Question

Ma puteti ajuta si pe mine cu rezolvarea acestor exerciti

Answer 1

Răspuns:

Explicație pas cu pas:

x°y=3-(x-3)(y-3)

a. 1°3=3-(1-3)(3-3)=3-0=3

b. x°e=x

3-(x-3)(e-3)=x

-(x-3)(e-3)=x-3

x-3+(x-3)(e-3)=0

(x-3)(1+e-3)=0

e-2=0

e=2

2. f(x)=eˣ+3x²+3

a. $\int\limits^2_1 {3x^2} \, dx =\frac{3x^3}{3}|^2_1 =x^3|_1^2=8-1=7$

b. $\int\limits^1_0 {x(e^x+3)} \, dx =\int\limits^1_0 {xe^x} \, dx+\int\limits^1_0 {3x} \, dx=xe^x|_0^1-\int\limits^1_0 {e^x} \, dx+\frac{3x^2}{2} |_0^1=e-e+1+\frac{3}{2} =\frac{5}{2}$

pt prima integrala facem prin parti

f=x          f'=1

g'=eˣ      g=eˣ

c. f'(x)=eˣ+6x

f(x)-f'(x)=eˣ+3x²+3-eˣ-6x=3x²-6x+3=3(x-1)²

$\int\limits^a_0 {\frac{a}{3(x-1)^2} } \, dx =\frac{1}{6} \\\\\frac{a}{3}\cdot \int\limits^a_0 {(x-1)^-^2} } \, dx =\frac{1}{6} \\\\\frac{a}{3}\cdot \frac{(x-1)^-^1}{-1} |_0^a=\frac{1}{6} \\\\\frac{-a}{3}\cdot \frac{1}{x-1}|_0^a =\frac{1}{6}$

$\frac{-a}{3}(\frac{1}{a-1} -1)=\frac{1}{6} \\\\-a(\frac{1}{a-1} -1)=\frac{1}{2}$

-a(1-a+1)=a-1

-2a-a²=a-1

a²+3a-1=0

Δ=9+4=13

$a_1=\frac{-3+\sqrt{13} }{2} \\\\a_2=\frac{-3-\sqrt{13} }{2}<0 \ nu$

3. x°y=3xy+3x+3y+2

a. x°y=3x(y+1)+3(y+1)-1

x°y=(y+1)(3x+3)-1

x°y=3(x+1)(y+1)-1

b. x°($-\frac{2}{3}$)=x

$3(x+1)(-\frac{2}{3}+1 )-1=(x+1)(-2+3)-1=(x+1)-1=x+1-1=x$