(x+y)^2+2(x+y)+1+2(y-1)^2+2013>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) \(C=4x-10-x^2=-\left(x^2-4x+10\right)\)

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

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

Có : x^2+y^2+z^2+4x-2y-4z+10

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

= (x+2)^2+(y-1)^2+(z-2)^2+1 >= 1

=> (x+2)^2+(y-1)^2+(z-2)^2 luôn dương với mọi x,y,z

\(x^2+y^2+z^2+4x-2y-4z+10\)

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

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

Vì  \(\hept{\begin{cases}\left(x+2\right)^2\ge0\\\left(y-1\right)^2\ge0\\\left(z-2\right)^2\ge0\end{cases}}\)\(\Leftrightarrow\)\(\left(x+2\right)^2+\left(y-1\right)^2+\left(z-2\right)^2\ge0\)

\(\Rightarrow\)\(\left(x+2\right)^2+\left(y-1\right)^2+\left(z-2\right)^2+1>0\) 

\(\Rightarrow\)\(đpcm\)

đâylà toán lớp 6, 7 rồi bạn

ta có \(A=2x^2-2xy+\frac{y^2}{2}+\frac{y^2}{2}-4y+8+7\)

\(=\frac{1}{2}\left[\left(4x^2-4xy+y^2\right)+\left(y^2-8y+18\right)\right]+7\)

\(=\frac{1}{2}\left[\left(2x-y\right)^2+\left(y-4\right)^2\right]+7\ge7\)

Vậy ta có A luôn dương

Bài 1

\(A=x^2-6x+15=x^2-2.3.x+9+6=\left(x-3\right)^2+6>0\forall x\)

\(B=4x^2+4x+7=\left(2x\right)^2+2.2.x+1+6=\left(2x+1\right)^2+6>0\forall x\)

Bài 2

\(A=-9x^2+6x-2021=-\left(9x^2-6x+2021\right)=-\left[\left(3x-1\right)^2+2020\right]=-\left(3x-1\right)^2-2020< 0\forall x\)

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

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

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

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) \(A=y^4+y^2+y^2+1=y^2\left(y^2+1\right)+\left(y^2+1\right)=\left(y^2+1\right)\left(y^2+1\right)=\left(y^2+1\right)>0\)với mọi y

b) \(B=\left(6x^2-2x^2\right)+\left(3xy-3xy\right)+\left(-2y^2+3y^2\right)+\left(-5+5\right)\)

\(=4x^2+y^2\ge0\)với mọi x, y

Q=x² - 2xy + y² - 12x + 12y + 36 + 5y² + 10y + 5 + 1976

Q=(x - y)² - 2.(x - y).6 + 6² + 5(y² + 2y + 1) + 1976

Q=(x - y - 6)² +5.(y + 1)² + 1976 (≥ 1976 > 0 ∀ x,y ∈ R)

Vậy biểu thức Q luôn nhận giá trị dương với mọi số thực x,y

Ta có: \(4x^2-28x+51=\left(2x\right)^2-2\cdot2x\cdot7+49+2\)

                                       \(=\left(2x-7\right)^2+2\)(*)

Vì \(\left(2x-7\right)^2\ge0\) với mọi x

=> (*)\(\ge1\)

 =>(*) luôn luôn dương với mọi x

ta có : \(4x^2-28x+51=\left(2x\right)^2-2.2x.7+7^2+51=\left(2x-7\right)^2+51\)

vì \(\left(2x-7\right)^2\ge0\) với mọi x 

\(\Rightarrow\left(2x-7\right)^1+51>0\) với mọi x  (đpcm)