\(\frac{2x}{x-2}-\frac{3x+10}{x^2-4}=\frac{x}{x+2}\left(x\ne\pm2\right)\)

\(\Leftrightarrow\frac{2x}{x-2}-\frac{3x+10}{\left(x-2\right)\left(x+2\right)}-\frac{x}{x+2}=0\)

\(\Leftrightarrow\frac{2x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{3x+10}{\left(x-2\right)\left(x+2\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{2x^2+4x}{\left(x-2\right)\left(x+2\right)}-\frac{3x+10}{\left(x-2\right)\left(x+2\right)}-\frac{x^2-2x}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{2x^2+4x-3x-10-x^2+2x}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{x^2+3x-10}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{x^2+5x-2x-10}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{\left(x+5\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

=> x+5=0

<=> x=-5(tmđk)

Vậy x=-5 là nghiệm của phương trình

\(\frac{2x}{x-2}-\frac{3x+10}{x^2-4}=\frac{x}{x+2}\) ( đkxđ : \(x\ne\pm2\))

\(\Leftrightarrow\frac{2x}{x-2}-\frac{3x+10}{\left(x+2\right)\left(x-2\right)}=\frac{x}{x+2}\)

\(\Leftrightarrow\frac{2x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3x+10}{\left(x+2\right)\left(x-2\right)}=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)

\(\Leftrightarrow2x^2+4x-3x-10=x^2-2x\)

\(\Leftrightarrow2x^2+4x-3x-10-x^2+2x=0\)

\(\Leftrightarrow x^2+3x-10=0\)

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

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

\(x\ne\pm2\)=> x = -5

ĐK: x > = 3

pt <=> \(x^2-5x+4+\left(\sqrt{2x+1}-3\right)+\left(\sqrt{x-3}-1\right)=0\)

<=> \(\left(x-1\right)\left(x-4\right)+\frac{2\left(x-4\right)}{\sqrt{2x+1}+3}+\frac{x-4}{\sqrt{x-3}+1}=0\)

<=> \(\left(x-4\right)\left(\left(x-1\right)+\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{x-3}+1}\right)=0\)

<=> x - 4 = 0  vì \(\left(x-1\right)+\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{x-3}+1}>0;\forall x\ge3\)

<=> x = 4  tm 

Vậy:...

 \(t=\sqrt{2x-3}=>\frac{t^2+3}{2}=x\)

\(=>P=\frac{t^2+3}{2}-2t=\frac{t^2-4t+3}{2}=\frac{\left(t-2\right)^2-1}{2}=\frac{\left(t-2\right)^2}{2}-\frac{1}{2}\)

ta có \(\frac{\left(t-2\right)^2}{2}\ge0\left(\forall t\right)\)

\(=>\frac{\left(t-2\right)^2}{2}-\frac{1}{2}\ge-\frac{1}{2}\left(\forall t\right)\)

minP=-1/2

dấu = xảy ra khi x=7/2

a) \(t=\sqrt{2x-3}\ge0\)

<=> \(t^2=2x-3\)

<=> \(x=\frac{t^2+3}{2}\)

=> \(P=\frac{t^2+3}{2}-2t\)

b) khi đó: \(P=\frac{t^2+3}{2}-2t=\frac{t^2-4t+3}{2}=\frac{\left(t-2\right)^2-1}{2}\ge-\frac{1}{2}\)

Dấu "=" xảy ra <=> t = 2  khi đó: x = 7/2

\(P=\sqrt{3a^2+2ab+3b^2}+...+\sqrt{3c^2+2ac+3a^2}\)

ta có:

\(\sqrt{3a^2+2ab+3b^2}\ge\sqrt{2}\left(a+b\right)\Leftrightarrow3a^2+2ab+3b^2\ge2a^2+4ab+2b^2\Leftrightarrow\left(a-b\right)^2\ge0\left(đ\right)\)

\(\text{tương tự suy ra:}P\ge2\sqrt{2}\left(a+b+c\right)\)