[tex] 4a) \text{Demonstr\u am prin metoda induc\c tiei matematice}\\
P(1):1+\frac{1}{1^2}+\frac{1}{(1+1)^2}=\big[1+\frac{1}{1\cdot(1+1)}\big]^2 \rightarrow 1+1+\frac{1}{4}=(1+\frac{1}{2})^2\rightarrow\\2+\frac{1}{4}=1+2\cdot\frac{1}{2}+\frac{1}{4}\text{ (A)}\\
\text{Presupunem }P(n):1+\frac{1}{n^2}+\frac{1}{(n+1)^2}=\big[1+\frac{1}{n\cdot(n+1)}\big]^2\text{ adev\u arat\u a. Demonstr\u am }\\P(n+1):1+\frac{1}{(n+1)^2}+\frac{1}{(n+1+1)^2}=\big[1+\frac{1}{(n+1)\cdot(n+1+1)}\big]^2\\ [/tex]
[tex] \rightarrow 1+\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}=\big[1+\frac{1}{(n+1)\cdot(n+2)}\big]^2\\
\text{ ad\u aug\u am }\frac{1}{n^2}\\
\rightarrow 1+\frac{1}{n^2}+\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}=\big[1+\frac{1}{(n+1)\cdot(n+2)}\big]^2+\frac{1}{n^2}\\\\
\text{Primii trei termeni din st\^anga seam\u an\u a cu cei din }P(n)\\
\rightarrow \big[1+\frac{1}{n(n+1)}\big]^2+\frac{1}{(n+2)^2}=\big[1+\frac{1}{(n+1)\cdot(n+2)}\big]^2+\frac{1}{n^2}\\\\
\text{Am separat termenii: } [/tex]
[tex] \big[1+\frac{1}{n(n+1)}\big]^2-\big[1+\frac{1}{(n+1)\cdot(n+2)}\big]^2=\frac{1}{n^2}-\frac{1}{(n+2)^2}\\\text{Folosim }a^2-b^2=(a-b)(a+b)\\
\big[1+\frac{1}{n(n+1)}-1-\frac{1}{(n+1)(n+2)}\big]\big[1+\frac{1}{n(n+1)}+1+\frac{1}{(n+1)(n+2)}\big]=\frac{1}{n^2}-\frac{1}{(n+2)^2}\\\\
\frac{n+2-n}{n(n+1)(n+2)}\cdot\big[2+\frac{n+2+n}{n(n+1)(n+2)}\big]=\frac{(n+2)^2-n^2}{n^2(n+2)^2}\\\\
\frac{2}{n(n+1)(n+2)}\cdot\frac{2n(n+1)(n+2)+2n+2}{n(n+1)(n+2)}=\frac{4n+4}{n^2(n+2)^2} [/tex]
[tex] \frac{2}{n(n+1)(n+2)}\cdot\frac{2n(n+1)(n+2)+2(n+1)}{n(n+1)(n+2)}=\frac{4n+4}{n^2(n+2)^2}\\\\
\frac{2}{n(n+1)(n+2)}\cdot\frac{2(n+1)[n(n+2)+1]}{n(n+1)(n+2)}=\frac{4n+4}{n^2(n+2)^2}\\\\
4\frac{n(n+2)+1}{n^2(n+1)(n+2)^2}=\frac{4n+4}{n^2(n+2)^2}\\\\
4\frac{n^2+2n+1}{n^2(n+1)(n+2)^2}=\frac{4n+4}{n^2(n+2)^2}\\\\4\frac{(n+1)^2}{n^2(n+1)(n+2)^2}=\frac{4n+4}{n^2(n+2)^2}\text{ (A), }\forall n\in\mathbb{N^*}. [/tex]
[tex] b) \text{Aici folose\c sti punctul a) sub fiecare radical.}\\
\sqrt{\big[1+\frac{1}{2\cdot3}\big]^2}+\sqrt{\big[1+\frac{1}{3\cdot4}\big]^2}+...+\sqrt{\big[1+\frac{1}{199\cdot200}\big]^2}=x+198\\
\big|1+\frac{1}{2\cdot3}\big|+\big|1+\frac{1}{3\cdot4}\big|+...+\big|1+\frac{1}{199\cdot200}\big|=x+198\\\\
1+\frac{1}{2\cdot3}+1+\frac{1}{3\cdot4}+...+1+\frac{1}{199\cdot200}=x+198\\\\
198+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{199\cdot200}=x+198\ [/tex]
[tex] \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{199\cdot200}=x\\\\
x=\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+...+\frac{200-199}{199\cdot200}=\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+\frac{4}{3\cdot4}-\frac{3}{3\cdot4}+...+\frac{200}{199\cdot200}-\frac{199}{199\cdot200}\\\\
=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}=\frac{1}{2}-\frac{1}{200}=\frac{100-1}{200}=\frac{99}{200}. [/tex]
[tex] 5.|2x-3|+\sqrt{4y^2-20y+25}+(2z-2\sqrt3)^2\leq0\\
|2x-3|+\sqrt{(2y-5)}+(2z-2\sqrt3)^2\leq0\\
|2x-3|+|2y-5|+(2z-2\sqrt3)^2\leq0\\
\text{Avem o sum\u a de trei termeni negativ\u a sau egal\u a cu 0. Pentru c\u a fiecare termen al }\\\text{sumei este pozitiv, \^inseamn\u a c\u a fiecare termen este nul.}\\
|2x-3|=0\rightarrow 2x-3=0\rightarrow 2x=3\rightarrow x=\frac{3}{2} \\
|2y-5|=0\rightarrow 2y-5=0\rightarrow 2y=-5\rightarrow y=\frac{-5}{2}\\
(2z-2\sqrt3)^2=0\rightarrow 2z-2\sqrt3=0\rightarrow z=\sqrt3 [/tex]
