2; x+y-1=3\Rightarrow x=4; y=0$

TH2: $x-2=2; x+y-1=-3\Rightarrow x=4; y=-6$

TH3: $x-2=-2; x+y-1=3\Rightarrow x=0; y=4$

TH4: $x-2=-2; x+y-1=-3\Rightarrow x=0; y=-2$

Nếu $(x-2)^=9\Rightarrow (x+y-1)^2=4$ (bạn cũng làm tương tự trên)

\(B=\left(x^2+y^2+4+2xy-4x-4y\right)+\left(x^2+z^2+1+2xz-2x-2z\right)+\left(y^2-4y+4\right)+4\)

\(B=\left(x+y-2\right)^2+\left(x+z-1\right)^2+\left(y-2\right)^2+4\ge4\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x+y-2=0\\x+z-1=0\\y-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\\z=1\end{matrix}\right.\)

Áp dụng Bunyakovsky, ta có :

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)

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

=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)

Mấy cái kia tương tự 

\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)

\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)

\(P=x\left(2x+1\right)-x^2\left(x+2\right)+x^3-x+3\\ P=2x^2+x-x^3-2x^2+x^3-x+3\\ P=3\left(đfcm\right)\)