mở bài là giới thiệu về cụ nha mn em viết lộn ạ 

thân bài là đóng góp ạ 

BC = 4,5 cm thì AB = ? AC = ?

Có thiếu đề ko bạn ?

Đây nhé '-'

Vẽ hơi méo,thông cảm

Tam giác này là Tam giác cân nhé

a: Xét ΔABC vuông tại A có 

\(BC^2=AB^2+AC^2\)

hay \(BC=\dfrac{9\sqrt{34}}{10}\left(cm\right)\)

Xét ΔABC vuông tại A có 

\(\sin\widehat{B}=\dfrac{AC}{BC}=\dfrac{5\sqrt{34}}{34}\)

\(\Leftrightarrow\widehat{B}\simeq59^0\)

\(\Leftrightarrow\widehat{C}=21^0\)

Thi tự làm ạ

\(1,\\ a,=3x^2y-2xy^2-2xy^2=3x^2y-4xy^2\\ b,=\left(x+2\right)^2:\left(x+2\right)=x+2\\ c,=\dfrac{2}{x}\\ 2,\\ 1,\\ a,=2\left(x^2-2x+1\right)=2\left(x-1\right)^2\\ b,=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\\ 2,\\ a,\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\\ 3,\\ a,ĐK:x\ne\pm1\\ b,A=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}\\ c,A=2\Leftrightarrow\dfrac{x+1}{x-1}=2\Leftrightarrow x+1=2x-2\Leftrightarrow x=3\left(tm\right)\)

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