Giải thích các bước giải:
Phiếu $6:$
Ta có:
$u_n-\dfrac12=\dfrac{n}{n^2+1}-\dfrac12$
$\to u_n-\dfrac12=\dfrac{2n-(n^2+1)}{2(n^2+1)}$
$\to u_n-\dfrac12=\dfrac{-(n^2-2n+1)}{2(n^2+1)}$
$\to u_n-\dfrac12=\dfrac{-(n-1)^2}{2(n^2+1)}\le0\quad\forall n\in N$
$\to u_n\le\dfrac12,\quad\forall n\in N$
Phiếu $7:$
Ta có:
$v_n-1=\dfrac{n^2+1}{2n}-1$
$\to v_n-1=\dfrac{n^2+1-2n}{2n}$
$\to v_n-1=\dfrac{(n-1)^2}{2n}\ge 0,\quad\forall x\in N^*$
$\to v_n\ge 1,\quad\forall x\in N^*$
