A=(x+2)^2 +3

B=(x-5)^2 +4

C=4(x+1/2)^2 +4

D=(x-1/2)^2 +19/4

E=2(x-3/4)^2 +95/8

\(A=3x^2-x+20=3\left(x^2-\dfrac{1}{3}x+\dfrac{20}{3}\right)=3\left(x^2-2.\dfrac{1}{6}x+\dfrac{1}{36}+\dfrac{239}{36}\right)\)

\(A=3\left[\left(x+\dfrac{1}{6}\right)^2+\dfrac{239}{36}\right]=3\left(x+\dfrac{1}{6}\right)^2+\dfrac{239}{12}\ge\dfrac{239}{12}\)

\(=>A>0\left(\forall x\right)\)

Ta có:A=3x²-x+20=2(x²-2x+1)+\(\left(x^2+2.\dfrac{3}{2}x+\dfrac{9}{4}\right)+\dfrac{73}{4}\)

            =\(2\left(x-1\right)^2+\left(x+\dfrac{3}{2}\right)^2+\dfrac{73}{4}\ge0\)

a) \(9x^2-6x+11=\left(3x\right)^2-2.3x+1+10=\left(3x-1\right)^2+10>0\forall x\)

b) \(3x^2-12x+81=3.\left(x^2-4x+9\right)=3.\left(x-2\right)^2+15>0\forall x\)

c) \(5x^2-5x+4=5.\left(x^2-x+\dfrac{4}{5}\right)=5.\left(x^2-x+\dfrac{1}{4}+\dfrac{11}{20}\right)=5.\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\forall x\)

d) \(2x^2-2x+9=2.\left(x^2-x+\dfrac{9}{2}\right)=2.\left(x-\dfrac{1}{2}\right)^2+\dfrac{17}{2}>0\forall x\)

a) = (3x-1)^2+10

Do (3x-1)^2>=0 với mọi x

--> (3x-1)^2+10>0 với mọi x

a) \(9x^2-6x+11=\left(3x-1\right)^2+10\ge10>0\)

b) \(3x^2-12x+81=3\left(x-2\right)^2+69\ge69>0\)

c) \(5x^2-5x+4=5\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\)

d) \(2x^2-2x+9=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{17}{2}\ge\dfrac{17}{2}>0\)

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

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

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

A = x(x - 6) + 10

A = x² - 6x + 10

A = x² - 2.3.x + 3² + 1

A = (x - 3)² + 1

Vì  (x - 3)² \(\ge\)0 với mọi x

=>  (x - 3)² + 1 \(\ge\)1 

=> A \(\ge\)1

=> A luôn dương với mọi x (Đpcm)

\(x^2-4x+8\\ =\left(x-2\right)^2+4\ge4>0\forall x\)

ta có x²+x+1= x²+x+1+x-x= (x+1)²-x

Vì (x+1)² \(\ge\)0   và (x+1)²>x 

nên x²+x+1 luôn luôn dương với mọi giá trị của x

xét x>0 suy ra biểu thúc có gi trị dương

xét x,0

ta có \(x^2\)>0

suy ra \(x^2\)+x > 0

suy ra \(x^2\)+x+1 luôn luôn  dương với mọi gi trị của x

nếu x =0 

=> x²+x+1 =1 (dương)

nếu x < 0

=>x²+x+1 >0 (dương)

nếu x >0 thì quá rõ ràng rồi

a) \(A=x^2+2x+2\)

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

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

b) \(B=4x^2-4x+11\)

\(=4x^2-4x+1+10\)

\(=\left(2x-1\right)^2+10>0\forall x\)

c) \(C=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

d) Ta có: \(D=x^2-2x+y^2+4y+6\)

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

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

e) Ta có: \(D=x^2-2xy+y^2+x^2-8x+20\)

\(=x^2-2xy+y^2+x^2-8x+16+4\)

\(=\left(x-y\right)^2+\left(x-4\right)^2+4>0\forall x,y\)

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

\(B=4x^2-4x+11=\left(2x-1\right)^2+10\ge10>0\left(\forall x\right)\)

\(C=x^2-x+1=x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(D=x^2-2x+y^2+4y+6=x^2-2x+1+y^2+4y+4+1\)

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

\(E=x^2-2xy+y^2+x^2-8x+16+4\)

\(=\left(x-y\right)^2+\left(x-4\right)^2+4\ge4>0\)\(\left(\forall x,y\right)\)

$x^2+2x+7$

$=x^2+2x+1+6$

$=(x+1)^2+6$

Vì $(x+1)^2 \ge 0$

$\Rightarrow (x+1)^2+6 \ge 6>0\forall x$

Hay $x^2+2x+7>0\forall x$

Ta có: \(x^2+2x+7\)

\(=x^2+2x+1+6\)

\(=\left(x+1\right)^2+6>0\forall x\)(đpcm)