Câu hỏi của nguyen lan mai

a: b: PTHĐGĐ là:-x^2-2x+3=0=>x^2+2x-3=0=>(x+3)(x-1)=0=>A(-3;-9); B(1;-1)\(OA=\sqrt{\left(-3\right)^2+\left(-9\right)^2}=3\sqrt{10}\)\(OB=\sqrt{1^2+\left(-1\right)^2}=\sqrt{2}\)\(AB=\sqrt{\left(1+3\right)^2+\left(-1+9\right)^2}=4\sqrt{5}\)\(cosO=\dfrac{90+2-80}{2\cdot3\sqrt{20}}=\dfrac{\sqrt{5}}{5}\)=>\(sinO=\dfrac{2\sqrt{5}}{5}\)\(S_{OAB}=\dfrac{1}{2}\cdot\dfrac{2\sqrt{5}}{5}\cdot3\sqrt{20}=6\)Câu trả lời: a: b: PTHĐGĐ là:-x^2-2x+3=0=>x^2+2x-3=0=>(x+3)(x-1)=0=>A(-3;-9); B(1;-1)\(OA=\sqrt{\left(-3\right)^2+\left(-9\right)^2}=3\sqrt{10}\)\(OB=\sqrt{1^2+\left(-1\right)^2}=\sqrt{2}\)\(AB=\sqrt{\left(1+3\right)^2+\left(-1+9\right)^2}=4\sqrt{5}\)\(cosO=\dfrac{90+2-80}{2\cdot3\sqrt{20}}=\dfrac{\sqrt{5}}{5}\)=>\(sinO=\dfrac{2\sqrt{5}}{5}\)\(S_{OAB}=\dfrac{1}{2}\cdot\dfrac{2\sqrt{5}}{5}\cdot3\sqrt{20}=6\)