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. [tex]\int\limits^2_1 {3x^2} \, dx =\frac{3x^3}{3}|^2_1 =x^3|_1^2=8-1=7[/tex]
b. [tex]\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}[/tex]
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
[tex]\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}[/tex]
[tex]\frac{-a}{3}(\frac{1}{a-1} -1)=\frac{1}{6} \\\\-a(\frac{1}{a-1} -1)=\frac{1}{2}[/tex]
-a(1-a+1)=a-1
-2a-a²=a-1
a²+3a-1=0
Δ=9+4=13
[tex]a_1=\frac{-3+\sqrt{13} }{2} \\\\a_2=\frac{-3-\sqrt{13} }{2}<0 \ nu[/tex]
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°([tex]-\frac{2}{3}[/tex])=x
[tex]3(x+1)(-\frac{2}{3}+1 )-1=(x+1)(-2+3)-1=(x+1)-1=x+1-1=x[/tex]
