Hạ CH vuông với AB tại H
Ta có :  \(HB=\frac{AB-CD}{2}=8\left(cm\right)\)

\(\Rightarrow BC^2=HB.AB=8.26\)

\(\Rightarrow BC=4\sqrt{3}\)

\(\Rightarrow HC=\sqrt{BC^2-HB^2}=12\)

\(\Rightarrow S_{ABCD}=\frac{HC.\left(AB+CD\right)}{2}=\frac{12.\left(26+10\right)}{2}=216\left(cm^2\right)\)

                  Ps : nhớ k ạ :33

                                                                                                                                                     # Aeri # 

BC=4\(\sqrt{ }\)13 chứ

HOK TỐT

Ta có: C = \(\frac{x+10}{\sqrt{x}+3}=\frac{x-9+19}{\sqrt{x}+3}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+19}{\sqrt{x}+3}=\sqrt{x}-3+\frac{19}{\sqrt{x}+3}\)

C = \(\sqrt{x}+3+\frac{19}{\sqrt{x}+3}-6\ge2.\sqrt{\left(\sqrt{x}+3\right)\cdot\frac{19}{\left(\sqrt{x}+3\right)}}-6\)(bđt cosi)

C \(\ge2\sqrt{19}-6\)

Dấu "=" xảy ra <=> \(\sqrt{x}+3=\frac{19}{\sqrt{x}+3}\) <=> \(\left(\sqrt{x}+3\right)^2=19\)

<=> \(\orbr{\begin{cases}\sqrt{x}+3=\sqrt{19}\\\sqrt{x}+3=-\sqrt{19}\left(vn\right)\end{cases}}\) <=> \(\sqrt{x}=\sqrt{19}-3\) <=> \(x=22-6\sqrt{19}\)

Vậy MinC = \(2\sqrt{19}-6\) <=> \(x=22-6\sqrt{19}\)

\(\hept{\begin{cases}\left(5x-4y\right)\left(3x+2y\right)=7y-2x\\\left(5y-4x\right)\left(3y+2x\right)=7x-2y\end{cases}}\)

<=>\(\hept{\begin{cases}15x^2-2xy-8y^2=7y-2x\left(1\right)\\15y^2-2xy-8x^2=7x-2y\end{cases}}\)

<=> \(15x^2-2xy-8y^2-15y^2+2xy+8x^2=7y-2x-7x+2y\)

<=> \(23x^2-23y^2-9y+9x=0\)

<=> \(23\left(x-y\right)\left(x+y\right)-9\left(x-y\right)=0\)

<=> \(\left(x-y\right)\left(23x+23y-9\right)=0\)

<=> \(\orbr{\begin{cases}x=y\\x+y=\frac{9}{23}\end{cases}}\)

Với x = y => thay vào pt (1)

<=> \(15x^2-2x^2-8x^2=7x-2x\)

<=> \(5x^2-5x=0\) <=> \(5x\left(x-1\right)=0\)

<=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\) => \(\orbr{\begin{cases}x=y=0\\x=y=1\end{cases}}\)

Với \(x+y=\frac{9}{23}\) => \(y=\frac{9}{23}-x\)thay vào pt (1)

Ta có: \(15x^2-2x\left(\frac{9}{23}-x\right)-8\left(\frac{9}{23}-x\right)^2=7\left(\frac{9}{23}-x\right)-2x\)

<=> \(15x^2-\frac{18}{23}x+2x^2-8\left(\frac{81}{529}-\frac{18}{23}x+x^2\right)=\frac{63}{23}-7x-2x\)

<=> \(17x^2-\frac{18}{23}x-\frac{648}{529}+\frac{144}{23}x-8x^2-\frac{63}{23}+9x=0\)

<=> \(9x^2+\frac{333}{23}x-\frac{2097}{529}=0\) (phần còn lại tự làm)

\(a,\sqrt{29+12\sqrt{5}}+2\sqrt{21-8\sqrt{5}}\)

\(\sqrt{29+6\sqrt{20}}+\sqrt{84-32\sqrt{5}}\)

\(\sqrt{\sqrt{20}^2+6\sqrt{20}+3^2}+\sqrt{84-16\sqrt{20}}\)

\(\sqrt{\left(\sqrt{20}+3\right)^2}+\sqrt{8^2-16\sqrt{20}+\sqrt{20}^2}\)

\(\left|\sqrt{20}+3\right|+\sqrt{\left(8-\sqrt{20}\right)^2}\)

\(\sqrt{20}+3+\left|8-\sqrt{20}\right|\)

\(\sqrt{20}+3+8-\sqrt{20}\)

\(=11\)

Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=\sqrt{5-4\sqrt{5}+4}-\sqrt{5}\)

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

Vayyj ...

Ta có : VT= \(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)

                =\(\sqrt{5-4\sqrt{5}+4}\)\(-\sqrt{5}\)

                =\(\sqrt{\left(\sqrt{5}\right)^2-2.2\sqrt{5}+2^2}\)\(-\sqrt{5}\)

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

                =\(\left|\sqrt{5}-2\right|-\sqrt{5}\)

                =\(\sqrt{5}-2-\sqrt{5}\)

                =\(-2\)=VP

Câu 11.12. 

Kẻ đường cao \(AH,BK\).

Do tam giác \(\Delta AHD=\Delta BKC\left(ch-gn\right)\)nên \(DH=BK\).

Đặt \(AB=AH=x\left(cm\right),x>0\).

Suy ra \(DH=\frac{10-x}{2}\left(cm\right)\)

Xét tam giác \(AHD\)vuông tại \(H\):

\(AD^2=AH^2+HD^2=x^2+\left(\frac{10-x}{2}\right)^2\)(định lí Pythagore) 

Xét tam giác \(DAC\)vuông tại \(A\)đường cao \(AH\):

\(AD^2=DH.DC=10.\left(\frac{10-x}{2}\right)\)

Suy ra \(x^2+\left(\frac{10-x}{2}\right)^2=10.\frac{10-x}{2}\)

\(\Leftrightarrow x=2\sqrt{5}\)(vì \(x>0\))

Vậy đường cao của hình thang là \(2\sqrt{5}cm\).

Câu 11.11. 

Kẻ \(AE\perp AC,E\in CD\).

Khi đó \(AE//BD,AB//DE\)nên \(ABDE\)là hình bình hành. 

Suy ra \(AE=BD=15\left(cm\right)\).

Kẻ đường cao \(AH\perp CD\)suy ra \(AH=12\left(cm\right)\).

Xét tam giác \(AEC\)vuông tại \(A\)đường cao \(AH\): 

\(\frac{1}{AH^2}=\frac{1}{AE^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{AC^2}=\frac{1}{AH^2}-\frac{1}{AE^2}=\frac{1}{12^2}-\frac{1}{15^2}=\frac{1}{400}\)

\(\Rightarrow AC=20\left(cm\right)\)

\(S_{ABCD}=\frac{1}{2}AC.BD=\frac{1}{2}.15.20=150\left(cm^2\right)\),

Khó quá, xé sách