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

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

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

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

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

\(#NqHahh\)

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

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

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

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 $x-2y+3z=y+z=z=0$

$\Leftrightarrow x=y=z=0$

a) Ta có: `a + b + c = 0`

`=> (a+b+c)^2 = 0`

`=> a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 0`

`=> 2 + 2 (ab+bc+ca) = 0`

`=> 1 + ab + bc + ca = 0`

`=> ab+bc+ca = -1`

`=> (ab+bc+ca)^2 = 1`

`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 2ab . bc + 2bc . ca + 2. ca. ab = 1`

`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 2abc(b+c+a) = 1`

`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 2abc . 0 = 1`

`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 0 = 1`

`=> a^2 b^2 + b^2 c^2 + c^2 a^2 = 1`

Mà `(a^2 + b^2 + c^2)^2 = 4`

`=> a^4 + b^4 + c^4 + 2a^2 b^2 + 2b^2 c^2 + 2c^2 a^2 = 4`

`=> a^4 + b^4 + c^4 + 2 (a^2 b^2 + b^2 c^2 + c^2 a^2) = 4`

`=>  a^4 + b^4 + c^4 + 2 . 1 = 4`

`=>  a^4 + b^4 + c^4 = 2`

Vậy `a^4 + b^4 + c^4 = 2`

--------------------------

b) Ta có: `a + b + c = 0`

`=> (a+b+c)^2 = 0`

`=> a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 0`

`=> 1 + 2 (ab+bc+ca) = 0`

`=> 0,5 + ab + bc + ca = 0`

`=> ab+bc+ca = -0,5`

`=> (ab+bc+ca)^2 = 0,25`

`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 2ab . bc + 2bc . ca + 2. ca. ab = 0,25`

`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 2abc(b+c+a) = 0,25`

`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 2abc . 0 = 0,25`

`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 0 = 0,25`

`=> a^2 b^2 + b^2 c^2 + c^2 a^2 = 0,25`

Mà `(a^2 + b^2 + c^2)^2 = 1`

`=> a^4 + b^4 + c^4 + 2a^2 b^2 + 2b^2 c^2 + 2c^2 a^2 = 1`

`=> a^4 + b^4 + c^4 + 2 (a^2 b^2 + b^2 c^2 + c^2 a^2) = 1`

`=>  a^4 + b^4 + c^4 + 2 . 0,25 = 1`

`=>  a^4 + b^4 + c^4 = 0,5`

Vậy `a^4 + b^4 + c^4 = 0,5`

\(\left(\dfrac{1}{2}xy^2+x^2-x^2y\right)-M=-xy^2+x^2y+1\)

=>\(M=\dfrac{1}{2}xy^2+x^2-x^2y+xy^2-x^2y-1\)

=>\(M=\dfrac{3}{2}xy^2+x^2-2x^2y-1\)

M= xy^2+x^y1-(1/2xy^2+x^2-x^2y)

M=xy^2+x^2y+1-1/2xy^2-x^2+x^2y

M=1/2xy^2+2x^2y-x^2-1

ok bẠN

rồi bn trả lời đi