Bài 1:
Xét tứ giác ABDC có \(\widehat{A}+\widehat{D}=180^0\)

nên ABDC là tứ giác nội tiếp

a: Đặt 1/x=a

1/(y-2)=b

Hệ phương trình trở thành:

\(\left\{{}\begin{matrix}2a+3b=4\\4a+b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a+6b=8\\4a+b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5b=7\\4a+b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{7}{5}\\a=-\dfrac{1}{10}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{-1}{10}\\\dfrac{1}{y-2}=\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y-2=\dfrac{5}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=\dfrac{19}{7}\end{matrix}\right.\)

b: \(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{2}{y}=0\\\dfrac{2}{x+1}+\dfrac{3}{y}=-1\end{matrix}\right.\)

=>Hệ phương trình vô nghiệm

a) \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y-2}=4.\\\dfrac{4}{x}+\dfrac{1}{y-2}=1.\end{matrix}\right.\) \(ĐK:x\ne0;y\ne2.\)

Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y-2}=b\left(a;b\ne0\right).\)

\(\Rightarrow\left\{{}\begin{matrix}2a+3b=4.\\4a+b=1.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{-1}{10}.\\b=\dfrac{7}{5}.\end{matrix}\right.\) (TM).

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{-1}{10}.\\\dfrac{1}{y-2}=\dfrac{7}{5}.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-10\left(TM\right).\\y=\dfrac{19}{7}\left(TM\right).\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(-10;\dfrac{19}{7}\right).\)

Bài 1: 

a: \(\dfrac{13\sqrt{2}-4\sqrt{6}}{24-4\sqrt{3}}=\dfrac{\sqrt{2}\left(13-4\sqrt{3}\right)}{4\sqrt{3}\left(2\sqrt{3}-1\right)}=\dfrac{\sqrt{2}\left(2\sqrt{3}-1\right)}{4\sqrt{3}}=\dfrac{\sqrt{6}\left(2\sqrt{3}-1\right)}{12}\)

b: \(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}=\sqrt{2}+1\)

\(tana=\dfrac{\sqrt{5}}{3}\)

=>\(\dfrac{sina}{cosa}=\dfrac{\sqrt{5}}{3}\)

=>\(sina=cosa\cdot\dfrac{\sqrt{5}}{3}\)

\(B=\dfrac{2\cdot cosa+\dfrac{\sqrt{5}}{3}\cdot cosa}{5\cdot cosa-3\cdot\dfrac{\sqrt{5}}{3}\cdot cosa}=\left(2+\dfrac{\sqrt{5}}{3}\right):\left(5-\sqrt{5}\right)\)

\(=\dfrac{6+\sqrt{5}}{3\left(5-\sqrt{5}\right)}=\dfrac{35+11\sqrt{5}}{40}\)

ai giúp mình với

làm ơn cứu tui với 😭😭

a: HB=4,5(cm)

BC=12,5(cm)

b: \(\widehat{B}=37^0\)

a: (d1): a=-3; b=1