a) |2x-2|=|2x+3|

TH1: 2x-2=2x+3

=> 2x-2=2x-2+5 ( vô lý )

=> Không tồn tại x

TH2: 2x-2=-2x-3

=> 2x+2x+3=2

=> 4x=-1

=> x=-1/4

Vậy: x=-1/4

b) \(A=\frac{1}{\sqrt{x-2}+3}\)

Để A đạt giá trị lớn nhất thì \(\sqrt{x-2}+3\) phải đạt giá trị nhỏ nhất

Có: \(\sqrt{x-2}\ge0\Rightarrow\sqrt{x-2}+3\ge3\)

Dấu = xảy ra khi x=2

Vậy: \(Max_A=\frac{1}{3}\) tại x=2

c) Có: \(\frac{2x+1}{x-2}< 2\Rightarrow\frac{2x+1}{x-2}-2< 0\)

\(\Rightarrow\frac{2x+1}{x-2}-\frac{2\left(x-2\right)}{x-2}< 0\)

\(\Rightarrow\frac{2x+1-2x+4}{x-2}< 0\)

\(\Rightarrow\frac{5}{x-2}< 0\)

\(\Rightarrow x< 2\)

a)

|2x-2| = |2x+3|

<=> \(\left[\begin{array}{nghiempt}2x-2=2x+3\\2x-2=-2x-3\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}0x=5\left(vl\right)\\4x=-1\end{array}\right.\)

<=> x = \(-\frac{1}{4}\)

\(A=\frac{3}{2x^2+2x+3}=\frac{3}{2x^2+2x+\frac{1}{2}+\frac{5}{2}}\)

\(=\frac{3}{2\left(x^2+x+\frac{1}{4}\right)+\frac{5}{2}}=\frac{3}{2\left(x+\frac{1}{2}\right)^2+\frac{5}{2}}\le\frac{3}{\frac{5}{2}}=\frac{6}{5}\)

Nên GTLN của A là \(\frac{6}{5}\) khi \(x=-\frac{1}{2}\)

Ta có: \(A=\frac{3}{2x^2+2x+3}\)

\(A=\frac{3}{2x^2+2x+\frac{1}{2}+\frac{5}{2}}\)

\(A=\frac{3}{2\left(x^2+x+\frac{1}{4}\right)+\frac{5}{2}}\)

\(A=\frac{3}{2\left(x+\frac{1}{2}\right)^2+\frac{5}{2}}\le\frac{3}{\frac{5}{2}}\)

\(A=\frac{6}{5}\)

Nên GTLN của A là \(\frac{6}{5}\) khi \(x=-\frac{1}{2}\)

\(A=\frac{3}{2x^2+2x+3}=\frac{3}{\left(2x^2+2x+\frac{1}{2}\right)+\frac{5}{2}}=\frac{3}{2\left(x^2+x+\frac{1}{4}\right)+\frac{5}{2}}\)

\(A=\frac{3}{2\left(x+\frac{1}{2}\right)^2+\frac{5}{2}}\le\frac{3}{\frac{5}{2}}=\frac{6}{5}\)

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

Vậy GTLN của \(A\) là \(\frac{6}{5}\) khi \(x=\frac{-1}{2}\)

Chúc bạn học tốt ~ 

a, \(M=\frac{3\left(x^2+1\right)}{\left(x^4+x^2\right)+\left(2x^3+2x\right)+\left(6x^2+6x\right)}=\frac{3\left(x^2+1\right)}{x^2\left(x^2+1\right)+2x\left(x^2+1\right)+6\left(x^2+1\right)}=\frac{3\left(x^2+1\right)}{\left(x^2+2x+6\right)\left(x^2+1\right)}=\frac{3}{x^2+2x+6}\)

b, ta có: \(M=\frac{3}{x^2+2x+6}=\frac{3}{\left(x^2+2x+1\right)+5}=\frac{3}{\left(x+1\right)^2+5}\)

Vì \(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2+5\ge5\Rightarrow\frac{1}{\left(x+1\right)^2+5}\le\frac{1}{5}\Rightarrow M=\frac{3}{\left(x+1\right)^2+5}\le\frac{3}{5}\)

Dấu "=" xảy ra <=>x+1=0 <=> x=-1

\(A=\frac{3x+1}{2x^2-x+3}\)

\(A=\frac{2x^2-x+3-2x^2+4x-2}{2x^2-x+3}\)

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

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

\(A=1-\frac{2\left(x-1\right)^2}{2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+\frac{23}{8}}\)

\(A=1-\frac{2\left(x-1\right)^2}{2\left(x-\frac{1}{4}\right)^2+\frac{23}{8}}\le1\)

Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(x-\frac{1}{4}\right)^2\ge0\forall x\end{cases}\Rightarrow\frac{2\left(x-1\right)^2}{2\left(x-\frac{1}{4}\right)^2+\frac{23}{8}}\ge0\forall x}\)

Dấu '' = '' xảy ra khi x = 1 

Vậy Max A =1 khi x = 1 .

Ta có: 

\(A=\frac{3x^2+6x+1}{x^2+2x+3}\)

   \(=\frac{3x^2+6x+9}{x^2+2x+3}+\frac{1}{x^2+2x+3}\)

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

    \(=3+\frac{1}{x^2+2x+3}\)

Lại có: \(x^2+2x+3=\left(x^2+2x+1\right)+2=\left(x+1\right)^2+2\ge2\)

\(\Rightarrow\frac{1}{x^2+2x+3}\le\frac{1}{2}\)

\(\Rightarrow A\le3+\frac{1}{2}=\frac{7}{2}\)

Dấu = xảy ra khi \(x^2+2x+3=2\Rightarrow x=-1\)

Vậy \(A_{Min}=\frac{7}{2}\Leftrightarrow x=-1\)

\(A=\frac{3x^2+6x+10}{x^2+2x+3}=\frac{3x^2+6x+9}{x^2+2x+3}+\frac{1}{x^2+2x+3}\)

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

Dấu "=" xảy ra <=> x=-1

Vậy GTLN của A=7/2 khi x=-1

\(x^2-2x+2=x^2-2x+1+1=\left(x-1\right)^2+1\ge1\)

\(\Rightarrow\sqrt{x^2-2x+2}\ge1\)

\(\Rightarrow2+\sqrt{x^2-2x+2}\ge2+1=3\)

\(\Rightarrow\frac{3}{2+\sqrt{x^2-2x+2}}\le\frac{3}{3}\)

\(\Rightarrow\frac{-3}{2+\sqrt{x^2-2x+2}}\ge\frac{-3}{3}=-1\)

vậy Amin = -1 khi x=1

Không có giá trị lớn nhất bạn nhé, hoặc là viết nhầm biểu thức hoặc nhầm câu hỏi. Chúc bạn may mắn.

Vì \(x^2-2x+2=\left(x-1\right)^2+1\ge1\)nên ta có :

 \(\Leftrightarrow\sqrt{\left(x-1\right)^2+1}\ge1\)

\(\Leftrightarrow2+\sqrt{x^2-2x+2}\ge3\)

\(\Leftrightarrow-\frac{3}{2+\sqrt{x^2-2x+2}}\le-\frac{3}{3}=-1\)

\(\Rightarrow A_{Max}=-1\)