Y = y^2/x^2-2 xy +y-e^x=0; y(a)=b - вопрос №203948212220 от dsokolovskiyp015em 10.06.2021 07:00

Question

Математика: Y = y^2/x^2-2 xy +y-e^x=0; y(a)=b

dsokolovskiyp015em · Answer

(4.3+3 1/2):3=(4 3/10+3 1/2):3=(43/10+35/10):3=(78/10):3=26/10=2.6

(3 9/20+2.75):2=(3 9/20+2 3/4):2=(3 9/20+2 16/20):2=(69/20+56/20):2=(125/20):2=(125/40)=3 5/40=3.125

(33.74-5 1/25):7=(33 74/100-5 4/100):7=(3374/100-504/100):7=(2870/100):7=(2870/100)*(1/7)=(2870/700)=4 70/700=4.1

(37 1/5-6.8):8=(37 2/10-6 8/10):8=(372/10-68/10):8=(304/10):8=(304/10)*(1/8)=38:10=3.8

(14.7+23 4/5):11=(14 7/10+23 8/10):11=(147/10+238/10):11=(385/100):11=(385/100)*(1/11)=(35/100)=0.35

(61.68-4 2/25):12=(61 68/100-4 8/100):12=(6168/100-408/100):12=(5760/100):12=(5760/100)*(1/12)=(480/100)=4.8

Y' = y^2/x^2-2 xy'+y-e^x=0; y(a)=b

Y' = y^2/x^2-2
xy'+y-e^x=0; y(a)=b