\(a,A=4x^2-20x+27=\left(2x\right)^2-2.2x.5+5^2+2\)\(=\left(2x-5\right)^2+2\)

Mà \(\left(2x-5\right)^2\ge0\Rightarrow\left(2x-5\right)^2+2>0\Rightarrow A>0\)

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

Mà \(\left(x-\frac{1}{4}\right)^2\ge0\Rightarrow\left(x-\frac{1}{4}\right)^2+\frac{3}{4}>0\Rightarrow B>0\)

\(c,C=x^2+4x+y^2-6y+15=x^2+4x+4+y^2-6y+9+2\)

\(\left(x+2\right)^2+\left(y-3\right)^2+2\)

Mà \(\left(x+2\right)^2+\left(y-3\right)^2\ge0\Rightarrow\left(x+2\right)^2+\left(y-3\right)^2+2>0\Rightarrow C>0\)

Bài 1: 

a) Ta có: \(A=-x^2-4x-2\)

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

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

\(=-\left(x+2\right)^2+2\le2\forall x\)

Dấu '=' xảy ra khi x=-2

b) Ta có: \(B=-2x^2-3x+5\)

\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)

c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)

\(=2x+8-x^2-4x\)

\(=-x^2-2x+8\)

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

\(=-\left(x^2+2x+1-9\right)\)

\(=-\left(x+1\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi x=-1

Bài 2: 
a) Ta có: \(=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)

b) Ta có: \(B=9x^2-6xy+2y^2+1\)

\(=9x^2-6xy+y^2+y^2+1\)

\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)

c) Ta có: \(E=x^2-2x+y^2-4y+6\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\forall x,y\)

a : x² + 4x + 7 = (x + 2)² + 3 > 0

b : 4x² - 4x + 5 = (2x - 1)² + 4 > 0

c : x² + 2y² + 2xy - 2y + 3 = (x + y)² + (y - 1)² + 2 > 0

d : 2x² - 4x + 10 = 2(x - 1)² + 8 > 0

e : x² + x + 1 = (x + 0,5)² + 0,75 > 0

f : 2x² - 6x + 5 = 2(x - 1,5)² + 0,5 > 0

a : x² + 4x + 7 = (x + 2)² + 3 > 0

b : 4x² - 4x + 5 = (2x - 1)² + 4 > 0

c : x² + 2y² + 2xy - 2y + 3 = (x + y)² + (y - 1)² + 2 > 0

d : 2x² - 4x + 10 = 2(x - 1)² + 8 > 0

e : x² + x + 1 = (x + 0,5)² + 0,75 > 0

f : 2x² - 6x + 5 = 2(x - 1,5)² + 0,5 > 0

1/

\(M=3x^2-4x+3=3\left(x^2-\frac{4}{3}x+1\right)=3\left(x^2-2x\cdot\frac{2}{3}+\frac{4}{9}\right)+\frac{5}{3}=3\left(x-\frac{2}{3}\right)^2+\frac{5}{3}\ge\frac{5}{3}>0\)

\(N=5x^2-10x+2018=5\left(x^2-2x+1\right)+2013=5\left(x-1\right)^2+2013\ge2013>0\)

\(P=x^2+2y^2-2xy+4y+7=\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)+3=\left(x-y\right)^2+\left(y+2\right)^2+3\ge3>0\)

2/

\(A=10x-6x^2+7=-6x^2+10x+7=-6\left(x^2-\frac{10}{6}x+\frac{25}{36}\right)-\frac{11}{6}=-6\left(x-\frac{5}{6}\right)^2-\frac{11}{6}\le-\frac{11}{6}< 0\)

\(B=-3x^2+7x+10=-3\left(x^2-\frac{7}{3}x+\frac{49}{36}\right)-\frac{311}{12}=-3\left(x-\frac{7}{6}\right)^2-\frac{311}{12}\le-\frac{311}{12}< 0\)

\(C=2x-2x^2-y^2+2xy-5=\left(2x-x^2-1\right)-\left(x^2-2xy+y^2\right)-4=-\left(x^2-2x+1\right)-\left(x-y\right)^2-4=-\left(x-1\right)^2-\left(x-y\right)^2-4\)\(\le-4< 0\)

Giups mik vs

\(A=2x^2-20x+7=2\left(x^2-10x+25\right)-43=2\left(x-5\right)^2-43\ge-43\left(\forall x\right)\)

=> Chưa thể khẳng định A dương

\(B=9x^2-6xy+2y^2+1\)

\(B=\left(9x^2-6xy+y^2\right)+y^2+1\)

\(B=\left(3x-y\right)^2+y^2+1\ge1>0\left(\forall x\right)\)

=> đpcm

\(C=x^2-2x+y^2+4y+6\)

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

\(C=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1>0\left(\forall x\right)\)

=> đpcm

\(D=x^2-2x+2=\left(x^2-2x+1\right)+1=\left(x-1\right)^2+1\ge1>0\left(\forall x\right)\)

=> đpcm

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

b) \(x^2+16x+64=\left(x+8\right)^2\)

c) \(x^3-8y^3=x^3-\left(2y\right)^3\)

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

d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)