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

Dấu "=" xảy ra khi \(x=2\)

Áp dụng BĐT cosi:

\(A=\sqrt{\left(2x+1\right)\left(x+2\right)}+2\sqrt{x+3}-2x\\ A\le\dfrac{2x+1+x+2}{2}+\dfrac{4+x+3}{2}-2x\\ A\le\dfrac{3x+3}{2}+\dfrac{x+7}{2}-2x=\dfrac{3x+3+x+7-4x}{2}=5\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2x+1=x+2\\4=x+3\end{matrix}\right.\Leftrightarrow x=1\)

\(P=\sqrt{\left(x+2\right)\left(2x+1\right)}+2\sqrt{x+3}-2x\)

\(P\le\dfrac{1}{2}\left(x+2+2x+1\right)+\dfrac{1}{2}\left(4+x+3\right)-2x=5\)

\(P_{max}=5\) khi \(x=1\)

hông biết mới học lớp 6 làm seo biết đc toán lớp 8 tự nghĩ đi nha

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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 = \(-\frac{1}{4}\)

\(x^2+2x+3\)

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

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

Do \(\left(x+1\right)^2\ge0\) với mọi x

\(\Rightarrow x^2+2x+3\ge2\)

Dấu = khi x=-1

Ta có :

\(\frac{x^2+2x+3}{x^2+2}=\frac{2x^2+4-x^2+2x-1}{x^2+2}=\frac{2\left(x^2+2\right)-\left(x-1\right)^2}{x^2+2}=2-\frac{\left(x-1\right)^2}{x^2+2}\le2\)

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