\(\left(cosa\ne1\right)A=\dfrac{sin^4a-cos^4a+cos^2a}{2\left(1-cosa\right)}=\dfrac{\left(sin^2a+cos^2a\right)\left(sin^2a-cos^2a\right)+cos^2a}{2\left(1-cosa\right)}=\dfrac{sin^2a}{2\left(1-cosa\right)}=\dfrac{1-cos^2a}{2\left(1-cosa\right)}=\dfrac{1+cosa}{2}\)

\(\left(cosa+cos3a+cos5a+cos7a\ne0\right)\Rightarrow B=\dfrac{sin7a+sina+sin5a+sin3a}{cos7a+cosa+cos5a+cos3a}=\dfrac{2sin\left(\dfrac{7a+a}{2}\right)cos\left(\dfrac{7a-a}{2}\right)+2sin\left(\dfrac{5a+3a}{2}\right)cos\left(\dfrac{5a-3a}{2}\right)}{2cos\left(\dfrac{7a+a}{2}\right)cos\left(\dfrac{7a-a}{2}\right)+2cos\left(\dfrac{5a+3a}{2}\right)cos\left(\dfrac{5a-3a}{2}\right)}=\dfrac{2sin4a.cos3a+2sin4a.cosa}{2cos4a.cos3a+2cos4a.cosa}=\dfrac{2sin4a\left(cos3a+cosa\right)}{2cos4a\left(cos3a+cosa\right)}=\dfrac{sin4a}{cos4a}=tan4a\)

\(C=\dfrac{1-2cos2x+cos4x}{1+2cos2x+cos4x}=\dfrac{1-2cos2x+2cos^22x-1}{1+2cos2x+2cos^22x-1}=\dfrac{cos^22x-cos2x}{cos^22x+cos2x}=\dfrac{cos2x-1}{cos2x+1}\)

\(D=\left(\dfrac{1+cos\beta}{1-cos\beta}\right)tan^2\dfrac{\beta}{2}+sin^2\beta-2\)

\(đặt:t=tan\dfrac{\beta}{2}\Rightarrow D=\left(\dfrac{1+\dfrac{1-t^2}{t^2+1}}{1-\dfrac{1-t^2}{t^2+1}}\right)t^2+\left(\dfrac{2t}{t^2+1}\right)^2-2=\left(\dfrac{\dfrac{2}{t^2+1}}{\dfrac{2t^2}{t^2+1}}\right)t^2+\left(\dfrac{2t}{t^2+1}\right)^2-2=\left(\dfrac{2t}{t^2+1}\right)^2-1=\left(\dfrac{2t}{t^2+1}+1\right)\left(\dfrac{2t}{t^2+1}-1\right)=\left(\dfrac{t^2+2t+1}{t^2+1}\right)\left(\dfrac{-\left(t^2-2t+1\right)}{t^2+1}\right)=\dfrac{-\left(t+1\right)^2\left(t-1\right)^2}{\left(t^2+1\right)^2}=-\left(\dfrac{t^2-1}{t^2+1}\right)^2=-\left(\dfrac{tan^2\dfrac{\beta}{2}-1}{tan^2\dfrac{\beta}{2}+1}\right)^2=-\dfrac{\left(tan^2\dfrac{\beta}{2}-1\right)^2}{\left(tan^2\dfrac{\beta}{2}+1\right)^2}\)

\(b;\Leftrightarrow cosx+sinx=sin^3x+sin^3x.cotx+sin^3x.cot^2x+sin^3x.cot^3x=sin^3x+sin^2x.cosx+sinx.cos^2x+cos^3x=\left(cosx+sinx\right)\left(sin^2x-sinx.cosx+cos^2x\right)+sinx.cosx\left(sin^2x+cos^2x\right)=\left(cosx+sinx\right)\left(1-sinx.cosx\right)+sinx.cosx=cosx+sinx\Rightarrowđúng\)

\(A=\dfrac{cosx+cos3x+cos5x}{sinx+sin3x+sin5x}=\dfrac{cos5x+cosx+cos3x}{sin5x+sinx+sin3x}=\dfrac{2cos\dfrac{5x+x}{2}cos\dfrac{5x-x}{2}+cos3x}{2sin\dfrac{5x+x}{2}cos\dfrac{5x-x}{2}+sin3x}=\dfrac{2cos3x.cos2x+cos3x}{2sin3x.cos2x+sin3x}=\dfrac{cos3x\left(2cos2x+1\right)}{sin3x\left(2cos2x+1\right)}=\dfrac{cos3x}{sin3x}=cot3x\)

\(B=sin\left(\dfrac{\pi}{2}-x\right)+cos\left(\pi-x\right)+tan\left(3\pi-x\right)+cot\left(\dfrac{\pi}{2}-x\right)=cosx-cosx-tanx+tanx=0\)

\(a;\sin\mu+cos\mu=\sqrt{2}.\left[\dfrac{1}{\sqrt{2}}sin\mu+\dfrac{1}{\sqrt{2}}cos\mu\right]=\sqrt{2}\left[sin\left(\dfrac{\pi}{4}\right).sin\mu+cos\left(\dfrac{\pi}{4}\right).cos\mu\right]=\sqrt{2}.sin\left(\mu+\dfrac{\pi}{4}\right)=\sqrt{2}.cos\left(\mu-\dfrac{\pi}{4}\right)\left(đpcm\right)\)

\(b;4sin\left(\alpha+\dfrac{\pi}{3}\right)sin\left(\alpha-\dfrac{\pi}{3}\right)=4\left[sin\alpha.cos\dfrac{\pi}{3}+cos\alpha.sin\dfrac{\pi}{3}\right].\left(sin\alpha.cos\left(\dfrac{\pi}{3}\right)-cos\alpha.sin\dfrac{\pi}{3}\right)=4\left[\dfrac{sin\alpha}{2}+\dfrac{\sqrt{3}cos\alpha}{2}\right]\left(\dfrac{sin\alpha}{2}-\dfrac{\sqrt{3}cos\alpha}{2}\right)=\left(sin\alpha+\sqrt{3}cos\alpha\right)\left(sin\alpha-\sqrt{3}cos\alpha\right)=sin^2\alpha-3cos^2\alpha=sin^2\alpha-3\left(1-sin^2\alpha\right)=4sin^2\alpha-3\)

\(c;cos\left(a+b\right).cos\left(a-b\right)=\left(cosa.cosb-sina.sinb\right)\left(cosa.cosb+sina.sinb\right)=cos^2a.cos^2b-sin^2a.sin^2b=cos^2a\left(1-sin^2b\right)-sin^2a\left(1-cos^2b\right)=cos^2a-cos^2a.sin^2b-sin^2b+sin^2b.cos^2b=cos^2a-sin^2a\)

\(A=\dfrac{1}{a^2+b^2}+\dfrac{1}{ab}+4ab=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}+8ab-4ab\ge\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{1}{2}.8}-\dfrac{4.\left(a+b\right)^2}{4}=\dfrac{4}{\left(a+b\right)^2}+4-\left(a+b\right)^2\ge4+4-1=7\Rightarrow minA=7\Leftrightarrow a=b=\dfrac{1}{2}\)

\(\Rightarrow2a+4b+6c+\dfrac{4}{a}+\dfrac{12}{b}+\dfrac{20}{c}=a+b+c+a+\dfrac{4}{a}+3b+\dfrac{12}{b}+5c+\dfrac{20}{c}\ge6+2\sqrt{4}+2\sqrt{3.12}+2\sqrt{5.20}=42\left(đpcm\right)\)

\(dấu"="\Leftrightarrow a=b=c=2\)

\(\Sigma\sqrt{\dfrac{a^3}{a^3+\left(b+c\right)^3}}=\Sigma\sqrt{\dfrac{1}{1+\left(\dfrac{b+c}{a}\right)^3}}\)\(\left(1\right)\)

\(đặt:\left(\left(\dfrac{b+c}{a}\right)^{ };\left(\dfrac{c+a}{b}\right)^{ };\left(\dfrac{a+b}{c}\right)^{ }\right)=\left(x;y;z\right)\)

\(\left(1\right)\Leftrightarrow\sqrt{\dfrac{1}{1+x^3}}+\sqrt{\dfrac{1}{1+y^3}}+\sqrt{\dfrac{1}{1+z^3}}=\sqrt{\dfrac{1}{\left(x+1\right)\left(x^2-x+1\right)}}+\sqrt{\dfrac{1}{\left(y+1\right)\left(y^2-y+1\right)}}+\sqrt{\left(z+1\right)\left(z^2-z+1\right)}\)

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

\(tương\) \(tự\Rightarrow\left(1\right)\ge\dfrac{2}{x^2+2}+\dfrac{2}{y^2+2}+\dfrac{2}{z^2+2}\)

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

\(bunhia\Rightarrow\left(b+c\right)^2\le2\left(b^2+c^2\right)\Rightarrow\dfrac{2a^2}{\left(b+c\right)^2+2a^2}\ge\dfrac{2a^2}{2\left(a^2+b^2\right)+2a^2}=\dfrac{a^2}{a^2+b^2+c^2}\)

\(tương\) \(tự\Rightarrow\left(1\right)\ge\dfrac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\left(đpcm\right)\)

\(pt\) \(hoành\) \(độ\) \(giao\) \(điểm:\dfrac{1}{2}x^2=\left(m+1\right)x-m\Leftrightarrow x^2-2\left(m+1\right)x+2m=0\)

\(\Delta'=\left(m+1\right)^2-2m=m^2+2m+1-2m=m^2+1>0\left(\forall m\right)\Rightarrowđpcm\)

\(b;\Rightarrow\left\{{}\begin{matrix}x1+x2=2\left(m+1\right)\\x1.x2=2m\end{matrix}\right.\)

\(điều\) \(kiện\) \(tồn\) \(tạicăn\Leftrightarrow0\le x1< x2\Leftrightarrow\left\{{}\begin{matrix}x1.x2\ge0\\x1+x2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\ge0\\m>-1\end{matrix}\right.\)

\(\Leftrightarrow m\ge0\)

\(\Rightarrow\sqrt{x1}+\sqrt{x2}=2\Leftrightarrow x1+x2-2\sqrt{x1x2}=4\)

\(\Leftrightarrow2\left(m+1\right)+2\sqrt{2m}=4\)

\(đặt:\sqrt{2m}=t\ge0\Rightarrow t^2+2t-2=0\Leftrightarrow\left[{}\begin{matrix}t=-1-\sqrt{3}\left(ktm\right)\\t=-1+\sqrt{3}\left(tm\right)\Leftrightarrow m=2-\sqrt{3}\left(tm\right)\end{matrix}\right.\)

\(\Leftrightarrow m=2-\sqrt{3}\left(tm\right)\)

\(17.\left\{{}\begin{matrix}\left(m-1\right)x-y=m+2\\2x-y=3\end{matrix}\right.\) \(\Leftrightarrow\dfrac{m-1}{2}\ne1\Leftrightarrow m\ne3\)

\(18..\sqrt{2}x^2-2\sqrt{6}x-\sqrt{10}=0\Rightarrow\left\{{}\begin{matrix}S=x1+x2=\dfrac{-b}{a}=\dfrac{2\sqrt{6}}{\sqrt{2}}=2\sqrt{3}\\P=x1x2=\dfrac{c}{a}=\dfrac{-\sqrt{10}}{\sqrt{2}}=-\sqrt{5}\end{matrix}\right.\)

\(đk\) \(sai:\) \(phải:a;b>0\)

\(a+b\ge2\sqrt{ab}\Rightarrow2\sqrt{ab}\le1\Leftrightarrow ab\le\dfrac{1}{4}\)

\(A=ab+\dfrac{1}{ab}=ab+\dfrac{1}{16ab}+\dfrac{15}{16ab}\ge2\sqrt{\dfrac{1}{16}}+\dfrac{15}{16.\dfrac{1}{4}}=\dfrac{17}{4}\Rightarrow minA=\dfrac{17}{4}\Leftrightarrow a=b=\dfrac{1}{2}\)

\(có:\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le a+b+c\)

\(b+1\ge2\sqrt{b}\Rightarrow-\dfrac{b\sqrt{a}}{1+b}\ge-\dfrac{b\sqrt{a}}{2\sqrt{b}}=-\dfrac{\sqrt{ab}}{2}\)

\(tương\) \(tự\Rightarrow-\dfrac{c\sqrt{b}}{1+c}\ge-\dfrac{\sqrt{bc}}{2};-\dfrac{a\sqrt{c}}{1+a}\ge-\dfrac{\sqrt{ac}}{2}\)

\(\Rightarrow P\ge\dfrac{2021}{a+b+c}-\left(\dfrac{\sqrt{ac}+\sqrt{bc}+\sqrt{ac}}{2}\right)\ge\dfrac{2021}{3}-\dfrac{a+b+c}{2}=\dfrac{2021}{3}-\dfrac{3}{2}=\dfrac{4033}{6}\)

\(\Rightarrow minP=\dfrac{4033}{6}\Leftrightarrow a=b=c=1\)