\(x^2+4y^2+z^2-2x-6z+8y+15\)

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

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

       \(x^2+5y^2+2x-4xy-10y+14\)

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

\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(y-3\right)^2+4\)

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

Chúc bạn học tốt.

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)

bạn cố tìm mọi cánh biến vế trái thành 1 dạng bình phương

rồi nó sẽ racau trả lời , gợi ý đó

sử dụng hằng đẳng thức 1.2

Chứng minh rằng:

a) A=9x^2-6x+11>0 với mọi x

b) (x^2-2xy+y^2)+1>0 với mọi x,y

c) (x-x^2+1)<0 với mọi x 

A) x²+4y²2+z²2-4x-6z+15>0 <=> (x²-2×2×x+2²)+4y²+(z²-2×3×z+3²) +(15 -2²-3²) >0

<=>(x-2)²+4y²2+(z-3)²

B) giải

(2X)²+ 2×2X×1 +1 >=0 với mọi X (   (2x+1)²  )

=> (2x+1)²+2 >0

\(A=4x^2+4x+11\)

\(=\left(4x^2+4x+1\right)+10\)

\(=\left(2x+1\right)^2+10\ge10\)

Min A = 10 khi:  2x + 1 = 0

                      <=> x = -1/2

jbdgvsvvsgvhvhb

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

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

a) \(\left(x\right)^2+2\left(x\right)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1\)

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

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

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

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

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

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

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