+) Nếu : \(x< 2\Rightarrow x-2< 2-2\Rightarrow x-2< 0\)

\(\Rightarrow\left|x-2\right|=-\left(x-2\right)=-x+2\)

\(\Rightarrow-x+2+x=3\Rightarrow-x+x=3-2\Rightarrow0=1\) (loại)

+) Nếu : \(x\ge2\Rightarrow x-2\ge2-2\Rightarrow x-2\ge0\)

\(\Rightarrow\left|x-2\right|=x-2\)

\(\Rightarrow x-2+x=3\Rightarrow x+x-2=3\)

\(\Rightarrow2x-2=3\Rightarrow2x=5\Rightarrow x=\dfrac{5}{2}\)

Vậy x = \(\dfrac{5}{2}\)

Với n = 0

\(\Rightarrow3.5^{2.0+1}+2^{3.0+1}=3.5+2=15+2=17⋮17\Rightarrow\)đúng với n = 0

Giả sử \(3.5^{2n+1}+2^{3n+1}\) đúng với n = k \(\in\) N*

\(\Rightarrow3.5^{2k+1}+2^{3k+1}⋮17\)

C/m : \(3.5^{2n+1}+2^{3n+1}\) đúng với n = k + 1 ( k \(\in\) N* )

Ta có :

\(3.5^{2n+1}+2^{3n+1}=3.5^{2\left(k+1\right)+1}+2^{3\left(k+1\right)+1}\)

\(=3.25.5^{2k+1}+8.3^{3k+1}=3.25.5^{2k+1}+25.2^{3k+1}-17.2^{3k+1}\)

\(=25\left(3.5^{2k+1}+2^{3k+1}\right)-17.2^{3k+1}\)

Vì : \(17.2^{3k+1}⋮17\) ; \(3.5^{2k+1}+2^{3k+1}⋮17\) theo phương pháp quy nạp

\(\Rightarrow3.5^{2\left(k+1\right)+1}+2^{3\left(k+1\right)+1}⋮17\)

Vậy ...

Vì : \(\left|x+2017\right|\ge0\forall x\)

\(\left|y-2017\right|\ge0\forall y\)

\(\Rightarrow\left|x+2017\right|+\left|y-2017\right|\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+2017=0\\y-2017=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2017\\y=2017\end{matrix}\right.\)

Vậy x = -2017 ; y = 2017

b, \(\left|2x-1\right|=\left|x+8\right|\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=x+8\\2x-1=-\left(x+8\right)\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x-x=8+1\\2x+x=-8+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=9\\3x=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{-7}{3}\end{matrix}\right.\)

Vậy x = 9

c, \(\left|3x-2\right|-\left|x+14\right|=0\Rightarrow\left|3x-2\right|=\left|x+14\right|\)

\(\Rightarrow\left[{}\begin{matrix}3x-2=x+14\\3x-2=-\left(x+14\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x=14+2\\3x+x=-14+2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=16\\4x=-12\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

Vậy ...

\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)

\(=\dfrac{9}{10}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)=\dfrac{9}{10}-\dfrac{9}{10}=0\)