Bài 5:

a. Gọi $d=ƯCLN(n-2, n+1)$

$\Rightarrow n-2\vdots d; n+1\vdots d$

$\Rightarrow (n+1)-(n-2)\vdots d$

$\Rightarrow 3\vdots d\Rightarrow d\in \left\{1; 3\right\}$
Để ps tối giản thì $n-2\not\vdots 3$

$\Leftrightarrow n\neq 3k+2$ với $k$ là số tự nhiên bất kỳ.

b.

Gọi $d=ƯCLN(n+5, n-2)$

$\Rightarrow n+5\vdots d; n-2\vdots d$

$\Rightarrow (n+5)-(n-2)\vdots d$

$\Rightarrow 7\vdots d$

$\Rightarrow d\in \left\{1; 7\right\}$

Để ps tối giản thì $n-2\not\vdots 7$

$\Rightarrow n\neq 7k+2$ với $k$ là số tự nhiên bất kỳ.

\(\left(-125\right)\cdot\left(-14\right)\cdot\left(-8\right)\cdot\left(-3\right)\)

\(=\left[\left(-125\right)\cdot\left(-8\right)\right]\cdot\left[\left(-14\right)\cdot\left(-3\right)\right]\)

\(=\left(125\cdot8\right)\cdot\left(14\cdot3\right)\)

\(=1000\cdot42\)

\(=42000\)

Lời giải:
$(8^{2017}-8^{2015}):(8^{2104}.8)=8^{2015}(8^2-1):8^{2105}$

$=\frac{8^2-1}{8^{2105-2015}=\frac{8^2-1}{8^90}}$

\(\dfrac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\dfrac{2^{10}.3^9\left(3-1\right)}{2^9.3^{10}}=\dfrac{2^{10}.3^9.2}{2^9.3^{10}}=\dfrac{2^9.2.3^9.2}{2^9.3.3^9}=\dfrac{4}{3}\)

\(\dfrac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\dfrac{2^{10}.3^9\left(3-1\right)}{2^9.3^{10}}=\dfrac{2^{10}.3^9.2}{2^9.3^{10}}=\dfrac{2^{10}.2.3^9.2}{2^9.3.3^9}=\dfrac{4}{3}\)

\(\dfrac{-11^5.13^7}{11^5.13^8}=\dfrac{-1.11^5.13^7}{11^5.13^8}=-\dfrac{1}{13}\)

\(\dfrac{-11^5\cdot13^8}{11^5\cdot13^7}\)

\(=\dfrac{-11^5\cdot13^7\cdot13}{11^5\cdot13^7}\)

\(=\dfrac{-11^5\cdot13}{11^5}\)

\(=\dfrac{11^5\cdot\left(-13\right)}{11^5}\)

\(=-13\)

\(P=2.5+5.8+8.11+...+101.104\)

\(\Rightarrow9P=2.5.9+5.8.9+8.11.9+...+101.104.9\)

\(\Rightarrow9P=2.5.9+5.8.\left(11-2\right)+8.11.\left(14-5\right)+...+101.104.\left(107-98\right)\)

\(\Rightarrow9P=2.5.9-2.5.8+5.8.11-5.8.11+8.11.14+...-98.101.104+101.104.107\)

\(\Rightarrow9P=2.5.9-2.5.8+101.104.107\)

\(\Rightarrow9P=1123938\)

\(\Rightarrow P=124882\)

Giúp em nhanh với

S = \(\dfrac{3}{1\times4}\) + \(\dfrac{3}{4\times7}\) + \(\dfrac{3}{7\times11}\) + \(\dfrac{3}{11\times14}\) + \(\dfrac{3}{14\times17}\)

S = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{4}{7.11}\) + \(\dfrac{3}{11.14}\) +\(\dfrac{3}{14.17}\) - \(\dfrac{1}{7.11}\)

S = \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{17}\) - \(\dfrac{1}{7.11}\)

S = \(\dfrac{1}{1}\) - \(\dfrac{1}{17}\) - \(\dfrac{1}{77}\)

S =  \(\dfrac{16}{17}\) - \(\dfrac{1}{77}\)

S = \(\dfrac{1215}{1309}\)

S = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.11}\) + \(\dfrac{3}{11.14}\) + \(\dfrac{3}{14.17}\)

S = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{4}{7.11}\) + \(\dfrac{3}{11.14}\) + \(\dfrac{3}{14.17}\) - \(\dfrac{1}{7.11}\)

S = \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\)  - \(\dfrac{1}{7}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{17}\) - \(\dfrac{1}{77}\)

S = \(\dfrac{1}{1}\) - \(\dfrac{1}{17}\) - \(\dfrac{1}{77}\)

S = \(\dfrac{16}{17}\) - \(\dfrac{1}{77}\)

S = \(\dfrac{1215}{1309}\)

S = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.11}\) + \(\dfrac{3}{11.14}\) + \(\dfrac{3}{14.17}\)

S = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{4}{7.11}\) + \(\dfrac{3}{11.14}\) + \(\dfrac{3}{14.17}\) - \(\dfrac{1}{7.11}\)

S = \(\dfrac{1}{1}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{17}\) - \(\dfrac{1}{77}\) 

S = 1 - \(\dfrac{1}{17}\) - \(\dfrac{1}{77}\)

S = \(\dfrac{16}{17}\) - \(\dfrac{1}{77}\)

S = \(\dfrac{1215}{1309}\)