\(A=X^2-4XY-2Y+2+5Y^2\)

\(=X^2-4XY+4Y^2+Y^2-2Y+1+1\)

\(=\left(X-2Y\right)^2+\left(Y-1\right)^2+1>0\)

\(A=x^2-4xy+4y^2+y^2-2y+1+1\)

\(\Leftrightarrow A=\left(x-2y\right)^2+\left(y-1\right)^2+1>0\)

Vậy biểu thức A luôn dương với mọi x,y

Ta có: \(M=\left(x^2-4xy+4y^2\right)+\left(y^2-2y+1\right)+2=\left(x-2y\right)^2+\left(y-1\right)^2+2\)

Vì \(\left(x-2y\right)^2,\left(y-1\right)^2>0\)với mọi x,y nên M luôn dương

Ta có điều phải chứng minh

\(M=5x^2+2y^2+4xy-2x+4y+6\)

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

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

Do \(\left(2x+y\right)^2\ge0\forall x;y\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)

\(\Rightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\forall x;y\)

\(\Rightarrow M\ge1>0\forall x;y\)

\(\left(đpcm\right)\)

\(x^2+4x-4=0\Leftrightarrow x^2+4x+4=8\Leftrightarrow\left(x+2\right)^2=8\)

\(\Leftrightarrow x+2=\sqrt{8}\Leftrightarrow x=\sqrt{8}-2\)

Bài 2 đề bn viết thiếu đấu + đó

Ta có M=x²+4xy+5y²-2y+3

=(x²+4xy+4y²)+(y²-2y+1)+2

=(x+2y)² +(y-1)²+2

Do \(\left(x+2y\right)^2+\left(y-1\right)^2\ge0\Rightarrow M\ge2\)

=> đpcm

\(A=\left(x^2-4xy+4y^2\right)+\left(x^2+10x+25\right)+\left(y^2-22y+121\right)+2\\ A=\left(x-2y\right)^2+\left(x+5\right)^2+\left(y-11\right)^2+2\ge2>0\)

a) \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

c) \(C=4x-10-x^2=-\left(x^2-4x+10\right)\)

\(=-\left(x^2-4x+4+6\right)=-\left[\left(x-2\right)^2+6\right]\)

\(=-\left(x^2-4x+4+6\right)=-\left[\left(x-2\right)^2\right]-6\le-6< 0\forall x\)