\(a,=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{19}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}>0\\ b,=-\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{7}{4}=-\left(x-\dfrac{5}{2}\right)^2-\dfrac{7}{4}\le-\dfrac{7}{4}< 0\)

a,=(x2+3x+94)+194=(x+32)2+194≥194>0b,=−(x2−5x+254)−74=−(x−52)2−74≤−74<0

a) MTC = (x -2)(x + 2). Ta rút gọn được M = 1 x − 2  

b) Gợi ý:  x 2 + 5 x + 6 = ( x + 2 ) ( x + 3 ) ; x 2 + x − 12 = ( x − 3 ) ( x + 4 )

Ta có  N = ( x + 2 ) ( x + 3 ) ( x − 3 ) ( x + 4 ) : ( x + 2 ) 2 x ( x − 3 ) = x ( x + 3 ) ( x + 2 ) ( x + 4 )

a) \(x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1>0\forall x\)

b) \(4x-x^2-5=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\le-1< 0\forall x\)

E=4x^​2​+5x+5>0 với mọi x

=(4x^​2 +4x+1)+4

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

Với mọi x thuộc R thì (2x+1)\(^2\)>=0

Suy ra(2x+1)\(^2\)+4>=4>0

Hay E>0 với mọi x thuộc R(đpcm)

F=5x²​-6x+7>0 với mọi x

=(5x\(^2\)-6x+\(\dfrac{36}{25}\))+\(\dfrac{139}{25}\)

=5\(\left(x-\dfrac{6}{5}\right)^2\)+\(\dfrac{139}{25}\)

Với mọi x thuộc R thì 5\(\left(x-\dfrac{6}{5}\right)^2\)>=0

Suy ra 5\(\left(x-\dfrac{6}{5}\right)^2\)+\(\dfrac{139}{25}\)>0

Hay F >0 với mọi x(đpcm)

G=-x^​2​​+5x -6<0 với mọi x​

=-(x^​2​​-5x+6,25)+0,25

=-(x-2,5)² +0,25

Với mọi x thuộc R thì -(x-2,5)² <=0

Suy ra -(x-2,5)² +0,25<0

Hay G<0 với mọi x (đpcm)

chúc bạn học tốt ạ

a. \(x^2+3x+5\)

\(=x^2+2.x^2.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

=> đpcm

b. \(4x^2+5x+7\)

\(=\left(2x\right)^2-2.2x.\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{87}{16}\)

= \(\left(2x+\dfrac{5}{4}\right)^2\) + \(\dfrac{87}{16}\) \(\ge\dfrac{87}{16}\)

=> đpcm

c. \(6x-9x^2-4\)

=> \(-9x^2+6x-4\)

\(=-\left(9x^2-6x+4\right)\)

\(=-[\left(3x\right)^2-2.3x.1+1-5]\)

\(=\left(3x-1\right)^2\le5\forall x\)

=> đpcm

a: Ta có: \(-x^2+4x-5\)

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

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

\(=-\left(x-2\right)^2-1< 0\forall x\)

b: Ta có: \(x^4\ge0\forall x\)

\(3x^2\ge0\forall x\)

Do đó: \(x^4+3x^2\ge0\forall x\)

\(\Leftrightarrow x^4+3x^2+3>0\forall x\)

c: Ta có: \(\left(x^2+2x+3\right)=\left(x+1\right)^2+2>0\forall x\)

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

Do đó: \(\left(x^2+2x+3\right)\left(x^2+2x+4\right)>0\forall x\)

\(\Leftrightarrow\left(x^2+2x+3\right)\left(x^2+2x+4\right)+3>0\forall x\)​

b: Ta có: \(x^4\ge0\forall x\)

\(3x^2\ge0\forall x\)

Do đó: \(x^4+3x^2\ge0\forall x\)

\(\Leftrightarrow x^4+3x^2+3>0\forall x\)

c: Ta có: \(\left(x^2+2x+3\right)=\left(x+1\right)^2+2>0\forall x\)

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

Do đó: \(\left(x^2+2x+3\right)\left(x^2+2x+4\right)>0\forall x\)

\(\Leftrightarrow\left(x^2+2x+3\right)\left(x^2+2x+4\right)+3>0\forall x\)

a) Ta có:

\(x^2-x+1\)

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

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

Mà: \(\left(x-\dfrac{1}{2}\right)^2\ge0\) và \(\dfrac{3}{4}>0\) nên

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

\(\Rightarrow x^2-x+1>0\forall x\)