Question

chứng minh các đẳng thức sau luôn dương với mọi giá trị của biến :

1)x^2+x+1

2)x^2+3x+3

3)x^2+y^2+2(x-2y)+6

Answer 1

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

2 ) \(x^2+3x+3=x^2+3x+\dfrac{9}{4}+\dfrac{3}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\left(đpcm\right)\)

3 ) \(x^2+y^2+2\left(x-2y\right)+6\)

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

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

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

Answer 2

1) x² +x+1

= (x² +2.x.1/2 +1/4) +3/4

= (x+1/2)²+3/4 \(\ge\dfrac{3}{4}\forall x\in R\left(Vì:\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\in R\right)\)

2) x² + 3x+3

= (x² +2.x.3/2 + 9/4)+ 3/4

= ( x+ 3/2)² + 3/4 \(\ge\dfrac{3}{4}\forall x\in R\left(Vì:\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\in R\right)\)