Bài 1:

\(D=\dfrac{5x^2-30x+53}{x^2-6x+10}=\dfrac{5\left(x^2-6x+10\right)+3}{x^2-6x+10}=5+\dfrac{3}{x^2-6x+10}\)

\(=5+\dfrac{3}{\left(x-3\right)^2+1}\)

Ta có: \(\left(x+3\right)^2+1\ge1\Rightarrow\dfrac{3}{\left(x-3\right)^2+1}\le3\)

\(\Rightarrow D\le3+5=8\)

Vậy max D= 8 <=> x=3

Bài 2: 

\(8\left(x-3\right)^3+x^3=6x^2-12x+8\)

\(\Leftrightarrow\left[2\left(x-3\right)^3\right]=-x^3+3.2x^2-3.2^2x+2^3\)

\(\Leftrightarrow\left(2x-6\right)^3=\left(2-x\right)^3\)

\(\Leftrightarrow2x-6=2-x\)

\(\Leftrightarrow3x=8\Leftrightarrow x=\dfrac{8}{3}\)

Vậy tập nghiệm : \(S=\left\{\dfrac{8}{3}\right\}\)

a) Ta có: \(x^3-9x^2+19x-11=0\)

\(\Leftrightarrow x^3-x^2-8x^2+8x+11x-11=0\)

\(\Leftrightarrow x^2\left(x-1\right)-8x\left(x-1\right)+11\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-8x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-8x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{5}+4\\x=-\sqrt{5}+4\end{matrix}\right.\)

Vậy: \(S=\left\{1;\sqrt{5}+4;-\sqrt{5}+4\right\}\)

           \(4x^2-12x+5=0\)

\(\Leftrightarrow\)\(4x^2-10x-2x+5=0\)

\(\Leftrightarrow\)\(2x\left(2x-5\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow\)\(\left(2x-1\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=0\\2x-5=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0,5\\x=2,5\end{cases}}\)

Vậy...

\(I\)\(don't\)\(know\)\(it\)!!!!!!!!!!!!!!!!

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

\(x^6-6x^4-64x^3+12x^2-8=0\)

\(\Leftrightarrow\left(x^2-4x-2\right)\left(x^4+4x^3+12x^2-8x+4\right)=0\)

\(\Leftrightarrow\left(x^2-4x-2\right)\left[\left(x^4+4x^3+4x^2\right)+\left(8x^2-8x+\frac{8}{4}\right)+2\right]=0\)

\(\Leftrightarrow\left(x^2-4x-2\right)\left[\left(x^2+2x\right)^2+8\left(x-\frac{1}{2}\right)^2+2\right]=0\)

\(\Leftrightarrow x^2-4x-2=0\)

\(\Leftrightarrow x=2\pm\sqrt{6}\)

a) Điều kiện xác định \(16x+8\ge0\Leftrightarrow x\ge-\frac{1}{2}.\)

Theo bất đẳng thức Cô-Si cho 4 số ta được 

\(4\sqrt[4]{16x+8}=4\sqrt[4]{2\cdot2\cdot2\cdot\left(2x+1\right)}\le2+2+2+2x+1=2x+7\)

Do vậy mà \(4x^3+4x^2-5x+9\le2x+7\Leftrightarrow\left(2x-1\right)^2\left(x+2\right)\le0\).

Vì \(x\ge-\frac{1}{2}\to x+2>0\to\left(2x-1\right)^2\le0\to x=\frac{1}{2}.\) 

b. Ta viết phương trình dưới dạng sau đây  \(9x^4-21x^3+27x^2+16x+16=0\Leftrightarrow3x^2\left(3x^2-7x+7\right)+4\left(x+2\right)^2=0\)

Vì \(3x^2-7x+7=\frac{36x^2-2\cdot6x\cdot7+49+35}{12}=\frac{\left(6x-7\right)^2+35}{12}>0\) nên vế trái dương, suy ra phương trinh vô nghiệm.