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)

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

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

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

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

\(=-\left(x-3\right)^2-5\le-5\)=>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. \(2x^2-4x+10=x^2-2x+1+x^2-2x+1+8=\left(x-1\right)^2+\left(x-1\right)^2+8=2\left(x-1\right)^2+8\)

Vì \(2\left(x-1\right)^2\ge0\Rightarrow2\left(x-1\right)^2+8\ge8\)

Vậy...

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

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Vậy..

c. \(2x^2-6x+5=x^2-4x+4+x^2-2x+1=\left(x-2\right)^2+\left(x-1\right)^2\)

Vì \(\hept{\begin{cases}\left(x-2\right)^2\ge0\\\left(x-1\right)^2\ge0\end{cases}}\Rightarrow\left(x-2\right)^2+\left(x-1\right)^2\ge0\)

Vậy...

a) Đặt  \(A=x^2+4x+7\)

\(A=\left(x^2+4x+4\right)+3\)

\(A=\left(x+2\right)^2+3\)

Mà  \(\left(x+2\right)^2\ge0\forall x\)

\(\Rightarrow A\ge3>0\)

b) Đặt  \(B=4x^2-4x+5\)

\(B=\left(4x^2-4x+1\right)+4\)

\(B=\left(2x-1\right)^2+4\)

Mà  \(\left(2x-1\right)^2\ge0\forall x\)

\(\Rightarrow B\ge4>0\)

c) Đặt  \(C=x^2+2y^2+2xy-2y+3\)

\(C=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+2\)

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

Mà  \(\left(x+y\right)^2\ge0\forall x;y\)

      \(\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow C\ge2>0\)

gọi ba số đó là x; x+1; x+2 

theo bài ra ta có PT \(x\left(x+1\right)+\left(x+1\right)\left(x+2\right)+x\left(x+2\right)=107\)

                           \(\Leftrightarrow x^2+x+x^2+3x+2+x^2+2x=107\)

                              \(\Leftrightarrow3x^2+6x-105=0\)

                         \(\Leftrightarrow3\left(x^2+2x-35\right)=0\)

                            \(\Leftrightarrow x^2+7x-5x-35=0\)

                            \(\Leftrightarrow x\left(x+7\right)-5\left(x+7\right)=0\)

                              \(\Leftrightarrow\left(x+7\right)\left(x-5\right)=0\)

\(\orbr{\begin{cases}x+7=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-7\left(loại\right)\\x=5\left(tm\right)\end{cases}}}\)

vậy số đó là 5,6,7

\(\left(a^2+b^2\right)\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\ge4\)

\(\Leftrightarrow1+\frac{a^2}{b^2}+\frac{b^2}{a^2}+1-4\ge0\)

\(\Leftrightarrow\frac{a^4+b^4-2a^2b^2}{a^2b^2}\ge0\)

\(\Leftrightarrow\frac{\left(a^2-b^2\right)^2}{a^2b^2}\ge0\) (luôn đúng)

Vậy...