Lời giải:
`1`/ Ta có :
`f(x)` `=` `10² -7x - 5`
`f(x)` `=` `10² - 15x + 8x - 12 + 7`
`f(x)` `=` `5x.(2x - 3) + 4(2x - 3) + 7`
Do `5x.(2x - 3) + 4(2x - 3)` chia hết cho `(2x - 3)``∀` `x`
`->` Để `f(x)` chia hết cho `(2x-3)`
Suy ra `7` chia hết cho `(2x-3)`
`⇒` \(\left[ \begin{array}{l}2x-3=7\\2x-3=-7\end{array} \right.\)
\(\left[ \begin{array}{l}2x-3=1\\2x-3=-1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}2x=10\\2x=-4\end{array} \right.\)
\(\left[ \begin{array}{l}2x=4\\2x=-2\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=5\\x=-2\end{array} \right.\)
\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
Vậy `x` `∈` `{5 ; 2 ; -2 ; -1}`
`2`/
Ta có :
`1/(2²)` `<` `1/(1.2)`
`1/(3²)` `<` `1/(2.3)`
`1/(4²)` `<` `1/(3.4)`
`.....`
`⇔` `1/(1.2)` `+` `1/(2.3)` `+` `....` `+` `n/((n-1)n)` `<` `1/(2²)` `+` `1/(3²)` `+` `....` `+` `1/(n²)`
`=` `1` `-` `1/2` `+` `1/2` `-` `1/3` `+``.....``+` `1/(n-1)` `-` `1/n`
`=` `1` - `1/n ``<` `1` `∀` `n`
`->` `1/(2²)` `+` `1/(3²)` `+` `....` `+` `1/(n²)` `<` `1` `∀` `n`
`->` `1` `+` `1/(2²)` `+` `1/(3²)` `+` `....` `+` `1/(n²)` `<` `2` `∀` `n`
`->` Đpcm
