\(25x^2+10x+4\\ =\left(25x^2+10x+1\right)+3\\ =\left[\left(5x\right)^2+2\cdot5x\cdot1+1^2\right]+3\\ =\left(5x+1\right)^2+3\)

Ta có: `(5x+1)^2>=0` với mọi x 

`=>(5x+1)^2+3>=3>0` với mọi x 

`=>` Đpcm 

\(a.4+12x+9x^2\\ =2^2+2\cdot2\cdot3x+\left(3x\right)^2\\ =\left(2+3x\right)^2\\ b.25+4a^2+10a\\ =5^2+2\cdot5\cdot2a+\left(2a\right)^2\\ =\left(5+2a\right)^2\\ c.14y+y^2+49\\ =y^2+2\cdot y\cdot7+7^2\\ =\left(y+7\right)^2\\ d.9x^2+y^2-6xy\\ =\left(3x\right)^2-2\cdot3x\cdot y+y^2\\ =\left(3x-y\right)^2\)

\(a.\left(x^2-2\right)^2=\left(x^2\right)^2-2\cdot x^2\cdot2+2^2=x^4-4x^2+4\\ b.\left(x+2y\right)^2=x^2+2\cdot x\cdot2y+\left(2y\right)^2=x^2+4xy+4y^2\\ c.\left(\dfrac{1}{2}-x\right)^2=\left(\dfrac{1}{2}\right)^2-2\cdot x\cdot\dfrac{1}{2}+x^2=\dfrac{1}{4}-x+x^2\\ d.\left(2a-1\right)^2=\left(2a\right)^2-2\cdot2a\cdot1+1^2=4a^2-4a+1\\ e.\left(\dfrac{1}{3}a+2\right)^2=\left(\dfrac{1}{3}a\right)^2+2\cdot\dfrac{1}{3}a\cdot2+2^2=\dfrac{1}{9}a^2+\dfrac{4}{3}a+4\\ f.\left(2a-\dfrac{2}{3}\right)^2=\left(2a\right)^2-2\cdot2a\cdot\dfrac{2}{3}+\left(\dfrac{2}{3}\right)^2=4a^2-\dfrac{8}{3}a+\dfrac{4}{9}\)

\(x^2+2x+10=\left(x^2+2x+1\right)+9\\ =\left(x+1\right)^2+9\ge9>0\forall x\inℝ\left(Vì:\left(x+1\right)^2\ge0\forall x\inℝ\right)\)

\(x^3-2x^2+5x-4\)

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

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

\(x^4+x^2+1=\left(x^2\right)^2+2.x^2.1+1^2=\left(x^2+1\right)^2\)

\(#NqHahh\)

\(x^4+x^2+1\)

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

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

Lời giải:

$A=x^2+2y^2+3z^2-2xy+2xz-2x-2y-8z+1998$

$2A=2x^2+4y^2+6z^2-4xy+4xz-4x-4y-16z+3996$

$=(x^2-4xy+4y^2)+(x^2+4xz+4z^2)+2z^2-4x-4y-16z+3996$

$=(x-2y)^2+(x+2z)^2-4x-4y-16z+2z^2+3996$

$=(x-2y)^2+2(x-2y)+1+(x+2z)^2-6(x+2z)+9+2z^2-4z+3986$

$=(x-2y+1)^2+(x+2z-3)^2+2(z^2-2z+1)+3984$

$=(x-2y+1)^2+(x+2z-3)^2+2(z-1)^2+3984\geq 3984$

$\Rightarrow A\geq 1992$

Vậy $A_{\min}=1992$
Giá trị này đạt tại $x-2y+1=x+2z-3=z-1=0$

$\Leftrightarrow x=y=z=1$

Lời giải:

$A=x^2+2y^2+3z^2-2xy+2xz-2x-2y-8z+1998$

$2A=2x^2+4y^2+6z^2-4xy+4xz-4x-4y-16z+3996$

$=(x^2-4xy+4y^2)+(x^2+4xz+4z^2)+2z^2-4x-4y-16z+3996$

$=(x-2y)^2+(x+2z)^2-4x-4y-16z+2z^2+3996$

$=(x-2y)^2+2(x-2y)+1+(x+2z)^2-6(x+2z)+9+2z^2-4z+3986$

$=(x-2y+1)^2+(x+2z-3)^2+2(z^2-2z+1)+3984$

$=(x-2y+1)^2+(x+2z-3)^2+2(z-1)^2+3984\geq 3984$

$\Rightarrow A\geq 1992$

Vậy $A_{\min}=1992$
Giá trị này đạt tại $x-2y+1=x+2z-3=z-1=0$

$\Leftrightarrow x=y=z=1$

Bạn kiểm tra lại đề giúp mình ạ !

đề cho thế á

Ta có: x³ + 3x² + 3x² + 9x + 2x +6

= x².(x+3) + 3x.(x+3) + 2.(x+3)

= (x²+3x+2). (x+3)

= ( x² + x +2x +2 ). ( x+3)

=(x+1).(x+2).(x+3)

TICK CHO MIK NHA

Bạn hãy xoá câu hỏi ở trên ik.

Lời giải:

$C=(x^2+4y^2+9z^2-4xy+6xz-12yz)+2y^2+5z^2+4yz$

$=(x-2y+3z)^2+2(y^2+z^2+2yz)+3z^2$

$=(x-2y+3z)^2+2(y+z)^2+3z^2\geq 0$ với mọi $x,y,z$

Vậy $C_{\min}=0$.

Giá trị này đạt tại $x-2y+3z=y+z=z=0$

$\Leftrightarrow x=y=z=0$