Question

CMR:

a) \(x^2-4x+5>0\) với mọi x

b) \(x^2-4xy+5y^2\ge0\) với mọi x, y

c) \(3-2x-x^2< 0\)với mọi x

Answer 1

a) \(x^2-4x+5\)

= \(\left(x^2-2.2x+4\right)+1\)

= \(\left(x-2\right)^2+1\)

Ta co: \(\left(x-2\right)^2>=0\)

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

b) \(x^2-4xy+5y^2\)

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

= \(\left(x-2y\right)^2+y^2\)

Ta co: \(\left(x-2y\right)^2>=0\)

            \(y^2>=0\)

=> \(\left(x-2y\right)^2+y^2>=0\)

c) \(3-2x-x^2\)

= \(-\left(x^2+2x\right)+3\)

= \(-\left(x^2+2.1x+1-1\right)+3\)

= \(-\left(x+1\right)^2+4\)

= 

Hình như câu này sai đề ...

Answer 2

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

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

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

b) \(x^2-4xy+5y^2\)

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

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

Dấu = xảy ra khi: \(x=y=0\)

c) \(-3-2x-x^2\)

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

\(=-2-\left(x+1\right)^2=-\left[2+\left(x+1\right)^2\right]< 0\)