a) \(A=x^2+6x+15\)
\(=x^2+6x+9+6\)
\(=\left(x+3\right)^2+6\)
Vì \(\left(x+3\right)^2\ge0\forall x\) nên \(\left(x+3\right)^2+6>0\forall x\)
Vậy ...
b) \(B=4x^2+4x+11\)
\(=4x^2+4x+1+10\)
\(=\left(2x+1\right)^2+10>0\forall x\) (trình bày như trên)
Vậy ...
a) \(-9x^2+12x-15\)
\(=-9x^2+12x-4-11\)
\(=-\left(3x-2\right)^2-11\)
Vì \(-\left(3x-2\right)^2\le0\forall x\) nên \(-\left(3x-2\right)^2-11< 0\forall x\)
Vậy ...
b) \(-5-\left(x-1\right)\left(x+2\right)\)
\(=-x^2-x+2-5\)
\(=-x^2-x-3\)
\(=-x^2-x-\dfrac{1}{4}-\dfrac{11}{4}\)
\(=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{11}{4}>0\forall x\)
Vậy ...
\(a,A=x^2+6x+15\)
\(=\left(x^2+6x+9\right)+6\)
\(=\left(x+3\right)^2+6\)
Ta có : ( x + 3 )2 ≥ 0 với mọi x
=> ( x + 3 )2 + 6 ≥ 6 > 0 với mọi x
=> A > 0 ( đpcm )
\(b,B=4x^2+4x+11\)
\(=\left(4x^2+4x+1\right)+10\)
\(=\left(2x+1\right)^2+10\ge10>0\forall x\left(đpcm\right)\)
( giải thích chi tiết thì tương tự câu a nhé bn Ttqminh2005
a, \(-9x^2+12x-15\)
\(=-\left(9x^2-12x+4\right)-11\)
\(=-\left(3x-2\right)^2-11\)
Ta có : \(-\left(3x-2\right)^2\le0\Rightarrow-\left(3x-2\right)^2-11\le-11< 0\forall x\) ( đpcm)
\(b,-5-\left(x-1\right)\left(x+2\right)\)
\(=-5-x^2-x+2\)
\(=-\left(x^2+x+3\right)\)
\(=-\left[\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{11}{4}\right]\)
\(=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{11}{4}\) < 0 ( đpcm )
