1
f`(x)=2x(x²-4)+2x(x²+5)+2/2√x=2x³-8x+2x³+10x+1/√x=4x³+2x+1/√x
2
f`(x)=2√2x-3(5x+1)-5(3x-2)-1/2√x=2√2x-15x-3-15x+10-1/2√x=(2√2-30)x-1/2√x
(5х²в -0,5в)²=(5х²в)²- 2*5х²в*0,5в +(0,5в)²= 25х⁴в² - 5х²в²+0,25в²
B1=3*1-1=2
b2=3*2-1=5
d=5-2=3
b16=b1+15d=2+45=47
s=(2+47)*16/2=49*8=392