f'(x)=17cosx+4
f'(x0)=17cosп/2+4=4
Ответ: 4
√67.6/(8.1*1.44)=√67600/(81*144)=260/(9*12)=260/108
1) а) (3х+9)-(ху+3у)=3(х+3)-у(х+3)=(3-у)(Х+3)