Ta có: 

\(A=\left(1+tan^2x\right)cos^2x-\left(1+cot^2x\right)\left(cos^2x-1\right)\)

\(=\frac{1}{cos^2x}.cos^2x-\frac{1}{sin^2x}.sin^2x\)

\(=1-1=0\)

\(B=tan72^o-cot18^o+sin^230^o+sin^260^o\)

\(=tan72^o-tan72^o+sin^230^o+cos^230^o\)

\(=1\)

\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{9-12\sqrt{5}+20}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)^2}=\sqrt{5-\sqrt{5}+1}=\sqrt{1}=1\)

a, Xét tam giác DEF vuông tại D, đường cao DH 

Áp dụng định lí Pytago : \(\text{EF}=\sqrt{DF^2+DE^2}=\sqrt{64+36}=10\)cm 

* Áp dụng hệ thức : \(DE^2=EH.BC\Rightarrow EH=x=\frac{DE^2}{BC}=\frac{36}{10}=\frac{18}{5}\)cm 

=> \(FH=y=10-\frac{18}{5}=\frac{50-18}{5}=\frac{32}{5}\)cm 

b, MNPH

Xét tam giác MNP vuông tại M, đường cao MH ta có : 

Áp dụng định lí Pytago : \(NP=\sqrt{MN^2+MP^2}=\sqrt{25+49}=\sqrt{74}\)cm 

* Áp dụng hệ thức : \(MH.NP=MN.MP\Rightarrow MH=x=\frac{MN.MP}{NP}=\frac{35}{\sqrt{74}}=\frac{35\sqrt{74}}{74}\)cm 

* Áp dụng hệ thức : \(MP^2=HP.NP\Rightarrow HP=y=\frac{MP^2}{NP}=\frac{49}{\sqrt{74}}=\frac{49\sqrt{74}}{74}\)cm 

Đk: x \(\ge\)0; y \(\ge\)0

Ta có: \(A=x\sqrt{x}+y\sqrt{y}=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\)

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

Với \(\sqrt{x}+\sqrt{y}=1\) => \(A=1^3-3\sqrt{xy}.1=1-3\sqrt{xy}\) (1)

Do \(\sqrt{xy}\le\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{4}\)(bđt cosi ) => \(1-3\sqrt{xy}\ge1-3\cdot\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{4}=1-\frac{3}{4}=\frac{1}{4}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\sqrt{x}=\sqrt{y}\\\sqrt{x}+\sqrt{y}=1\end{cases}}\) <=> \(x=y=\frac{1}{4}\)

Vậy MinA = 1/4 <=> x = y = 1/4

Lại có: \(\sqrt{x}+\sqrt{y}=1\) => \(\sqrt{y}=1-\sqrt{x}\le1\)  => \(\sqrt{y}-1\le0\)

          => \(\sqrt{x}=1-\sqrt{y}\le1\) ==> \(\sqrt{x}-1\le0\)

=> \(\left(\sqrt{x}-1\right)\left(\sqrt{y}-1\right)\ge0\) <=> \(xy-\left(\sqrt{x}+\sqrt{y}\right)+1\ge0\)

<=> \(xy-1+1\ge0\) <=> \(xy\ge0\) <=> \(\sqrt{xy}\ge\)0

Do đó: \(A=1-3\sqrt{xy}\le1-3.0=1\)

Dấu "=" xảy ra<=> \(\hept{\begin{cases}xy=0\\\sqrt{x}+\sqrt{y}=1\end{cases}}\) <=> \(\hept{\begin{cases}x=0\\y=1\end{cases}}\) hoặc \(\hept{\begin{cases}x=1\\y=0\end{cases}}\)

Vậy MaxA = 1 <=> \(\hept{\begin{cases}x=1\\y=0\end{cases}}\)hoặc \(\hept{\begin{cases}x=0\\y=1\end{cases}}\)

Đặt A = \(\sqrt{5-\sqrt{21}}-\sqrt{6-\sqrt{32}}\)

\(\sqrt{2}A=\sqrt{10-2\sqrt{7.3}}-\sqrt{2}\sqrt{6-2\sqrt{4.2}}\)

\(=\sqrt{7}-\sqrt{3}-\sqrt{2}\left(\sqrt{4}-\sqrt{2}\right)\)

\(=\sqrt{7}-\sqrt{3}-2\sqrt{2}+2\)

Vậy \(A=\frac{\sqrt{7}-\sqrt{3}-2\sqrt{2}+2}{\sqrt{2}}=\frac{\sqrt{14}-\sqrt{6}-8+2\sqrt{2}}{2}\)

\(A=\sqrt{5-\sqrt{21}}-\sqrt{6-4\sqrt{2}}\)

nên \(A=\sqrt{\frac{7}{2}-2\sqrt{\frac{7}{2}.\frac{3}{2}}+\frac{3}{2}}-\sqrt{4-4\sqrt{2}+2}=\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(A=\frac{\sqrt{14}-\sqrt{6}}{2}-2+\sqrt{2}\)