Bài 1: 

Ta có: \(D=\sqrt{16x^4}-2x^2+1\)

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

\(=2x^2+1\)

\(A=\dfrac{2x-6+7}{x-3}=2+\dfrac{7}{x-3}\)

Do 2 không đổi nên A lớn nhất khi \(\dfrac{7}{x-3}\) lớn nhất

\(\Rightarrow x-3\) là số dương nhỏ nhất

\(\Rightarrow x-3=1\) (do \(x-3\) nguyên)

\(\Rightarrow x=4\)

Khi đó: \(A=2+\dfrac{7}{4-3}=9\)

\(D=\dfrac{15}{3\left|2x+1\right|+5}\)

Ta có:

\(\left\{{}\begin{matrix}15>0\\3\left|2x+1\right|\ge5\forall x\end{matrix}\right.\)Nên:

\(\Rightarrow D=\dfrac{15}{3\left|2x-1\right|+5}\le3\left(=\dfrac{15}{5}\right)\forall x\) 

Dấu "=" xảy ra:

\(\dfrac{15}{3\left|2x+1\right|+5}=3\)

\(\Rightarrow3\left|2x+1\right|+5=5\)

\(\Rightarrow3\left|2x+1\right|=0\)

\(\Rightarrow\left|2x+1\right|=0\)

\(\Rightarrow2x+1=0\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=-\dfrac{1}{2}\)

Vậy: \(D_{max}=3\) khi \(x=-\dfrac{1}{2}\)

D = \(\dfrac{15}{3.\left|2x-1\right|+5}\)  vì |2\(x\) - 1| ≥ 0 ∀ \(x\) ⇒3.|2\(x-1\)| + 5 ≥ 5 ∀ \(x\)

⇒D = \(\dfrac{15}{3.\left|2x-1\right|+5}\) ≤ \(\dfrac{15}{5}\) = 3 dấu bằng xảy ra khi 2\(x\) - 1 =0 ⇒ \(x=\dfrac{1}{2}\)

Kết luận D_min = 3 ⇔ \(x\) = \(\dfrac{1}{2}\)

\(P=\dfrac{3\left(x^2+2x+3\right)+1}{x^2+2x+3}=3+\dfrac{1}{x^2+2x+3}=3+\dfrac{1}{\left(x+1\right)^2+2}\le3+\dfrac{1}{2}=\dfrac{7}{2}\)

\(P_{max}=\dfrac{7}{2}\) khi \(x=-1\)

\(M=\dfrac{2\left(x^2+3x+3\right)+1}{x^2+3x+3}=2+\dfrac{1}{x^2+3x+3}=2+\dfrac{1}{\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}}\le2+\dfrac{1}{\dfrac{3}{4}}=\dfrac{10}{3}\)

\(M_{max}=\dfrac{10}{3}\) khi \(x=-\dfrac{3}{2}\)

a: |2x-3|=1

=>2x-3=1 hoặc 2x-3=-1

=>x=1(nhận) hoặc x=2(loại)

KHi x=1 thì \(A=\dfrac{1+1^2}{2-1}=2\)

b: ĐKXĐ: x<>-1; x<>2

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

\(A=0,6+\left|\dfrac{1}{2}-x\right|\\ Vì:\left|\dfrac{1}{2}-x\right|\ge\forall0x\in R\\ Nên:A=0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\forall x\in R\\ Vậy:min_A=0,6\Leftrightarrow\left(\dfrac{1}{2}-x\right)=0\Leftrightarrow x=\dfrac{1}{2}\)

\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\\ Vì:\left|2x+\dfrac{2}{3}\right|\ge0\forall x\in R\\ Nên:B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\forall x\in R\\ Vậy:max_B=\dfrac{2}{3}\Leftrightarrow\left|2x+\dfrac{2}{3}\right|=0\Leftrightarrow x=-\dfrac{1}{3}\)

\(\Leftrightarrow Mx^2+M=4x-3\\ \Leftrightarrow Mx^2-4x+M+3=0\\ \text{PT có nghiệm nên }\Delta'=4-M\left(M+3\right)\ge0\\ \Leftrightarrow4-M^2-3M\ge0\\ \Leftrightarrow-4\le M\le1\)

Vậy \(M_{max}=1\Leftrightarrow\dfrac{4x-3}{x^2+1}=1\Leftrightarrow x^2+1-4x+3=0\Leftrightarrow x=2\)