Bài 2: 

a: \(=-\left(x^2+4x-10\right)\)

\(=-\left(x^2+4x+4-14\right)=-\left(x+2\right)^2+14< =14\)

Dấu = xảy ra khi x=-2

b: \(=-2\left(x^2-2x+\dfrac{5}{2}\right)\)

\(=-2\left(x^2-2x+1+\dfrac{3}{2}\right)\)

\(=-2\left(x-1\right)^2-3< =-3\)

Dấu = xảy ra khi x=1

c: \(=x^2-2x+1-2\left(x^2+6x+9\right)+20\)

\(=x^2-2x+21-2x^2-12x-18\)

\(=-x^2-14x+3\)

\(=-\left(x^2+14x-3\right)\)

\(=-\left(x^2+14x+49-52\right)=-\left(x+7\right)^2+52< =52\)

Dấu = xảy ra khi x=-7

\(1,\dfrac{1}{1+x}=1-\dfrac{1}{1+y}+1-\dfrac{1}{1+z}=\dfrac{y}{1+y}+\dfrac{z}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)

Cmtt: \(\dfrac{1}{1+y}\ge2\sqrt{\dfrac{xz}{\left(1+x\right)\left(1+z\right)}};\dfrac{1}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)

Nhân VTV

\(\Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge8\sqrt{\dfrac{x^2y^2z^2}{\left(1+x\right)^2\left(1+y\right)^2\left(1+z\right)^2}}\\ \Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge\dfrac{8xyz}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\\ \Leftrightarrow8xyz\le1\Leftrightarrow xyz\le\dfrac{1}{8}\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{2}\)

\(2,\\ a,2x^2+y^2-2xy=1\\ \Leftrightarrow\left(x-y\right)^2+x^2=1\\ \Leftrightarrow\left(x-y\right)^2=1-x^2\ge0\\ \Leftrightarrow x^2\le1\Leftrightarrow\sqrt{x^2}\le1\Leftrightarrow\left|x\right|\le1\)

a) |x| + |-5| = |-37|

<=> |x| + 5 = 37

<=> |x| = 37 - 5 = 32

=> x \(\in\) {32 ; -32}

b)|-6| . |x| = |54|

<=> 6 . |x| = 54

|x| = 54 : 6 = 9

=> x \(\in\){9;-9}

c) |x| > 21

Có |x| \(\ge\) 0 > 21

=> |x| \(\in\) { 22 ; 23 ; 24 ; 25 ; ....}

=> x \(\in\) { 22; -22 ; 23; -23; 24; -24; 25; -25; ....}

a) ( 2x - 1 )( 2x + 1 ) - 4( x² + x ) = 16

⇔ 4x² - 1 - 4x² - 4x = 16

⇔ -4x - 1 = 16

⇔ -4x = 17

⇔ x = -17/4

b) 5x( x - 2013 ) - x + 2013 = 0

⇔ 5x( x - 2013 )  - ( x - 2013 ) = 0

⇔ ( x - 2013 )( 5x - 1 ) = 0

⇔ \(\orbr{\begin{cases}x-2013=0\\5x-1=0\end{cases}}\)

⇔ \(\orbr{\begin{cases}x=2013\\x=\frac{1}{5}\end{cases}}\)

a) \(\left(2x-1\right)\left(2x+1\right)-4.\left(x^2+x\right)=16\)

\(4x^2-1-4x^2-4x=16\)

\(-1-4x=16\)

\(-4x=17\)

\(x=-\frac{17}{4}\)

b) \(5x\left(x-2013\right)-x+2013=0\)

\(\left(x-2013\right)\left(5x-1\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-2013=0\\5x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=2013\\x=\frac{1}{5}\end{cases}}}\)

a) \(\left(2x-1\right)\left(2x+1\right)-4\left(x^2+x\right)=16\)

\(\Leftrightarrow4x^2-1-4x^2-4x=16\)

\(\Leftrightarrow-1-4x=16\)\(\Leftrightarrow4x=-17\)\(\Leftrightarrow x=-\frac{17}{4}\)

Vậy \(x=-\frac{17}{4}\)

b) \(5x\left(x-2013\right)-x+2013=0\)

\(\Leftrightarrow5x\left(x-2013\right)-\left(x-2013\right)=0\)

\(\Leftrightarrow\left(x-2013\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2013=0\\5x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2013\\5x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2013\\x=\frac{1}{5}\end{cases}}\)

Vậy \(x=2013\)hoặc \(x=\frac{1}{5}\)