a) \(x^2+xy+y^2+1\)

\(=x^2+xy+\dfrac{y^2}{4}-\dfrac{y^2}{4}+y^2+1\)

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

mà \(\left\{{}\begin{matrix}\left(x+\dfrac{y}{2}\right)^2\ge0,\forall x;y\\\dfrac{3y^2}{4}\ge0,\forall x;y\end{matrix}\right.\)

\(\Rightarrow\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0,\forall x;y\)

\(\Rightarrow dpcm\)

b) \(...=x^2-2x+1+4\left(y^2+2y+1\right)+z^2-6z+9+1\)

\(=\left(x-1\right)^2+4\left(y^{ }+1\right)^2+\left(z-3\right)^2+1>0,\forall x.y\)

\(\Rightarrow dpcm\)

b.

$x^2+4y^2+z^2-2x-6z+8y+15=(x^2-2x+1)+(4y^2+8y+4)+(z^2-6z+9)+1$

$=(x-1)^2+(2y+2)^2+(z-3)^2+1\geq 0+0+0+1>0$ với mọi $x,y,z$

Ta có đpcm.

Câu 29:

a: \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)

\(\Leftrightarrow-a^2+2ab-b^2\le0\)

\(\Leftrightarrow-\left(a-b\right)^2\le0\)(luôn đúng)

Hả lơp 1 ????????

\(14,P=x^2+xy+y^2-3x-3y+3\\ P=\left(x^2+xy+\dfrac{1}{4}y^2\right)-3\left(x+\dfrac{1}{2}y\right)+\dfrac{3}{4}y^2-\dfrac{3}{2}y+3\\ P=\left(x+\dfrac{1}{2}y\right)^2-3\left(x+\dfrac{1}{2}y\right)+\dfrac{9}{4}+\dfrac{3}{4}\left(y^2-2y+1\right)\\ P=\left(x+\dfrac{1}{2}y-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2\ge0\)

đây là lớp 4 ư

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) \(x^2+y^2+z^2\ge xy+yz+xz\)

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

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

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

\(ĐTXR\Leftrightarrow x=y=z\)

\(x^2+4y^2+z^2-2x-6z+8y+14=0\\\Leftrightarrow (x^2-2x+1)+(4y^2+8y+4)+(z^2-6z+9)=0\\\Leftrightarrow (x^2-2\cdot x\cdot1+1^2)+[(2y)^2+2\cdot2y\cdot 2+2^2]+(z^2-2\cdot z\cdot3+3^2)=0\\\Leftrightarrow (x-1)^2+(2y+2)^2+(z-3)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(2y+2\right)^2\ge0\forall y\\\left(z-3\right)^2\ge0\forall z\end{matrix}\right.\)

\(\Rightarrow\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\ge0\forall x;y;z\)

Mặt khác: \(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2=0\)

nên ta được: 

\(\left\{{}\begin{matrix}x-1=0\\2y+2=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=3\end{matrix}\right.\)

Vậy: ...

\(x^2+4y^2+z^2-2x-6z+8y+14=0\)

\(\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)=0\)

\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2=0\) (1)

Do \(\left(x-1\right)^2\ge0;\left(2y+2\right)^2\ge0;\left(z-3\right)^2\ge0\)

\(\left(1\right)\Rightarrow\) \(\left(x-1\right)^2=0;\left(2y+2\right)^2=0;\left(z-3\right)^2=0\)

*) \(\left(x-1\right)^2=0\)

\(x-1=0\)

\(x=1\)

*) \(\left(2y+2\right)^2=0\)

\(2y+2=0\)

\(2y=-2\)

\(y=-1\)

*) \(\left(z-3\right)^2=0\)

\(z-3=0\)

\(z=3\)

Vậy x = 1; y = -1; z = 3

Latex của mình có chút lỗi nên xin phép đánh máy hoàn toàn nhé
x^2+4y^2+z^2-2x-6z+8y+14=0
<=> (x^2-2x+1) +4(y^2+2y+1) +(z^2-6z+9)=0
<=> (x-1)^2+4(y+1)^2+(z-3)^2=0
Do (x-1)^2, (y+1)^2>=0, (z-3)^2>=0
Nên x-1=0, y+1=0, z-3=0
<=> x=1, y=-1, z=3

làm tắt ko hiểu thì hỏi 

a) \(=x^2+2.xy.\frac{1}{2}+\frac{1}{4}y^2-\frac{1}{4}y^2+y^2+1\)

\(=\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2+1>0\)

b) \(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6x+9\right)+1\)

\(=\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1>0\)

Bài làm:

a) Ta có: \(-4x^2-4x-2=-\left(4x^2+4x+1\right)-1\)

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

=> đpcm

b) \(x^2+4y^2+z^2-2x-6z+8y+15\)

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

\(=\left(x-1\right)^2+4\left(y-1\right)^2+\left(z-3\right)^2+1\ge1>0\left(\forall x\right)\)

=> đpcm

a) Ta có: \(-4x^2-4x-2=-\left(4x^2+4x+1\right)-1\)

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

    Vì \(-\left(2x+1\right)^2\le0\forall x\)\(\Rightarrow\)\(-\left(2x+1\right)^2-1\le-1\forall x\)

              \(\Rightarrow\)\(-\left(2x+1\right)^2-1< 0\forall x\)

              \(\Rightarrow\)\(-4x^2-4x-2< 0\forall x\)( ĐPCM )

b) Ta có: \(x^2+4y^2+z^2-2x-6z+8y+15\)

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

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

    Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(2y+2\right)^2\ge0\forall y\\\left(z-3\right)^2\ge0\forall z\end{cases}}\)\(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\ge0\forall x,y,z\)

          \(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1\ge1\forall x,y,z\)

          \(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1>0\forall x,y,z\)( ĐPCM )

a) Ta có : -4x² - 4x - 2 = -(4x² + 4x + 1) - 1 = -(2x + 1)² - 1 < 0 (đpcm)

b) x² + 4y² + z² - 2x - 6z + 8y + 15

= (x² - 2x + 1) + (z² - 6z + 9) + (4y² + 8y + 4) + 1

= (x - 1)² + (z - 3)² + 4(y + 1)² + 1 > 0 (đpcm)

\(x\) mũ bao nhiêu thì cô và các bạn mới giúp được chứ em?

7) Chứng minh rằng: x^2 +4y^2 + z^2- 2x -6z +8y + 15 > 0 với mọi x, y, z.

Để được trợ giúp nhanh chóng thì lần sau nhớ ghi đề bài cẩn thận em nhé.

A = \(x^2\) + 4y² + z² - 2\(x\) - 6z + 8y + 15

A = (\(x^2\) - 2\(x\) + 1) + (4y² + 8y + 4) + (z² - 6z + 9) + 1

A = (\(x\) -1)² + (2y+2)² + (z-3)² + 1

Vì (\(x-1\))² ≥ 0 ∀ \(x\) ;  (2y +2)² ≥ 0 ∀ y; (z-3)² ≥ 0 ∀ z

⇒ A = (\(x\) - 1)² + (2y+2)² + (z-3)² + 1 ≥ 1 > 0 ∀ \(x\); y;z (đpcm)