Question

Answer 1

câu 4:

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

thay x=87, y=13 ta có:

\(x^2-y^2=\left(x-y\right)\left(x+y\right)=\left(87-13\right)\left(87+13\right)=74.100=7400\)

Câu 5:

\(x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\)

Vì \(\left(x-3\right)^2\ge0;1>0\Rightarrow\left(x-3\right)^2+1>0\Rightarrow x^2-6x+10>0\)

 

 

Answer 2

Câu 5: 

Ta có: \(x^2-6x+10\)

\(=x^2-6x+9+1\)

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