Bài 1:

\(a,A=2x^2+2x+1=\left(x^2+2x+1\right)+x^2=\left(x+1\right)^2+x^2\\ Mà:\left(x+1\right)^2\ge0\forall x\in R\\ \Rightarrow\left(x+1\right)^2+x^2>0\forall x\in R\\ Vậy:A>0\forall x\in R\)

2:

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

=-(x^2-3x+9/4-5/4)

=-(x-3/2)^2+5/4 chưa chắc <0 đâu bạn

b: =-2(x^2+3/2x+3/2)

=-2(x^2+2*x*3/4+9/16+15/16)

=-2(x+3/4)^2-15/8<0 với mọi x

Bài 1:

\(B=4+x^2+x=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{15}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}\forall x\in R\\ Vậy:B>0\forall x\in R\)

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

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

Δ=(-m)^2-4(2m-4)

=m^2-8m+16=(m-4)^2>=0

=>Phương trình luôn có hai nghiệm

a: x1^2+x2^2=13

=>(x1+x2)^2-2x1x2=13

=>m^2-2(2m-4)-13=0

=>m^2-4m-5=0

=>m=5 hoặc m=-1

b: x1^3+x2^3=9

=>(x1+x2)^3-3*x1x2(x1+x2)=9

=>m^3-3*(2m-4)*m=9

=>m^3-6m^2+12m-9=0

=>m=3

a) Ta có: \(\dfrac{x^2+2x+1}{x^2+x}\)

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

\(=\dfrac{x+1}{x}\)

b) Ta có: \(\dfrac{x^2-4x+3}{x^2-x}\)

\(=\dfrac{\left(x-1\right)\left(x-3\right)}{x\left(x-1\right)}\)

\(=\dfrac{x-3}{x}\)

a: x^2+10x+100

=x^2+10x+25+75=(x+5)^2+75>0 với mọi x

b: -x^2+4x-100

=-(x^2-4x+100)

=-(x^2-4x+4+96)

=-(x-2)^2-96<0 với mọi x

c: x^2-5x+6

=x^2-5x+25/4-1/4

=(x-5/2)^2-1/4 chưa chắc lớn hơn 0 đâu nha bạn

a) \(x^2+4x+10\)

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

\(=\left(x+2\right)^2+6\)

Mà: \(\left(x+2\right)^2+6>0\forall x\)

\(\Rightarrow x^2+4x+10>0\forall x\)

b) \(x^2-6x+15\)

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

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

Mà: \(\left(x-3\right)^2+6>0\forall x\)

\(\Rightarrow x^2-6x+16>0\forall x\)

c) \(-x^2+2x-5\)

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

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

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

Mà: \(-\left(x-1\right)^2-4< 0\forall x\)

\(\Rightarrow-x^2+2x-5< 0\forall x\)

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

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

Câu 9: D

Câu 10: A

`A=x(x-6)+10=x^2-6x+10`

`=x^2 -2.x .3 + 3^2 + 1`

`=(x-3)^2+1 >0 forall x`

`B=x^2-2x+9y^2-6y+3`

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

`=(x-1)^2+(3y-1)^2+1 > 0 forall x,y`.