a, Ta có: \(x^2-2x+1999=x^2-x-x+1+1998\)
\(=x\left(x-1\right)-\left(x-1\right)+1998\)
\(=\left(x-1\right)\left(x-1\right)+1998=\left(x-1\right)^2+1998\)
Do \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+1998\ge1998>0\)
Vậy đa thức \(x^2-2x+1999\) vô nghiệm
b, Ta có: \(x^2+3x+5=x^2+\dfrac{3}{2}.2x+\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\)
Vậy đa thức \(x^2+3x+5\) vô nghiệm
a) x2 - 2x + 1999
= x2 - x - x + 1 + 1998
= x(x-1) -1(x-1) + 1998
= (x-1).(x-1) + 1998
= (x-1)2 + 1998. Vì (x-1)2\(\ge\)0 \(\forall\) x
\(\Rightarrow\)(x-1)2+1998 > 0 \(\forall\) x
Vậy đa thức trên vô nghiệm
b) x2 + 3x + 5
= x2 + \(\dfrac{3}{2}\)x + \(\dfrac{3}{2}\)x + \(\dfrac{9}{4}\)+ \(\dfrac{11}{4}\)
= x(x+\(\dfrac{3}{2}\)) + \(\dfrac{3}{2}\)(x+\(\dfrac{3}{2}\)) + \(\dfrac{11}{4}\)
= (x+\(\dfrac{3}{2}\)).(x+\(\dfrac{3}{2}\))+\(\dfrac{11}{4}\)
= (x+\(\dfrac{3}{2}\))2+\(\dfrac{11}{4}\). Vì (x+\(\dfrac{3}{2}\))2 \(\ge\) 0 \(\forall\) x
\(\Rightarrow\)(x+\(\dfrac{3}{2}\))2 +\(\dfrac{11}{4}\)> 0 \(\forall\) x
Vậy đa thức trên vô nghiệm.
chúc bn thi tốt!!!!!!!!!
ai giúp mik vs mai mik thi rồi
