a) \(x^2+y^2-2x+4y+6=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\)

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

b) \(2x^2+2x+3=2\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{5}{2}\)

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

c) \(x^2+y^2+z^2\ge xy+yz+xz\)

\(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2xz\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(x^2+2xz+z^2\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\left(đúng\right)\)

\(ĐTXR\Leftrightarrow x=y=z\)

a) Với \(x\ne0\) , ta rút gọn :

\(A=\left(6x^3+12x^2\right):2x-2x\left(x+1\right)+5\)

\(A=3x^2+6x-2-2x+5\)

\(A=3x^2+6x+3\)

\(A=3\left(x^2+2x+1\right)\)

\(A=3\left(x+1\right)^2\)

Vậy sau khi rút gọn kết quả là : \(A=3\left(x+1\right)^2\)

b) Ta thấy \(x\ne0\Rightarrow x+1\ne1\)

\(\Rightarrow\left(x+1\right)^2\ge1;\forall x\ne0\)

\(\Rightarrow3\left(x+1\right)^2\ge3>1;\forall x\ne0\)

Vậy \(3\left(x+1\right)^2>1\Leftrightarrow A>1\) với \(\forall x\ne0\) \(\left(ĐPCM\right)\)

a) Với \(x\ne0\) , ta rút gọn :

\(A=\left(6x^3+12x^2\right):2x-2\left(x+1\right)+5\)

\(A=3x^2+6x-2x-2+5\)

\(A=3x^2+4x+3\)

Vậy sau khi rút gọn kết quả là : \(A=3x^2+4x+3\)

b) \(A=3x^2+4x+3\)

\(A=3x^2+4x+\frac{4}{3}+\frac{5}{3}\)

\(A=3\left(x^2+\frac{4}{3}x+\frac{4}{9}\right)+\frac{5}{3}\)

\(A=3\left(x+\frac{2}{3}\right)^2+\frac{5}{3}\)

Ta thấy : \(\left(x+\frac{2}{3}\right)^2\ge0;\forall x\)

\(\Rightarrow3\left(x+\frac{2}{3}\right)^2\ge0;\forall x\)

\(\Rightarrow3\left(x+\frac{2}{3}\right)^2+\frac{5}{3}\ge\frac{5}{3}>0\)

\(\Leftrightarrow A>0\)

Vậy \(A>0\) với \(\forall x\ne0\) \(\left(ĐPCM\right)\)

⇒(x−1)^2+4(y+1)^2+(z−3)^2≥0

x^2+4y^2+z^2-2x-6z+8y+15

=x^2+4y^2+z^2-2x-6z+8y+1+1+4+9

=(x^2-2x+1)+(4y^2+8y+4)+(z^2-6z+9)+1

=(x-1)^2+4(y+1)^2+(z-3^)2+1

Ta thấy:(x−1)^2≥0

              4(y+1)^2≥0

             (z−3)^ 2≥0

{(x−1)^24(y+1)^2(z−3)^2≥0

⇒(x−1)^2+4(y+1)^2+(z−3)^2≥0

⇒(x−1)2+4(y+1)2+(z−3)2+1≥0+1=1>0

\(x^2+xy+y^2+1.=x^2+2.x.\dfrac{y}{2}+\left(\dfrac{y}{2}\right)^2+\dfrac{3}{4}y^2+1.\\ =\left(x+\dfrac{y}{2}\right)^2+\dfrac{3}{4}y^2+1>0\forall x;y\in R.\\ \Rightarrow x^2+xy+y^2+10\forall x;y\in R.\)

Kkk

câu b sai đề bb ơi ,-,

a/ \(-x^2+4x-9=-\left(x^2-4x+4\right)-5=-\left(x-2\right)^2-5\)

Có: \(\left(x-2\right)^2\ge0\forall x\Rightarrow-\left(x-2\right)^2\le0\Rightarrow-\left(x-2\right)^2-5\le-5\left(đpcm\right)\)

b/ \(x^2-2x+90=\left(x^2-2x+1\right)+89=\left(x-1\right)^2+89\)

Có: \(\left(x-1\right)^2\ge0\forall x\Rightarrow\left(x-1\right)^2+89\ge89\left(đpcm\right)\)

P/s: b tui sửa đề nhes

a/ \(x^2-6x+10=x^2-2.x.3+3^2+1=\left(x-3\right)^2+1\)

Với mọi x ta có :

\(\left(x-3\right)^2\ge0\)

\(\Leftrightarrow\left(x-3\right)^2+1>0\)

\(\Leftrightarrow x^2-6x+10>0\)

b/ \(x^2-4x+7=x^2-2.x.2+2^2+3=\left(x-2\right)^2+3\)

Với mọi x ta có :

\(\left(x-2\right)^2\ge0\)

\(\Leftrightarrow\left(x-2\right)^2+3\ge3\)

\(\Leftrightarrow x^2-4x+7\ge3\left(đpcm\right)\)

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

Với mọi x ta có :

\(\left(x+\dfrac{1}{2}\right)^2\ge0\)

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

\(\Leftrightarrow x^2+x+1>0\left(đpcm\right)\)

d/ \(x^2+y^2+4x-6y+15=\left(x^2+4x+2^2\right)+\left(y^2-6y+3^2\right)+2=\left(x+2\right)^2+\left(y-3\right)^2+2\)

Với mọi x,y ta có :

\(\left\{{}\begin{matrix}\left(x+2\right)^2\ge0\\\left(y-3\right)^2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left(x+2\right)^2+\left(y-3\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)^2+\left(y-3\right)^2+2\ge0\)

\(\Leftrightarrow x^2+y^2+4x-6y+15>0\left(đpcm\right)\)

2/ Ta có :

\(\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2=\left(a-b\right)^2\)

Vậy \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\left(đpcm\right)\)

3/ \(x^2+y^2=x^2+y^2+2xy-2xy=\left(x+y\right)^2-2xy\)

Mà \(x+y=7;xy=-3\)

\(\Leftrightarrow x^2+y^2=7^2-2.\left(-3\right)=49+6=55\)

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:

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

hơi ngán dạng này :((((

a, \(x^2-3x+5=x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+5=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\forall x\)

b,

\(x^2-\frac{1}{3}x+\frac{5}{4}=x^2-2.\frac{1}{6}+\frac{1}{36}-\frac{1}{36}+\frac{5}{4}=\left(x-\frac{1}{6}\right)^2+\frac{11}{9}>0\forall x\)

c,

\(x-x^2-3=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\frac{1}{4}-3=-\left(x-\frac{1}{2}\right)^2-\frac{11}{4}< 0\forall x\)d,

\(x-2x^2-\frac{5}{2}=-2\left(x^2-\frac{1}{2}x+\frac{5}{4}\right)=-2\left(x^2-2.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}+\frac{5}{4}\right)=-2\left[\left(x-\frac{1}{4}\right)^2+\frac{19}{16}\right]=-2\left(x-\frac{1}{4}\right)^2-\frac{19}{8}< 0\forall x\)P/s : ko chắc lém :)))