a) Để \(\overline{7a5b1}⋮3\) thì \(7+a+5+b+1⋮3\)
\(\Rightarrow13+a+b⋮3\)
mà \(a+b< 18\)
\(\Rightarrow a+b\in\left\{2;5;8;11;14;17\right\}\)
Lại có: `a-b=4`
\(\Rightarrow a+b\notin\left\{2;5;11;17\right\}\) do \(a+b< 4;a+b⋮̸2\)
\(\Rightarrow a+b\in\left\{8;14\right\}\)
`TH1: a+b=8`
\(\left\{{}\begin{matrix}a+b=8\\a-b=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{8+4}{2}=6\\b=8-6=2\end{matrix}\right.\)
`TH1: a+b=14`
\(\left\{{}\begin{matrix}a+b=14\\a-b=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{14+4}{2}=9\\b=14-9=5\end{matrix}\right.\)
Vậy `(a,b)=(6,2);(9,5)`
`(a):` Ta có : `a-b=4=>a=b+4`
Xét tổng các chữ số của `\overline{7a5b1}` :
`7+a+5+b+1=13+a+b=13+b+4+b=17+2b`
Để `\overline{7a5b1}\vdots 3`
Thì : `17+2b\vdots 3`
Vì : `b` là STN có `1` chữ số `=>2b\vdots 2;0\le 2b\le 18`
`=>2b \in {4;10;16}`
`=>b\in {2;5;8}\ (TM)`
`+)\ b=2=>a=2+4=6\ (TM)`
`+)\ b=5=>a=5+4=9\ (TM)`
`+)\ b=8=>a=8+4=12\ (KTM)`
Vậy `(a;b)=(6;2);(9;5)`
`(b):\ 1/2^{2}<(1)/(1.2);(1)/(3^{2})<(1)/(2.3);(1)/(4^{2})<(1)/(3.4);...;(1)/(10^{2})<(1)/(9.10)`
`=>D< (1)/(1.2)+(1)/(2.3)+(1)/(3.4)+(1)/(4.5)+...+(1)/(9.10)`
`=>D< 1-(1)/(2)+(1)/(2)-(1)/(3)+(1)/(3)-(1)/(4)+(1)/(4)-(1)/(5)+...+(1)/(9)-(1)/(10)`
`=>D<1-(1)/(10)<1` (DPCM)
b) Có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{10^2}< \dfrac{1}{9.10}\)
\(\Rightarrow D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)
\(\Rightarrow D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow D< 1-\dfrac{1}{10}< 1\left(đpcm\right)\)
