a, \(-x^2-4x-2015\)
\(=-\left(x^2+4x+4+2011\right)\)
\(=-\left[\left(x+2\right)^2+2011\right]\)
\(=-\left(x+2\right)^2-2011\le-2011< 0\)
\(\Rightarrow\)Đa thức trên vô nghiệm ( đpcm )
Vậy...
b, \(x\left(x-1\right)+1=x^2-x+1\)
\(=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
\(\Rightarrow\)Đa thức trên vô nghiệm
Vậy...
a)\(-x^2-4x-2015\)
\(=-x^2-4x-4-2011\)
\(=-\left(x^2+4x+4\right)-2011\)
\(=-\left(x+2\right)^2-2011< 0\) (vô nghiệm)
b)\(x\left(x-1\right)+1\)
\(=x^2-x+1\)
\(=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) (vô nghiệm)
a) \(-x^2-4x-2015=-\left(x^2+4x+2015\right)\)
\(=-\left(x^2+2x+2x+2015\right)\)
\(=-\left(x^2+2x+2x+4+2011\right)\)
\(=-\left[\left(x^2+2x\right)+\left(2x+4\right)+2011\right]\)
\(=-\left[x\left(x+2\right)+2\left(x+2\right)+2011\right]\)
\(=-\left[\left(x+2\right)^2+2011\right]\)
\(=-\left(x+2\right)^2-2011\)
Vì \(-\left(x+2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x+2\right)^2-2011\le-2011\)
\(\Rightarrow-\left(x+2\right)^2-2011< 0\)ư
\(\Rightarrow\) đa thức này vô nghiệm.
b) \(x\left(x-1\right)+1=x^2-x+1\)
\(=x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+1\)
\(=x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x^2-\dfrac{1}{2}x\right)-\left(\dfrac{1}{2}x-\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=x\left(x-\dfrac{1}{2}\right)-\dfrac{1}{2}\left(x-\dfrac{1}{2}\right)+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Do \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
\(\Rightarrow\) đa thức này vô nghiệm.