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

\(P_{max}=\dfrac{7}{2}\) 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 \(x=-\dfrac{3}{2}\)

\(P=\dfrac{2\left(x^2+2\right)+x^2-4x+4}{x^2+2}=2+\dfrac{\left(x-2\right)^2}{x^2+2}\ge2\)

\(P=\dfrac{5\left(x^2+2\right)-2x^2-4x-2}{x^2+2}=5-\dfrac{2\left(x+1\right)^2}{x^2+2}\le5\)

+) Giá trị nhỏ nhất

Ta có: \(A=\dfrac{6x+8}{x^2+1}=\dfrac{-\left(x^2+1\right)+x^2+6x+9}{x^2+1}\) \(=-1+\dfrac{\left(x+3\right)^2}{x^2+1}\ge-1\)

  Dấu bằng xảy ra \(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

+) Giá trị lớn nhất 

Ta có: \(A=\dfrac{6x+8}{x^2+1}=\dfrac{9\left(x^2+1\right)-9x^2+6x-1}{x^2+1}\) \(=9-\dfrac{\left(3x-1\right)^2}{x^2+1}\ge9\)

  Dấu bằng xảy ra \(\Leftrightarrow3x-1=0\Leftrightarrow x=\dfrac{1}{3}\)

  Vậy \(P_{Min}=-1\) khi \(x=-3\)

         \(P_{Max}=9\) \(\Leftrightarrow x=\dfrac{1}{3}\)

a: Ta có: \(x^2=3-2\sqrt{2}\)

nên \(x=\sqrt{2}-1\)

Thay \(x=\sqrt{2}-1\) vào A, ta được:

\(A=\dfrac{\left(\sqrt{2}+1\right)^2}{\sqrt{2}-1}=\dfrac{3+2\sqrt{2}}{\sqrt{2}-1}=7+5\sqrt{2}\)

a. Ta có : \(A=\frac{8x^2-9}{x^2+3}=\frac{8x^2+24-33}{x^2+3}=8-\frac{33}{x^2+3}\)

Để Amin thì \(\frac{33}{x^2+3}_{max}\) mà \(\frac{33}{x^2+3}\le11\)

Dấu "=" xảy ra \(\Leftrightarrow x^2+3=3\Leftrightarrow x=0\)

Vậy Amin = 8 - 11 = - 3 <=> x = 0

b. Ta có : \(B=\frac{3x^2-6x+40}{x^2-2x+5}=\frac{3\left(x^2-2x+5\right)+25}{x^2-2x+5}=3+\frac{25}{x^2-2x+5}\)

Để Bmax thì \(\frac{25}{x^2-2x+5}=\frac{25}{\left(x-1\right)^2+4}_{max}\)

mà \(\frac{25}{\left(x-1\right)^2+4}\le\frac{25}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)^2+4=4\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy Bmax \(=3+\frac{25}{4}=\frac{37}{4}\)  <=> x = 1

a) Để biểu thức vô nghĩa thì \(\dfrac{3x-2}{5}-\dfrac{x-4}{3}=0\)

\(\Leftrightarrow\dfrac{3x-2}{5}=\dfrac{x-4}{3}\)

\(\Leftrightarrow3\left(3x-2\right)=5\left(x-4\right)\)

\(\Leftrightarrow9x-6=5x-20\)

\(\Leftrightarrow9x-5x=-20+6\)

\(\Leftrightarrow4x=-14\)

\(\Leftrightarrow x=-\dfrac{7}{2}\)

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 : \(S=\left\{\dfrac{8}{3}\right\}\)