Question

Chứng minh
a,A=a^2-4a+10>0 với mọi a
b,B=(x-3)(x-5)+4>0 với mọi x

Answer 1

`A=a^2 - 4a + 10`

`=a^2 - 4a + 4 + 6`

`=(a-2)^2 + 6`

Mà : `(a-2)^2 \(\ge0\forall a\)

`=> (a-2)^2 + 6`\(\ge0\forall a\)

`=>a^2 - 4a + 10`\(\ge0\forall a\left(đfcm\right)\)

__________________________

`B=(x-3).(x-5)+4`

`=x^2 -8x+15+4`

`=x^2 - 8x + 19`

`=x^2-8x + 16 + 3`

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

Mà : `(x-4)^2`\(\ge0\forall x\)

`=> (x-4)^2 + 3`\(\ge0\forall x\)

Answer 2

a: \(=a^2-4a+4+6=\left(a-2\right)^2+6>0\)

b: \(=x^2-8x+15+4\)

\(=x^2-8x+16+3=\left(x-4\right)^2+3>0\)