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

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

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

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

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

gọi số tự nhiên liên tiếp đó là x; x+1;x+2;x+3

theo bài ra ta có PT \(\left(x+2\right)\left(x+3\right)-x\left(x+1\right)=50\)

                         \(\Leftrightarrow x^2+5x+6-x^2-x=50\)

                            \(\Leftrightarrow4x+6=50\)

                          \(\Leftrightarrow4x=44\Leftrightarrow x=11\)

vậy các số đó là 11,12,13

mình xin lỗi mình nhầm rồi gọi ba số là x;x+1 ;x+2 

ta có pt \(\left(x+1\right)\left(x+2\right)-x\left(x+1\right)=50\)

bạn tự giải pt nhé mình xin lỗi

a)  \(\frac{1}{9}x^4-2x^2y+9y^2=\left(\frac{1}{3}\right)^2\left(x^2\right)^2-2x^2y+\left(3y\right)^2\)

\(=\left(\frac{1}{3}x^2\right)^2-2\frac{1}{3}x^23y+\left(3y\right)^2\)

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

b)  \(25x^2-20xy+4y^2=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)

\(=\left(5x-2y\right)^2\)

\(\frac{1}{9}x^4-2x^2y+9y^2\)

\(=\left(\frac{1}{3}x^2\right)^2-2\times\frac{1}{3}x^2\times3y+\left(3y\right)^2\)

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

\(25x^2-20xy+4y^2\)

\(=\left(5x\right)^2-2\times5x\times2y+\left(2y\right)^2\)

\(=\left(5x-2y\right)^2\)