a, Vì M là trung điểm AB 

=> \(OM\perp AB\)

b, Ta có : R = 4 = OA = OB = 4 cm 

Theo định lí Pytago tam giác AOB vuông tại O

\(AB=\sqrt{OA^2+OB^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}\)cm 

Theo định lí Pytago tam giác AOM vuông tại M

\(OM=\sqrt{OA^2-AM^2}\)lại có : \(AM=\frac{1}{2}AB=\frac{1}{2}.4\sqrt{2}=2\sqrt{2}\)cm 

\(=\sqrt{16-8}=\sqrt{8}=2\sqrt{2}\)cm 

ta có :

\(A=\frac{sin^3x+cos^3x}{sin^3x-cos^3x}=\frac{\frac{sin^3x}{cos^3x}+1}{\frac{sin^3x}{cos^3x}-1}=\frac{tan^3x+1}{tan^3x-1}=\frac{\left(\frac{\sqrt{2}}{5}\right)^3+1}{\left(\frac{\sqrt{2}}{5}\right)^3-1}\)

áp dụng cô si  ta có :

\(a^3+abc\ge2\sqrt{a^3\cdot abc}=2a^2\sqrt{bc}\)

\(b^3+abc\ge2\sqrt{b^3\cdot abc}=2b^2\sqrt{ac}\)

\(c^3+abc\ge2\sqrt{c^3\cdot abc}=2c^2\sqrt{ab}\)

\(\Rightarrow a^3+b^3+c^3+3abc\ge2\left(a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}\right)\)

mà có \(a^3+b^3+c^3\ge3\sqrt[3]{a^3\cdot b^3\cdot c^3}=3abc\)

\(\Rightarrow2\left(a^3+b^3+c^3\right)\ge2\left(a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}\right)\)

\(\Rightarrow a^3+b^3+c^3\ge a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}\)

dấu = xảy ra khi a = b = c

a) Đặt \(sinx+cosx=t\left(\left|t\right|\le\sqrt{2}\right)\Rightarrow sinx.cosx=\frac{t^2-1}{2}\)

=> pt có dạng: \(t=\sqrt{2}\left(t^2-1\right)\Leftrightarrow\sqrt{2}t^2-t-\sqrt{2}=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=\frac{-\sqrt{2}}{2}\\t=\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}sinx+cosx=\frac{-\sqrt{2}}{2}\\sinx+cosx=\sqrt{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}sin\left(x+\frac{\pi}{4}\right)=\frac{-1}{2}\\sin\left(x+\frac{\pi}{4}\right)=1\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x+\frac{\pi}{4}=\frac{-\pi}{6}+2k\pi\\x+\frac{\pi}{4}=\frac{7\pi}{6}+2k\pi\\x+\frac{\pi}{4}=\frac{\pi}{2}+2k\pi\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{-5\pi}{12}+2k\pi\\x=\frac{11\pi}{12}+2k\pi\\x=\frac{\pi}{4}+2k\pi\end{cases}}\left(k\inℤ\right)}\)

\(A=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-x}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)

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

\(A=\left(\frac{x+2\sqrt{x}+1+x-\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(A=\frac{2x+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{4\sqrt{x}}\)

\(A=\frac{2x+1}{4\sqrt{x}}\)

c, \(A=\frac{2x+1}{4\sqrt{x}}=\frac{\sqrt{x}}{2}+\frac{1}{4\sqrt{x}}\)

ap dụng cô si ta có \(\frac{\sqrt{x}}{2}+\frac{1}{4\sqrt{x}}\ge2\sqrt{\frac{\sqrt{x}}{2}\cdot\frac{1}{4\sqrt{x}}}=\frac{\sqrt{2}}{2}\)

dấu = xảy ra khi \(\frac{\sqrt{x}}{2}=\frac{1}{4\sqrt{x}}\Leftrightarrow x=\frac{1}{2}\) (tm)