Toán $\left \{ {{(60-3x)^2+(80-3y)^2=70^2} \atop {(60-5x)^2+(80-5y)^2=50^2}} \right.$ 18/07/2021 By Aubrey $\left \{ {{(60-3x)^2+(80-3y)^2=70^2} \atop {(60-5x)^2+(80-5y)^2=50^2}} \right.$
$\left \{ {{(60-3x)^2+(80-3y)^2=70^2(1)} \atop {(60-5x)^2+(80-5y)^2=50^2(2)}} \right.$ Lấy 25.(1)-9.(2) ta được: 60-3,6x-4,8y=0⇔y=$\frac{50-3x}{4}$ Thế vào (1) ta được: $(60-3x)^2+(80-3.\frac{50-3x}{4})^2=70^2$ ⇔$\frac{225}{16}x^2-\frac{675}{4}x+\frac{2025}{4}=0$ ⇔ x=6⇒y=8 Trả lời
(60−3x)2+(80−3y)2=702(1)(60−5x)2+(80−5y)2=502(2){(60−3x)2+(80−3y)2=702(1)(60−5x)2+(80−5y)2=502(2) Lấy 25.(1)-9.(2) ta được: 60-3,6x-4,8y=0⇔y=50−3x450−3×4 Thế vào (1) ta được: (60−3x)2+(80−3.50−3x4)2=702(60−3x)2+(80−3.50−3×4)2=702 ⇔22516x2−6754x+20254=022516×2−6754x+20254=0 ⇔ x=6⇒y=8 Trả lời
$\left \{ {{(60-3x)^2+(80-3y)^2=70^2(1)} \atop {(60-5x)^2+(80-5y)^2=50^2(2)}} \right.$
Lấy 25.(1)-9.(2) ta được: 60-3,6x-4,8y=0⇔y=$\frac{50-3x}{4}$
Thế vào (1) ta được: $(60-3x)^2+(80-3.\frac{50-3x}{4})^2=70^2$
⇔$\frac{225}{16}x^2-\frac{675}{4}x+\frac{2025}{4}=0$
⇔ x=6⇒y=8
(60−3x)2+(80−3y)2=702(1)(60−5x)2+(80−5y)2=502(2){(60−3x)2+(80−3y)2=702(1)(60−5x)2+(80−5y)2=502(2)
Lấy 25.(1)-9.(2) ta được: 60-3,6x-4,8y=0⇔y=50−3x450−3×4
Thế vào (1) ta được: (60−3x)2+(80−3.50−3x4)2=702(60−3x)2+(80−3.50−3×4)2=702
⇔22516x2−6754x+20254=022516×2−6754x+20254=0
⇔ x=6⇒y=8