\(1\) \(cách\) \(khác\) \(\sqrt{2019a+bc}=\sqrt{\left(a+b+c\right)a+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\le\dfrac{a+b+a+c}{2}=\dfrac{2a+b+c}{2}\)

\(\dfrac{a}{a+\sqrt{2019a+bc}}=\dfrac{a\left(a-\sqrt{2019a+bc}\right)}{a^2-\left(a+b+c\right)a-bc}=\dfrac{a\left(a-\sqrt{2019a+bc}\right)}{-\left(ab+ca+bc\right)}=\dfrac{a\sqrt{\left(a+b\right)\left(b+c\right)}-a^2}{ab+bc+ca}\le\dfrac{a\left(\dfrac{2a+b+c}{2}\right)-a^2}{ab+bc+ca}=\dfrac{\dfrac{2a^2+ab+ac}{2}-a^2}{ab+bc+ca}=\dfrac{\dfrac{ab+ac}{2}}{ab+bc+ca}\)

\(tương\) \(tự:\dfrac{b}{b+\sqrt{2019b+ca}}\le\dfrac{\dfrac{ba+bc}{2}}{ab+bc+ca};\dfrac{c}{c+\sqrt{2019c+ab}}\le\dfrac{\dfrac{ca+cb}{2}}{ab+bc+ca}\Rightarrow P\le\dfrac{1}{2}\dfrac{2\left(ab+bc+ca\right)}{ab+bc+ca}=1\)

\(\sqrt{2019a+bc}=\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{ab}+\sqrt{ac}\left(bunhia\right)\)

\(\Rightarrow\dfrac{a}{a+\sqrt{2019a+bc}}\le\dfrac{a}{a+\sqrt{ab}+\sqrt{ac}}=\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

\(tương\) \(tự\Rightarrow P\le\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\) \(dấu"="\Leftrightarrow a=b=c=673\)

a; \(OA=OB=R\left(1\right);OC\) \(chung\left(2\right)\)

\(OC\perp AB;\Delta OAB\) \(cân\) \(tạiO\Rightarrow OC\) \(là\) \(phân\) \(giácAOB\Rightarrow góc\) \(AOC=góc\) \(BOC\left(3\right)\)

\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow\Delta AOC=\Delta BOC\left(cgc\right)\Rightarrow góc\) \(OAC=góc\) \(OBC=90^o\left(4\right)\)

\(\left(1\right)\left(4\right)\Rightarrowđpcm\)

\(b;gọi\) \(giao\) \(điểm\) \(OC\) \(và\) \(AB\)  \(tại\) \(I\Rightarrow I\) \(là\) \(trung\) \(điểmAB\Rightarrow IB=\dfrac{AB}{2}=12cm\Rightarrow OI=\sqrt{OB^2-IB^2}=9cm\Rightarrow OB^2=OI.OC\Rightarrow OC=\dfrac{OB^2}{OI}=25cm\)

\(a;\left(đk:a,b>0\right)\sqrt{\dfrac{a^2}{b}}+\sqrt{\dfrac{b^2}{a}}=\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\ge\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}=\sqrt{a}+\sqrt{b}\left(đpcm\right)\)

\(b;\) \(bđt\) \(bunhia\Rightarrow a+b\le\sqrt{2\left(a^2+b^2\right)}=\sqrt{2c^2}=c\sqrt{2}\)

\(gọi:BM=h;AC=AB=x\Rightarrow Sabc=\dfrac{1}{2}hAC=\dfrac{1}{2}hx\)

\(sina=\dfrac{h}{BC}\)

\(\)\(AB^2=AC^2+BC^2-2AC.BC.cosa\Leftrightarrow2x.BC.cosa=BC^2\Leftrightarrow2x.cosa=BC\Leftrightarrow cosa=\dfrac{BC}{2x}\)

\(\Rightarrow S=\dfrac{h^2}{4sina.cosa}=\dfrac{h^2}{4\dfrac{h}{BC}.\dfrac{BC}{2x}}=\dfrac{1}{2}hx=\dfrac{1}{2}h.AC\left(đpccm\right)\)

\(gọi:AB=x;AC=y\)

\(\Rightarrow x^2+y^2-2xy.cos135^o=BC^2=25\left(1\right)\)

\(\Rightarrow\dfrac{BC}{sin135^o}=\dfrac{5}{sin135^o}=5\sqrt{2}=\dfrac{x}{sinC}=\dfrac{y}{sinB}\)

\(\Leftrightarrow\dfrac{x}{\dfrac{AH}{y}}=\dfrac{y}{\dfrac{AH}{x}}=5\sqrt{2}\Leftrightarrow2xy=5\sqrt{2}\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\left\{{}\begin{matrix}x^2+y^2-2xy.cos135^o=25\\2xy=5\sqrt{2}\end{matrix}\right.\)

hệ đối xứng loại 1 dễ xử lí bằng cách đặt \(\left\{{}\begin{matrix}x+y=S\\xy=P\end{matrix}\right.\)\(\left(S^2\ge4P\right)\)

\(đặt:AB=c;BC=a;AC=b\)

\(2cosB-sinC=\dfrac{1}{2}\Leftrightarrow\dfrac{2c}{a}-\dfrac{c}{a}=\dfrac{1}{2}\Leftrightarrow\dfrac{c}{a}=\dfrac{1}{2}\Leftrightarrow2c=a\left(1\right)\)

\(gọi\) \(I\) \(trung\) \(điểmBC\Rightarrow AI=BM=\dfrac{BC}{2}=\dfrac{a}{2}\Leftrightarrow2AI=a\left(2\right)\)

\(\left(1\right)\left(2\right)\Leftrightarrow AI=c=BI\Rightarrow ABI\) đều \(\Rightarrow B=60^o;C=30^o\)

\(a;AB^2=BH.BC\Rightarrow BC=\dfrac{AB^2}{BH}=25cm\)

\(\Rightarrow AC=\sqrt{BC^2-AB^2}=20cm\)

\(\Rightarrow AH.BC=AB.AC\Rightarrow AH=\dfrac{AB.AC}{BC}=12cm\)

\(b;\Rightarrow AM=BM=MC=\dfrac{1}{2}BC=12,5cm;AC^2=CH.BC\Rightarrow CH=16cm\Rightarrow HM=CH-CM=3,5cm\Rightarrow Sahm=\dfrac{1}{2}AH.HM=21cm^2\)

\(a;\sqrt{x-3}+\sqrt{15-x}\le\sqrt{2\left(x-3+15-x\right)}=2\sqrt{6}\)\(\left(đk:3\le x\le15\right)\)

\(dấu"="\) \(xảy\) \(ra\Leftrightarrow\sqrt{x-3}=\sqrt{15-x}\Leftrightarrow x-3=15-x\Leftrightarrow x=9\left(tm\right)\)

\(b;\sqrt{\dfrac{x}{2x-1}}+\sqrt{\dfrac{2x-1}{x}}\ge2\sqrt{\sqrt{\dfrac{x}{2x-1}}.\sqrt{\dfrac{2x-1}{x}}}=2\)

\(dấu"="\Leftrightarrow\sqrt{\dfrac{x}{2x-1}}=\sqrt{\dfrac{2x-1}{x}}\Leftrightarrow x=1\) thhuer vào thấy thỏa mãn

\(a^2=8+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}=8+2\sqrt{16-10-2\sqrt{5}}=8+2\sqrt{5-2\sqrt{5}+1}=8+2\sqrt{\left(\sqrt{5}+1\right)^2}=8+2\left(\sqrt{5}+1\right)=5+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\Rightarrow a=\sqrt{5}+1\Rightarrow A=a^2-2a+12=\left(a-1\right)^2+11=5+11=16\)