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ĐKXĐ: x khác 2 và -2 

Ta có : \(\frac{x-2}{x+2}\)- \(\frac{x+2}{x-2}\)= \(\frac{-24}{5}\)

<=> \(\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)- \(\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)= \(\frac{-24}{5}\)

<=> \(\frac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)= \(\frac{-24}{5}\)

<=> \(\frac{\left(x-2+x+2\right)\left(x-2-x-2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{-24}{5}\)

<=> \(\frac{2x.\left(-4\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{-24}{5}\)

<=> -40x= -24(x^2-4)

<=> -40x= -24x^2+96

<=> 24x^2-40x-96=0

<=> 24x^2-72x+32x-96=0

<=> 24x(x-3)+32(x-3)=0

<=> (x-3)(24x+32)=0

=> \(\orbr{\begin{cases}x-3=0\\24x+32=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=3\\x=\frac{-4}{3}\end{cases}}\)

Vậy S=\(\hept{\begin{cases}\\\end{cases}}3;\frac{-4}{3}\)

\(\Rightarrow x^2+y^2\ge2\sqrt{x^2y^2}=2xy\)

\(\Rightarrow1\ge2xy\)

\(\Rightarrow\frac{1}{2}\ge xy\)

Có \(x+y\ge2\sqrt{xy}\ge2\sqrt{\frac{1}{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)

Vậy \(Min_{x+y}=\sqrt{2}\)

Làm tương tự với max

Thêm đk: x,y>0

Tìm max:

Áp dụng BĐT bunhiacopxki ta có:

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

\(\Leftrightarrow2\ge\left(x+y\right)^2\)

\(\Leftrightarrow\sqrt{2}\ge x+y\)

Dấu " = " xảy ra <=> x=y

KL:...............................

\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)ĐKXĐ : \(x\ne\pm4\)

\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)

\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)

\(\Leftrightarrow5x^2-2x^2-3x^2-11x+9x=4-4+80-96\)

\(\Leftrightarrow-2x=-16\)

\(\Leftrightarrow x=8\)( t/m )

Vậy....