4.
a) Pentru a verifica relatia, vom aduce la acelasi numitor membrul drept :
[tex] \frac{1}{x^{2}}-\frac{1}{(x+1)^{2}}=\frac{(x+1)^{2}-x^{2}}{x^{2}(x+1)^{2}}= \frac{x^{2}+2x+1-x^{2}}{x^{2}(x+1)^{2}}=\frac{2x+1}{x^{2}(x+1)^{2}} [/tex] astfel am verificat relatia.
b) Pentru a calcula suma S vom folosi relatia de la punctul a:
[tex] \frac{2*1+1}{1^{2}*2^{2}}}=\frac{1}{1^{2}}-\frac{1}{2^{2}} [/tex]
[tex] \frac{2*2+1}{2^{2}+3^{2}}=\frac{1}{2^{2}}-\frac{1}{3^{2}} [/tex]
.
.
[tex] \frac{4035}{2017^{2}*2018^{2}}=\frac{1}{2017^{2}}-\frac{1}{2018^{2}} [/tex]
dupa reducerea termenilor S=1-1/2018²=(2018²-1)/2018²
5. [tex] \sqrt{2-\sqrt{3}}= [/tex] √(√3/√2-√1/√2)²=√3/√2-√1/√2
[tex] \sqrt{2+\sqrt{3}}= [/tex] √(√3/√2+√1/√2)²=√3/√2+√1/√2
Suma va fi :
√2+√3/√2-√1/√2+√3/√2+√1/√2=√2+2√3/√2=√2+√6=√2(1+√3)
