\(-5-\left(x-1\right)\left(x+2\right)=-5-\left(x^2+x-2\right)=-5-x^2-x+2\)

\(=-x^2-x-3=-\left(x+\frac{1}{2}\right)^2-\frac{11}{4}< 0,\forall x\inℝ\)

a,\(-\left(x^2-3x+4\right)\)

   \(-\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]\)

\(\Leftrightarrow-\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\le-\frac{7}{4}\)(luôn âm)

b\(-2\left(x^2-5x+\frac{15}{2}\right)\)

\(-2\left[\left(x-\frac{5}{2}\right)^2+\frac{5}{4}\right]\)

\(-2\left(x-\frac{5}{4}\right)^2-\frac{5}{2}\le-\frac{5}{2}\)(luôn âm)

c,\(-\left[\left(4x^2-4x+1\right)+\left(2y^2-6y+5\right)\right]\)

 \(=-\left[\left(2x-1\right)^2+2\left(y^2-3y+\frac{5}{2}\right)\right]\)

\(=-\left[\left(2x-1\right)^2+2\left(y-\frac{3}{2}\right)^2+\frac{1}{4}\right]\)

\(=-\left[\left(2x-1\right)^2+2\left(y-\frac{3}{2}\right)^2\right]-\frac{1}{4}\le-\frac{1}{4}\)(luôn âm)

\(A=-x^2+3x-7\)

\(=-\left(x^2-3x+7\right)\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{19}{4}\right)\)

\(=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{19}{4}< 0\forall x\)

\(3x-7-x^2=-\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{19}{4}=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{19}{4}\le-\dfrac{19}{4}< 0\)

\(-9x^2+12x-15\)

\(=-\left[\left(3x\right)^2-2.3x.2+2^2\right]-11\)

\(=-\left(3x-2\right)^2-11\)

Ta có: \(\left(3x-2\right)^2\ge0\forall x\)

\(\Rightarrow-\left(3x-2\right)^2\le0\forall x\)

\(\Rightarrow-\left(3x-2\right)^2-11\le-11\forall x\)

\(\Rightarrow-\left(3x-2\right)^2-11< 0\forall x\)

\(\Rightarrow-9x^2+12x-15< 0\forall x\)

                                           đpcm

Tham khảo nhé~

\(a,-x^2+6x-16\)

\(=-x^2+3x+3x-9-5\)

\(=-x\left(x-3\right)+3\left(x-3\right)-5\)

\(=\left(3-x\right)\left(x-3\right)-5\)

\(=-\left(x-3\right)^2-5\le-5\)=>Luôn âm

\(c,-1+x-x^2\)

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

\(=-\left(x^2-x+\frac{1}{2}+\frac{1}{2}\right)\)

\(=-\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\le\frac{-1}{2}\)=>Luôn âm

-a + a - 3 = -3

=>  luôn âm

ta có \(-a^2+a-3=-\left(a^2-\frac{2a.1}{2}+\frac{1}{4}\right)+\frac{1}{4}-3\)

  = \(-\left(a-\frac{1}{2}\right)^2-2.75\)

vì \(-\left(a-\frac{1}{2}\right)^2\le0\)với mọi a 

nên biểu thức luôn âm

\(-a^2+a-3\)

\(=-\left(a^2-a+3\right)\)

\(=-\left(a^2-2.\frac{1}{2}a+\frac{1}{4}-\frac{1}{4}+3\right)\)

\(=-\left[\left(a-\frac{1}{2}\right)^2+\frac{11}{4}\right]\)

Vì \(\left(a-\frac{1}{2}\right)^2\ge0\)

\(\Rightarrow\left(a-\frac{1}{2}\right)^2+\frac{11}{4}>0\)

\(\Rightarrow-\left[\left(a-\frac{1}{2}\right)^2+\frac{11}{4}\right]< 0\)

\(\Leftrightarrow-a^2+a-3< 0\)\(\left(đpcm\right)\)

a: ta có: \(A=x^2-3x+10\)

\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{31}{4}\)

\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}>0\forall x\)

b: Ta có: \(B=x^2-5x+2021\)

\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{8015}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{8015}{4}>0\forall x\)