Có \(x^3=3+2\sqrt{2}-3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right)-\left(3-2\sqrt{2}\right)\)

\(\Leftrightarrow x^3=4\sqrt{2}-3x\) \(\Leftrightarrow x^3+3x=4\sqrt{2}\) (1)

Có \(y^3=17+12\sqrt{2}-3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}-\sqrt[3]{17-12\sqrt{2}}\right)-\left(17-12\sqrt{2}\right)\)

\(\Leftrightarrow y^3=24\sqrt{2}-3y\) \(\Leftrightarrow y^3+3y=24\sqrt{2}\) (2)

Từ (1) (2)\(\Rightarrow x^3+3x-y^3-3y=-20\sqrt{2}\)

Có \(M=\left(x-y\right)^3+3\left(x-y\right)\left(xy+1\right)=\left(x-y\right)\left[\left(x-y\right)^2+3\left(xy+1\right)\right]\)

\(=\left(x-y\right)\left(x^2+xy+y^2+3\right)=x^3-y^3+3\left(x-y\right)=-20\sqrt{2}\)

Vậy \(M=-20\sqrt{2}\)

Ta chứng minh công thức sau: \(S_{ABC}=\dfrac{1}{2}.sinA.AB.AC\)

Kẻ \(BF\perp AC\)\(\Rightarrow sinA=\dfrac{BF}{AB}\)

Có \(S_{ABC}=\dfrac{1}{2}.BF.AC=\dfrac{1}{2}.AC.AB.sinA\) (đpcm)

\(\Rightarrow S_{ABC}=\dfrac{1}{2}.sin60^0.AB.AC=\dfrac{\sqrt{3}}{4}.AB.AC\)\(\le\dfrac{\sqrt{3}}{4}.\left(\dfrac{AB+AC}{2}\right)^2=\dfrac{9\sqrt{3}}{4}\) (cm²) (Theo bđt AM-MG với hai số dương)

Dấu = xảy ra khi AB=AC=3 (cm)

Vậy \(S_{max}=\dfrac{9\sqrt{3}}{4}\) cm²

3)\(sin6x.sin2x=sin5x.sinx\)

\(\Leftrightarrow\dfrac{1}{2}\left(cos4x-cos8x\right)=\dfrac{1}{2}\left(cos4x-cos6x\right)\)

\(\Leftrightarrow cos8x=cos6x\)

\(\Leftrightarrow\left[{}\begin{matrix}8x=6x+k2\pi\\8x=-6x+k2\pi\end{matrix}\right.\) (\(k\in Z\)) \(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{7}\end{matrix}\right.\)(\(k\in Z\))

Vậy...

13)\(cosx.cos3x-sin2x.sin6x-sin4x.sin6x=0\)

\(\Leftrightarrow\dfrac{1}{2}.\left(cos2x+cos4x\right)-\dfrac{1}{2}\left(cos4x-cos8x\right)-\dfrac{1}{2}\left(cos2x-cos10x\right)=0\)

\(\Leftrightarrow cos8x+cos10x=0\)

\(\Leftrightarrow2.cos9x.cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos9x=0\\cosx=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{18}+\dfrac{k\pi}{9}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\) (\(k\in Z\))

Vậy...

Giả sử \(BH\perp AD\)

Gọi  \(O=AC\cap BD\)

Có \(S_{ABCD}=\dfrac{1}{2}AC.BD=BH.AD\)

\(\Leftrightarrow\left\{{}\begin{matrix}AC.BD=2S_{ABCD}\\BH=\dfrac{S_{ABCD}}{AD}\end{matrix}\right.\)

Có \(\dfrac{1}{AC^2}+\dfrac{1}{BD^2}=\dfrac{AC^2+BD^2}{AC^2.BD^2}=\dfrac{4\left(OA^2+OD^2\right)}{\left(2S_{ABCD}\right)^2}\)\(=\dfrac{4AD^2}{4S_{ABCD}}=\dfrac{1}{BH^2}\) 

Vậy \(\dfrac{1}{BH^2}=\dfrac{1}{AC^2}+\dfrac{1}{BD^2}\)

Có \(\widehat{ADH}=\widehat{AEH}=90^0\)

=> Tứ giác ADHE nội tiếp đt

=> \(\widehat{AED}=\widehat{AHD}\) mà \(\widehat{AHD}=\widehat{ABC}\) (vì cùng phụ với \(\widehat{HAB}\))

=> \(\widehat{AED}=\widehat{ABC}\) mà \(\widehat{DAE}=\widehat{BDH}=90^0\)

=> \(\Delta ADE\sim\Delta DHB\left(g.g\right)\) \(\Rightarrow\dfrac{C_{ADE}}{C_{DHB}}=\dfrac{AD}{DH}\)

CM tt: \(\dfrac{C_{ADE}}{C_{ECH}}=\dfrac{AE}{EH}=\dfrac{DH}{AD}\)

\(\Rightarrow\dfrac{C_{ADE}}{C_{ECH}}.\dfrac{C_{ADE}}{C_{DHB}}=1\Rightarrow\)\(\dfrac{\left(C_{ADE}\right)^2}{ab}=1\)\(\Leftrightarrow C_{ADE}=\sqrt{ab}\)

\(A=cos\left(\dfrac{\pi}{3}+\alpha\right)+cos\left(\dfrac{\pi}{3}-\alpha\right)\)

\(=cos\dfrac{\pi}{3}.cos\alpha-sin\dfrac{\pi}{3}.sin\alpha+cos\dfrac{\pi}{3}.cos\alpha+sin\dfrac{\pi}{3}.sin\alpha\)

\(=2.cos\dfrac{\pi}{3}.cos\alpha=cos\alpha\)

\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) (đk: \(x\ge2\))

\(=\sqrt{\left(x-2\right)+2.\sqrt{2}\sqrt{x-2}+2}+\sqrt{\left(x-2\right)-2\sqrt{2}\sqrt{x-2}+2}\)

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

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

\(=\left[{}\begin{matrix}\sqrt{x-2}+\sqrt{2}+\sqrt{x-2}-\sqrt{2};\sqrt{x-2}\ge2\\\sqrt{x-2}+\sqrt{2}-\left(\sqrt{x-2}-\sqrt{2}\right);\sqrt{x-2}< 2\end{matrix}\right.\)

\(=\left[{}\begin{matrix}2\sqrt{x-2};x\ge6\\2\sqrt{2};2\le x< 6\end{matrix}\right.\)

Vậy...

\(P=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^2}{\left(b-c\right)\left(b-a\right)}+\dfrac{c^2}{\left(c-b\right)\left(c-a\right)}\)

\(=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{-b^2}{\left(b-c\right)\left(a-b\right)}+\dfrac{c^2}{\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{a^2b-a^2c-ab^2+b^2c+c^2a-bc^2}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)\(=\dfrac{ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{\left(a-b\right)\left(ab-c\left(a+b\right)+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=1\)