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: \(x^2-5x+10\)

\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{15}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{15}{4}>0\forall x\)

b: \(2x^2+8x+15\)

\(=2\left(x^2+4x+\dfrac{15}{2}\right)\)

\(=2\left(x^2+4x+4+\dfrac{7}{2}\right)\)

\(=2\left(x+2\right)^2+7>0\forall x\)

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

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

A= x² + x + 1

A = x² + 2. \(\dfrac{1}{2}\). x + (\(\dfrac{1}{2}\))² +\(\dfrac{3}{4}\)

A = ( x + \(\dfrac{1}{2}\))² + \(\dfrac{3}{4}\) ≥ \(\dfrac{3}{4}\)

Vậy, x² + x + 1>0 với mọi x

Đúng thì like giúp mik nha. Thx bạn

         \(x^2+x+1\)

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

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

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

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

Vì \(\left(x+\dfrac{1}{2}\right)^2\) luôn dương với mọi \(x\)     ( 1 )

mà cộng thêm 1 lượng \(\dfrac{3}{4}\)  luôn dương   ( 2 )

Từ ( 1 ) và ( 2 ):     ⇒    \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\) luôn dương

⇒   \(x^2+x+1\) luôn dương với mọi giá trị của x

a, \(E=4x^2+6x+5=4\left(x^2+\frac{2.3}{4}x+\frac{9}{16}-\frac{9}{16}\right)+5\)

\(=4\left(x+\frac{3}{4}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\forall x\)

Vậy ta có đpcm 

b, \(F=2x^2-3x+7=2\left(x^2-\frac{2.3}{4}x+\frac{9}{16}-\frac{9}{16}\right)+7\)

\(=2\left(x-\frac{3}{4}\right)^2+\frac{47}{8}\ge\frac{47}{8}>0\forall x\)

Vậy ta có đpcm 

c, \(K=5x^2-4x+1=5\left(x^2-\frac{2.2}{5}x+\frac{4}{25}-\frac{4}{25}\right)+1\)

\(=5\left(x-\frac{2}{5}\right)^2+\frac{1}{5}\ge\frac{1}{5}>0\forall x\)

Vậy ta có đpcm 

d, \(Q=3x^2+2x+5=3\left(x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9}\right)+5\)

\(=3\left(x+\frac{1}{3}\right)^2+\frac{14}{3}\ge\frac{14}{3}>0\forall x\)

Vậy ta có đpcm 

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

b) \(B=2x^2+2x+1=2\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{1}{2}=2\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}>0\)

\(B=x^2+x+5=\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)+\frac{19}{4}=\left(x+\frac{1}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}>0\)

=>B luôn dương

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\(D=\left(x-3\right)\left(x-5\right)+4=x^2-8x+15+4=\left(x^2-2.x.4+16\right)+3\)

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

=>D luôn dương

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