Ta có : 

\(\hept{\begin{cases}AC^2=AD^2+DC^2\\BD^2=AD^2+AB^2\end{cases}}\) vì \(CD>AB\Rightarrow AC>BD\)

b. ta có \(AC^2-BD^2=\left(AD^2+DC^2\right)-\left(AD^2+AB^2\right)=CD^2-AB^2\)

Ta có : \(\hept{\begin{cases}BD^2=BA^2+AD^2\\BC^2=\left(DC-AB\right)^2+AD^2\end{cases}\Leftrightarrow\hept{\begin{cases}225=BA^2+AD^2\\169=\left(14-AB\right)^2+AD^2\end{cases}}}\)

lấy hiệu ta có : \(56=28AB-196\Leftrightarrow AB=9\Rightarrow AD=\sqrt{225-9^2}=12\)

Diện tích hình thang là :\(\frac{1}{2}\left(AB+CD\right)\times AD=\frac{1}{2}\left(9+14\right)\times12=138cm^2\)

Mik có lớp 4 thôi seo giúp đc . Seo bn zô zuyên hế . ^~^

xin lỗi nhé

ta có :

\(P=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}+1-\sqrt{xy}}{1-\sqrt{xy}}\right):\left(\frac{\sqrt{xy}-1-\sqrt{xy}-\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)

\(=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{x}+1}{1-\sqrt{xy}}\right):\left(\frac{1+\sqrt{x}}{1-\sqrt{xy}}-\frac{1+\sqrt{x}}{\sqrt{xy}+1}\right)\)

\(=\left(\frac{1}{\sqrt{xy}+1}+\frac{1}{1-\sqrt{xy}}\right):\left(\frac{1}{1-\sqrt{xy}}-\frac{1}{\sqrt{xy}+1}\right)\)

\(=\frac{2}{1-xy}:\text{​​}\frac{2\sqrt{xy}}{1-xy}=\frac{1}{\sqrt{xy}}\)

b.ta có \(P=\frac{1}{\sqrt{xy}}\le\frac{\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)^2}{4}=\frac{36}{4}=9\)

Vậy GTLN của P =9

oh no oh no oh no oh no no no no no

oh no oh no oh no no no no no

\(ĐKXĐ:x\ge3\)

\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=0\)

\(2\sqrt{9\left(x-3\right)}-\frac{1}{5}\sqrt{25\left(x-3\right)}-\frac{1}{7}\sqrt{49\left(x-3\right)}=0\)

\(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=0\)

\(4\sqrt{x-3}=0\)

\(x-3=0\)

\(x=3\left(TM\right)\)

\(\sqrt{x-3}-2\sqrt{x^2-9}=0\)ĐK : \(x\le-3;x\ge3\)

\(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x+3}\right)=0\)

TH1 : \(\sqrt{x-3}=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)(tm)

TH2 : \(1-2\sqrt{x+3}=0\Leftrightarrow2\sqrt{x+3}=1\)

\(\Leftrightarrow4\left(x+3\right)=1\Leftrightarrow x+3=\frac{1}{4}\Leftrightarrow x=-\frac{11}{4}\)(ktm)