Đáp án:

Giải thích các bước giải:

 a) Ta có \(\left(x - \frac{1}{2}\right)^2 \geq 0\) $\forall$ \(x\)

                \(\left\vert{}y + \frac{3}{4}\right\vert{} \geq 0\) `AA` \(y\)

⇒ có: \(\left(x-\frac{1}{2}\right)^{2}+\left|{}y+\frac{3}{4}\right|{}\ge 0\quad \text{với mọi }x,y\)

Mà theo đề bài ta có:

                             \(\left(x-\frac{1}{2}\right)^{2}+\left|{}y+\frac{3}{4}\right|{}\le 0\)

⇒  \(\begin{cases}\left(x-\frac{1}{2}\right)^{2}=0\\ \left|{}y+\frac{3}{4}\right|{}=0\end{cases}\)

⇒  \(\begin{cases}x-\frac{1}{2}=0\\ y+\frac{3}{4}=0\end{cases}\implies \begin{cases}x=\frac{1}{2}\\ y=-\frac{3}{4}\end{cases}\)

b) Ta có: $\frac{1}{2^2} < \frac{1}{1.2}$

$\frac{1}{3^2} < \frac{1}{2.3}$

$\dots$

$\frac{1}{100^2} < \frac{1}{99.100}$

Suy ra: $\frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{100^2} < \frac{1}{1.2} + \frac{1}{2.3} + \dots + \frac{1}{99.100}$

$\frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{100^2} < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \dots + \frac{1}{99} - \frac{1}{100}$

$\frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{100^2} < 1 - \frac{1}{100}$

$\frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{100^2} < \frac{99}{100} < \frac{100}{100} = 1$

Vậy $\frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{100^2} < 1.$