giúp em với ạ
chiều nay em phải nộp rồi ạ
a) \(x^2+6x+10=\left(x^2+6x+9\right)+1=\left(x+3\right)^2+1\) (luôn dương)
b) \(x^2-8x+18=\left(x^2-8x+16\right)+2=\left(x-4\right)^2+2\) (luôn dương)
c) \(x^2-10x+35=\left(x-5\right)^2+10\)(luôn dương)
d) \(9x^2-6x+5=\left(3x-1\right)^2+4\) (luôn dương)
e) \(4x^2-6x+7=\left(2x-\dfrac{3}{2}\right)^2+\dfrac{19}{4}\) (luôn dương)
f) \(4x^2-10x+35=\left(2x+\dfrac{5}{2}\right)^2+\dfrac{115}{4}\) (luôn dương)
a)=(x+3)^2 +1 d)=(3x-1)^2 +4
b)=(x-4)^2 +2 e)=(2x-6/4)^2 +19/4
c)=(x-5)^2 +10 f)=(2x-5/2)^2 +25/4
g) \(-x^2-5x-7=-\left(x^2+5x+7\right)=-\left[\left(x+\dfrac{5}{2}\right)^2+\dfrac{3}{4}\right]\) (luôn âm)
h) \(-4x^2+4x-5=-\left(4x^2-4x+5\right)=-\left[\left(2x-1\right)^2+4\right]\) (luôn âm)
i) \(12x-9x^2-5=-\left(9x^2-12x+5\right)=-\left[\left(3x-2\right)^2+1\right]\) (luôn âm)
k) \(2x-16x^2-3=-\left(16x^2-2x+3\right)=-\left[\left(4x-\dfrac{1}{4}\right)^2+\dfrac{47}{16}\right]\) (luôn âm)
m) \(-x^2+x-1=-\left(x^2-x+1\right)=-\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\) (luôn âm)
n) \(3x-x^2-4=-\left(x^2-3x+4\right)=-\left[\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}\right]\) (luôn âm)
g. \(-x^2-5x-7=-\left(x^2+5x+7\right)=-\left[\left(x+\dfrac{5}{2}\right)^2+\dfrac{3}{2}\right]\le-\dfrac{3}{2}< 0\)
h. \(-4x^2+4x-5=-\left(4x^2-4x+5\right)=-\left[\left(2x-1\right)^2+4\right]\le-4< 0\)
i. \(12x-9x^2-5=-\left(9x^2-12x+5\right)=-\left[\left(3x-2\right)^2+1\right]\le-1< 0\)
k. \(2x-16x^2-3=-\left(16x^2-2x+3\right)=-\left[\left(4x-\dfrac{1}{4}\right)^2+\dfrac{47}{16}\right]\le-\dfrac{47}{16}< 0\)
m. \(-x^2+x-1=-\left(x^2-x+1\right)=-\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\le-\dfrac{3}{4}< 0\)
n. \(3x-x^2-4=-\left(x^2-3x+4\right)=-\left[\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}\right]\le-\dfrac{7}{4}< 0\)
m: Ta có: \(-x^2+x-1\)
\(=-\left(x^2-x+1\right)\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}< 0\forall x\)
