a: Đặt \(B=\sqrt{a+\sqrt{b}}\pm\sqrt{a-\sqrt{b}}\)

\(B^2=a+\sqrt{b}+a-\sqrt{b}\pm2\sqrt{\left(a+\sqrt{b}\right)\left(a-\sqrt{b}\right)}\)

\(=2a\pm2\sqrt{a^2-b}=2\left(a\pm\sqrt{a^2-b}\right)\)

=>\(B=\sqrt{2\left(a\pm\sqrt{a^2-b}\right)}\)

b: Đặt \(A=\sqrt{\dfrac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}}\)

=>\(A^2=\dfrac{a+\sqrt{a^2-b}}{2}+\dfrac{a-\sqrt{a^2-b}}{2}\pm2\sqrt{\dfrac{a^2-\left(\sqrt{a^2-b}\right)^2}{4}}\)

\(=\dfrac{2a}{2}\pm2\cdot\dfrac{\sqrt{a^2-a^2+b}}{2}\)

\(=a\pm\sqrt{b}\)

=>\(A=\sqrt{a\pm\sqrt{b}}\)

Min P em có thể tự tìm đơn giản bằng AM-GM

Min R cũng khá đơn giản:

Đặt \(\left(\sqrt[3]{a};\sqrt[3]{b};\sqrt[3]{c}\right)=\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}0\le x;y;z\le1\\x^3+y^3+z^3=\dfrac{9}{8}\end{matrix}\right.\)

\(R=\dfrac{1}{1+x}+\dfrac{1}{1+y}+\dfrac{1}{1+z}\ge\dfrac{9}{3+x+y+z}\ge\dfrac{9}{3+\sqrt[3]{9\left(x^3+y^3+z^3\right)}}=\dfrac{6}{2+\sqrt[3]{3}}\)

Xét \(Q=x+y+z\)

Do \(\left(x+y+z\right)^3\ge x^3+y^3+z^3=\dfrac{9}{8}\Rightarrow x+y+z\ge\sqrt[3]{\dfrac{9}{8}}>1\Rightarrow Q-1>0\)

\(x^3+y^3+z^3=\left(x+y+z\right)^3-3\left(x+y+z\right)\left(xy+yz+zx\right)+3xyz\)

\(\Rightarrow\dfrac{9}{8}=Q^3-3Q\left(xy+yz+zx\right)+3xyz\)

\(\Rightarrow\dfrac{9}{8}=Q^3-3\left(Q-1\right)\left(xy+yz+zx\right)-3\left(xy+yz+zx-xyz\right)\)

Do \(0\le x;y;z\le1\Rightarrow\left(1-x\right)\left(1-y\right)\left(1-z\right)\ge0\)

\(\Rightarrow xy+yz+zx-xyz\ge Q-1\)  (1)

\(\Rightarrow xy+yz+zx\ge xyz+Q-1\ge Q-1\) (2)

(1);(2)\(\Rightarrow\dfrac{9}{8}\le Q^3-3\left(Q-1\right)\left(Q-1\right)-3\left(Q-1\right)\)

\(\Rightarrow8Q^3-24Q^2+24Q-9\ge0\)

\(\Rightarrow\left(2Q-3\right)\left(4Q^2-6Q+3\right)\ge0\)

Do \(4Q^2-6Q+3=4\left(Q-\dfrac{3}{4}\right)^2+\dfrac{3}{4}>0;\forall Q\)

\(\Rightarrow2Q-3\ge0\Rightarrow Q\ge\dfrac{3}{2}\)

\(Q_{min}=\dfrac{3}{2}\) khi \(\left(x;y;z\right)=\left(0;1;\dfrac{1}{2}\right)\) và hoán vị hay \(\left(a;b;c\right)=\left(0;1;\dfrac{1}{8}\right)\) và hoán vị

Xét ΔABC vuông tại A có \(sinB=\dfrac{AC}{BC}\)

=>\(\dfrac{6}{BC}=sin30=\dfrac{1}{2}\)

=>\(BC=6\cdot2=12\left(cm\right)\)

\(\left|A+B\right|< =\left|A\right|+\left|B\right|\)

=>\(\left(\left|A+B\right|\right)^2< =\left(\left|A\right|+\left|B\right|\right)^2\)

=>\(A^2+B^2+2AB< =A^2+B^2+2\left|AB\right|\)

=>2AB<=2|AB|

=>AB<=|AB|(luôn đúng)

Dấu '=' xảy ra khi AB>=0

xét ΔABC vuông tại A có \(cosB=\dfrac{AB}{BC}\)

=>\(\dfrac{6}{BC}=cos30=\dfrac{\sqrt{3}}{2}\)

=>\(BC=6\cdot\dfrac{2}{\sqrt{3}}=4\sqrt{3}\left(cm\right)\)

BD+CD=BC

=>BC=15+20=35(cm)

Xét ΔABC có AD là phân giác

nên \(\dfrac{AB}{BD}=\dfrac{AC}{CD}\)

=>\(\dfrac{AB}{15}=\dfrac{AC}{20}\)

=>\(\dfrac{AB}{3}=\dfrac{AC}{4}=k\)

=>AB=3k; AC=4k

Xét ΔABC vuông tại A có \(AB^2+AC^2=BC^2\)

=>\(\left(3k\right)^2+\left(4k\right)^2=35^2\)

=>\(25k^2=1225\)

=>\(k^2=49\)

=>k=7

=>\(AB=3\cdot7=21\left(cm\right);AC=4\cdot7=28\left(cm\right)\)

Xét ΔABC vuông tại A có AH là đường cao

nên \(AH\cdot BC=AB\cdot AC\)

=>\(AH=\dfrac{21\cdot28}{35}=21\cdot\dfrac{4}{5}=16,8\left(cm\right)\)

BD+CD=BC

=>BC=15+20=35(cm)

Xét ΔABC có AD là phân giác

nên \(\dfrac{AB}{BD}=\dfrac{AC}{CD}\)

=>\(\dfrac{AB}{15}=\dfrac{AC}{20}\)

=>\(\dfrac{AB}{3}=\dfrac{AC}{4}=k\)

=>AB=3k; AC=4k

Xét ΔABC vuông tại A có \(AB^2+AC^2=BC^2\)

=>\(\left(3k\right)^2+\left(4k\right)^2=35^2\)

=>\(25k^2=1225\)

=>\(k^2=49\)

=>k=7

=>\(AB=3\cdot7=21\left(cm\right);AC=4\cdot7=28\left(cm\right)\)

Xét ΔABC có AD là phân giác

nên \(AD=\dfrac{2\cdot AB\cdot AC}{AB+AC}\cdot cos\left(\dfrac{BAC}{2}\right)\)

\(=\dfrac{2\cdot21\cdot28}{21+28}\cdot cos45=12\sqrt{2}\left(cm\right)\)

Ta có \(BC=BD+CD=35\left(cm\right)\)

Áp dụng định lý phân giác:

\(\dfrac{AB}{AC}=\dfrac{BD}{CD}=\dfrac{15}{20}=\dfrac{3}{4}\)

\(\Rightarrow AB=\dfrac{3}{4}AC\)

Áp dụng định lý Pitago:

\(AB^2+AC^2=BC^2\Rightarrow\dfrac{9}{16}AC^2+AC^2=35^2\)

\(\Rightarrow AC^2=784\Rightarrow AC=28\left(cm\right)\)

\(\Rightarrow AB=\dfrac{3}{4}AC=21\left(cm\right)\)

Áp dụng hệ thức lượng trong tam giác vuông ABC:

\(AB^2=BH.BC\Rightarrow BH=\dfrac{AB^2}{BC}=\dfrac{63}{5}\left(cm\right)\)

\(\Rightarrow AH=\sqrt{AB^2-BH^2}=\dfrac{84}{5}\left(cm\right)\)

\(HD=BD-BH=\dfrac{12}{5}\left(cm\right)\)

Áp dụng Pitago trong tam giác vuông ADH:

\(AD=\sqrt{AH^2+HD^2}=12\sqrt{2}\left(cm\right)\)

a.

Ta có: \(\widehat{BAE}+\widehat{BAC}+\widehat{CAF}=180^0\)

\(\Rightarrow\widehat{BAE}+90^0+\widehat{CAF}=180^0\)

\(\Rightarrow\widehat{BAE}+\widehat{CAF}=90^0\) (1)

Lại có \(BE\perp d\Rightarrow\Delta BAE\) vuông tại E

\(\Rightarrow\widehat{BAE}+\widehat{ABE}=90^0\) (2)

(1);(2) \(\Rightarrow\widehat{CAF}=\widehat{ABE}\)

Xét hai tam giác ABE và CAF có:

\(\left\{{}\begin{matrix}\widehat{ABE}=\widehat{CAF}\\\widehat{AEB}=\widehat{CFA}=90^0\end{matrix}\right.\)

\(\Rightarrow\Delta ABE\sim\Delta CAF\left(g.g\right)\)

\(\Rightarrow\dfrac{AE}{CF}=\dfrac{BE}{AF}\Rightarrow AE.AF=BE.CF\)

b.

\(S_{ABC}=\dfrac{1}{2}AB.AC\Rightarrow AC=\dfrac{2S_{ABC}}{AB}=\dfrac{2.24}{6}=8\left(cm\right)\)

Áp dụng hệ thức lượng:

\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\Rightarrow AH=\dfrac{AB.AC}{\sqrt{AB^2+AC^2}}=\dfrac{6.8}{\sqrt{6^2+8^2}}=4,8\left(cm\right)\)