Bài 2:
a: Ta có: \(5x\left(x-1\right)+10x-10=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
b: Ta có: \(\left(x+2\right)\left(x+3\right)-2x=6\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
c: Ta có: \(\left(x-1\right)\left(x-2\right)-2=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
\(1,\widehat{D}=360-\widehat{A}-\widehat{B}-\widehat{C}=360-120-50-90=100\)
\(2,\widehat{D}+\widehat{C}=360-\widehat{A}-\widehat{B}=360-50-110=200\\ \Rightarrow4\widehat{D}=200\Rightarrow\widehat{D}=50\Rightarrow\widehat{C}=50\cdot3=150\)
\(3,\\ \widehat{D}+\widehat{C}=360-\widehat{A}-\widehat{B}=160\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\dfrac{\widehat{D}}{5}=\dfrac{\widehat{C}}{3}=\dfrac{\widehat{D}+\widehat{C}}{5+3}=\dfrac{160}{8}=20\\ \Rightarrow\left\{{}\begin{matrix}\widehat{D}=100\\\widehat{C}=60\end{matrix}\right.\)
hình tự vẽ đi :(((
vì tổng các góc của tứ giác là 360độ
=> góc D= 360-(A+B+C).
=> góc D=360-(50+120+90)=100
Vậy góc D=100 độ
