a, Kẻ OH \(\perp\)AB 

=> OH là đường trung tuyến 

=> \(AH=\frac{AB}{2}=\frac{24}{2}=12\)cm 

Theo định lí Pytago tam giác OHA vuông tại H 

\(OH=\sqrt{AO^2-AH^2}=5\)cm 

Điều kiện : \(\hept{\begin{cases}1-x\ge0\\x+1\ge0\end{cases}\Leftrightarrow x\in\left[-1,1\right]}\)

Đặt : \(a=\sqrt{1-x}+\sqrt{x+1}\Rightarrow a^2=2+2\sqrt{1-x^2}\)

vậy ta có :\(a+a^2-2=4\Leftrightarrow a^2+a-6=0\Leftrightarrow\orbr{\begin{cases}a=2\\a=-3\end{cases}}\)

mà hiển nhiên a nhận giá trị dương nên : \(a=2\Rightarrow a^2=4=2+2\sqrt{1-x^2}\Leftrightarrow\sqrt{1-x^2}=1\Leftrightarrow x=0\)

\(ĐK:-1\le x\le1\)

áp dụng bunhiakopxki ta có : 

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

\(\Leftrightarrow\left(\sqrt{1-x}+\sqrt{x+1}\right)^2\le4\)

\(\Leftrightarrow\sqrt{1-x}+\sqrt{x+1}\le2\)

có \(-x^2\le0\Leftrightarrow1-x^2\le1\Leftrightarrow2\sqrt{1-x^2}\le2\)

\(\Rightarrow VT\le4\) 

dấu = xảy ra khi \(\frac{\sqrt{1-x}}{1}=\frac{\sqrt{x+1}}{1}\) và \(x^2=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

\(ĐK:x\ge1\)

\(\sqrt{x+2\sqrt{x-1}}=2\)

\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)

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

\(\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)

\(\Leftrightarrow\sqrt{x-1}+1=2\)

\(\Leftrightarrow x=2\left(tm\right)\)

Sửa đề: \(\sqrt{x+2\sqrt{x-1}}=2\left(ĐKXĐ:x\ge1\right)\) 

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

\(\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)

Vì \(\sqrt{x-1}+1>0\)

\(\Rightarrow\sqrt{x-1}+1=2\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(TMĐK\right)\) 

Vậy \(S=\left\{2\right\}\)

ta có :

\(P=\frac{\sqrt{x}+4}{1-7\sqrt{x}}+\frac{\sqrt{x}-2}{\sqrt{x}+1}+\frac{24\sqrt{x}}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}\)

\(\frac{-\left(\sqrt{x}+4\right)\left(\sqrt{x}+1\right)+\left(\sqrt{x}-2\right)\left(7\sqrt{x}-1\right)+24\sqrt{x}}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}=\frac{6x+4\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)}\)

\(=\frac{6\sqrt{x}+2}{7\sqrt{x}-1}\)

Để \(P\ge-6\Leftrightarrow\frac{6\sqrt{x}+2}{7\sqrt{x}-1}\ge-6\Leftrightarrow\frac{48\sqrt{x}-4}{7\sqrt{x}-1}\ge0\)

\(\Leftrightarrow\orbr{\begin{cases}0\le\sqrt{x}\le\frac{1}{12}\\\sqrt{x}>\frac{1}{7}\end{cases}}\Leftrightarrow\orbr{\begin{cases}0\le x\le\frac{1}{144}\\x>\frac{1}{49}\end{cases}}\)

Điều kiện: x>0

\(\frac{\sqrt{x}+1}{\sqrt{x}-1}>\frac{1}{2}\)

\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{1}{2}>0\)

\(\Leftrightarrow\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-1}{2\left(\sqrt{x}-1\right)}>0\)

\(\Leftrightarrow\frac{\sqrt{x}+1-\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}>0\)

\(\Leftrightarrow\frac{2}{2\left(\sqrt{x}-1\right)}>0\)

\(\Leftrightarrow\frac{1}{\sqrt{x}-1}>0\)

mà 1>0

nên \(\sqrt{x}-1>0\)

\(\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)