Bài 1: 

Ta có: \(D=\sqrt{16x^4}-2x^2+1\)

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

\(=2x^2+1\)

a: Ta có: \(A=16x^2+8x+5\)

\(=16x^2+8x+1+4\)

\(=\left(4x+1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{4}\)

b: Ta có: \(B=2x^2-5x\)

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

\(=2\left(x^2-2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{25}{16}\right)\)

\(=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{25}{8}\ge-\dfrac{25}{8}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{4}\)

A = 16x² - 8x + 5

A = [ ( 4x )² - 2 . 4x . 1 + 1 ] + 4

A = ( 4x - 1 )² + 4

Vì ( 4x - 1 )² \(\ge\) 0 \(\forall\)x

=> ( 4x - 1 )² + 4 \(\ge\)4 \(\forall\)x

=> A \(\ge\)4 \(\forall\)x

=> A = 4 <=> ( 4x - 1 )² = 0

              <=> 4x - 1 = 0

              <=> 4x = 1

              <=> x = \(\frac{1}{4}\)

Vậy A min = 4 <=> x = \(\frac{1}{4}\)

\(P=-3x^2-4x\sqrt{y}+16x-2y+12\sqrt{y}+1998\)

\(\Leftrightarrow3P=-9x^2-12x\sqrt{y}-4y+16\left(3x+2\sqrt{y}\right)-64-\left(2y-4\sqrt{y}+2\right)+6060\)

\(=-\left(3y+2\sqrt{y}-8\right)^2-2\left(\sqrt{y}-1\right)^2+6060\le6060\)

=> P \(\le2020\) 

"=" khi \(\left\{{}\begin{matrix}3x+2\sqrt{y}=8\\\sqrt{y}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy Min P = 2020 khi x = 2 ; y = 1

\(a=\sqrt{25x^2-10x+1+16}=\sqrt{\left(5x-1\right)^2+16}\ge\sqrt{16}=4\)

\(a_{min}=4\) khi \(5x-1=0\Leftrightarrow x=\frac{1}{5}\)

\(b=\sqrt{x^2-10x+25+5}=\sqrt{\left(x-5\right)^2+5}\ge\sqrt{5}\)

\(b_{min}=\sqrt{5}\) khi \(x=5\)

\(c=\sqrt{-16x^2-8x-1+4}=\sqrt{4-\left(4x+1\right)^2}\le\sqrt{4}=2\)

\(c_{max}=2\) khi \(x=-\frac{1}{4}\)

a) \(\sqrt{1-8x+16x^2}=\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{1^2-2\cdot4x\cdot1+\left(4x\right)^2}=\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{\left(4x-1\right)^2}=\dfrac{1}{3}\)

\(\Leftrightarrow\left|4x-1\right|=\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1=\dfrac{1}{3}\left(ĐK:x\ge\dfrac{1}{4}\right)\\4x-1=\dfrac{1}{3}\left(ĐK:x< \dfrac{1}{4}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{4}{3}\\4x=\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\x=\dfrac{1}{6}\left(tm\right)\end{matrix}\right.\)

b) \(\sqrt{16x-32}+\sqrt{25x-50}=18+\sqrt{9x-18}\) (ĐK: \(x\ge2\)) 

\(\Leftrightarrow\sqrt{16\left(x-2\right)}+\sqrt{25\left(x-2\right)}=18+\sqrt{9\left(x-2\right)}\)

\(\Leftrightarrow4\sqrt{x-2}+5\sqrt{x-2}=18+3\sqrt{x-2}\)

\(\Leftrightarrow6\sqrt{x-2}=18\)

\(\Leftrightarrow\sqrt{x-2}=3\)

\(\Leftrightarrow x-2=9\)

\(\Leftrightarrow x=9+2\)

\(\Leftrightarrow x=11\left(tm\right)\)