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

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) \(C=4x-10-x^2=-\left(x^2-4x+10\right)\)

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

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

${\mathrm{A=x}}^2-2x+2$

$\mathrm{=x2-2x+1+1}$

$\mathrm{=(x2-2x+1)+1}$

=(x-1)²+1

vì (x-1)²\(\ge0\forall x\)

=>(x-1)²+1\(\ge1\)

vậy A luôn dương với mọi x

B=x${}^{}+y^2+2x-4y+6$

=x²+2x+1+y²-4y+4+1

=(x²+2x+1)+(y²-4y+4)+1

=(x+1)²+(y-2)²+1

do (x+1)²\(\ge0\forall x\)

(y-2)²\(\ge0\forall y\)

=>(x+1)²+(y-2)²\(\ge0\)

=>(x+1)²+(y-2)²+1\(\ge1\)

=>B\(\ge1\)

vậy B luôn dương với mọi x;y

C= $x^2+y^2+z^2+4x-2y-4z+10$

$\mathrm{=x2+4x+4+y2-2y+1+z2-4z+4+1}$

=(x²+4x+4)+(y²-2y+1)+(z²-4z+4)+1

=(x+2)²+(y-1)²+(z-2)²+1

do (x+2)²\(\ge0\forall x\)

(y-1)²\(\ge0\forall y\)

(\(\)z-2)²\(\ge0\forall z\)

=>(x+2)²+(y-1)²+(z-2)²\(\ge0\)

=>(x+2)²+(y-1)²+(z-2)²+1\(\ge1\)

=>C\(\ge1\)

vậy C luôn dương với mọi x;y;z

bài 2: tìm x

a)\(x^2+y^2-2x+4y+5=0\)

\(\Leftrightarrow x^2+y^2-2x+4y+1+4=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy x=1; y=-2

b)\(5x^2+9y^2-12xy-6x+9=0\)

\(\Leftrightarrow\left(4x^2-12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2.3-3.y=0\\x=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=3\end{matrix}\right.\)

Vậy x=2; y=3

x^2 -2x+2= x^2-2x+1+1=(x-1)^2+1> 0

vậy...

B= x^2+y^2+2x-4y+6

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

=(x+1)^2+(y-2)^2 >0

C, tương tự

a : x² + 4x + 7 = (x + 2)² + 3 > 0

b : 4x² - 4x + 5 = (2x - 1)² + 4 > 0

c : x² + 2y² + 2xy - 2y + 3 = (x + y)² + (y - 1)² + 2 > 0

d : 2x² - 4x + 10 = 2(x - 1)² + 8 > 0

e : x² + x + 1 = (x + 0,5)² + 0,75 > 0

f : 2x² - 6x + 5 = 2(x - 1,5)² + 0,5 > 0

a : x² + 4x + 7 = (x + 2)² + 3 > 0

b : 4x² - 4x + 5 = (2x - 1)² + 4 > 0

c : x² + 2y² + 2xy - 2y + 3 = (x + y)² + (y - 1)² + 2 > 0

d : 2x² - 4x + 10 = 2(x - 1)² + 8 > 0

e : x² + x + 1 = (x + 0,5)² + 0,75 > 0

f : 2x² - 6x + 5 = 2(x - 1,5)² + 0,5 > 0

a) 

\(=x^2+2.1,5x+1.5^2+0,75\)

\(=\left(x+1.5\right)^2+0,75\)

Vì (x+1.5)^2 luôn dương và 0,75 dương nên biểu thức luôn dương

b) 

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

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

Lập luận tương tự câu a), được biểu thức luôn dương

c)

\(=x^2+2xy+y^2+x^2-2x+1+1\)

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

Lập luận tương tự

\(a.\)

\(A=9x^2-6xy+2y^2+1\)

\(A=\left(3x\right)^2-2\cdot3x\cdot y+y^2+y^2+1\)

\(A=\left(3x-y\right)^2+\left(y^2+1\right)\ge0\)

\(b.\)

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

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

\(B=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)

\(c.\)

\(C=x^2-2x+2\)

\(C=x^2-2x+1+1\)

\(C=\left(x-1\right)^2+1\ge1\)

a) A=9x²-6xy+2y²+1

    A=(3x)²-2.3x.y+y²+y²+1

    A=(3x-y)²+(y²+1)≥0

Câu b, c tương tự câu a

a) Ta có: \(A=9x^2-6xy+2y^2+1\)

\(=\left(3x-1\right)^2+2y^2\ge0\forall x,y\)

b) Ta có: \(B=x^2-2x+y^2+4y+6\)

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

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

c) Ta có: \(C=x^2-2x+2\)

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

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

1/

\(M=3x^2-4x+3=3\left(x^2-\frac{4}{3}x+1\right)=3\left(x^2-2x\cdot\frac{2}{3}+\frac{4}{9}\right)+\frac{5}{3}=3\left(x-\frac{2}{3}\right)^2+\frac{5}{3}\ge\frac{5}{3}>0\)

\(N=5x^2-10x+2018=5\left(x^2-2x+1\right)+2013=5\left(x-1\right)^2+2013\ge2013>0\)

\(P=x^2+2y^2-2xy+4y+7=\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)+3=\left(x-y\right)^2+\left(y+2\right)^2+3\ge3>0\)

2/

\(A=10x-6x^2+7=-6x^2+10x+7=-6\left(x^2-\frac{10}{6}x+\frac{25}{36}\right)-\frac{11}{6}=-6\left(x-\frac{5}{6}\right)^2-\frac{11}{6}\le-\frac{11}{6}< 0\)

\(B=-3x^2+7x+10=-3\left(x^2-\frac{7}{3}x+\frac{49}{36}\right)-\frac{311}{12}=-3\left(x-\frac{7}{6}\right)^2-\frac{311}{12}\le-\frac{311}{12}< 0\)

\(C=2x-2x^2-y^2+2xy-5=\left(2x-x^2-1\right)-\left(x^2-2xy+y^2\right)-4=-\left(x^2-2x+1\right)-\left(x-y\right)^2-4=-\left(x-1\right)^2-\left(x-y\right)^2-4\)\(\le-4< 0\)

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

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

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