1. 

$(5^{1986}-5^{1985}):5^{1985}=5^{1985}(5-1):5^{1985}=5-1=4$

2.

\((7^{846}+7^{847}):7^{846}=7^{846}(1+7):7^{846}=1+7=8\)

3.

\((9^{2018}-3^{4036}):6^{2006}=[(3^2)^{2018}-3^{4036}]:6^{2006}\)

$=(3^{4036}-3^{4036}):6^{2006}=0:6^{2006}=0$

4.

$(7^{80}.8^{70}-56^{70}):56^{70}$

$=[7^{10}(7.8)^{70}-56^{70}]:56^{70}$

$=[7^{10}.56^{70}-56^{70}]:56^{70}$

$=56^{70}(7^{10}-1):56^{70}=7^{10}-1$

5.

$4^{4016}:(4^{4017}-4^{4016})=4^{4016}:[4^{4016}(4-1)]$

$=4^{4016}:4^{4016}:3=1:3=\frac{1}{3}$

6.

$(12^{206}.2^{207}-24^{206}):24^{206}$

$=(12^{206}.2^{206}.2-24^{206}):24^{206}$

$=[(12.2)^{206}.2-24^{206}]:24^{206}$

$=(24^{206}.2-24^{206}):24^{206}$

$=24^{206}(2-1):24^{206}=2-1=1$

7.

$(5^2-24)^{8946}+4^{30}:2^{60}=1^{8946}+(2^2)^{30}:2^{60}$

$=1+2^{60}:2^{60}=1+1=2$

8.

$(37.8^{1007}-7.2^{3021}):8^{1007}=[37.8^{1007}-7.(2^3)^{1007}]:8^{1007}$

$=[37.8^{1007}-7.8^{1007}]:8^{1007}$

$=8^{1007}(37-7):8^{1007}=37-7=30$

P=24(7^2+1)(7^4+1)(7^8+1)(7^16+1)

=> 2P = 48(7^2+1)(7^4+1)(7^8+1)(7^16+1)

      = (7^2 - 1)(7^2+1)(7^4+1)(7^8+1)(7^16+1)

      = (7^4 - 1)(7^4+1)(7^8+1)(7^16+1)

      = (7^8 - 1)(7^8+1)(7^16+1)

      = (7^16 - 1)(7^16+1)

      = 7^32 - 1

 => P =  (7^32 - 1) / 2

Lời giải:

a.

$\frac{5}{15}-\frac{1}{6}\times \frac{2}{5}=\frac{5}{15}-\frac{1}{15}=\frac{4}{15}$

b.

$\frac{8}{24}+\frac{3}{4}:\frac{1}{8}=\frac{1}{3}+6=\frac{19}{3}$

c.

$\frac{1}{7}: \frac{2}{8}-\frac{1}{7}=\frac{1}{7}\times 4-\frac{1}{7}$

$=\frac{1}{7}\times (4-1)=\frac{1}{7}\times 3=\frac{3}{7}$

a) \(\dfrac{2}{5}\cdot\dfrac{1}{7}+\dfrac{2}{5}\cdot\dfrac{5}{7}+\dfrac{2}{5}\)

\(=\dfrac{2}{5}\left(\dfrac{1}{7}+\dfrac{5}{7}+1\right)\)

\(=\dfrac{2}{5}\cdot\dfrac{13}{7}=\dfrac{26}{35}\)

b) \(\dfrac{1}{5}+\dfrac{2}{8}+\dfrac{4}{5}+\dfrac{7}{8}-\dfrac{1}{8}\)

\(=\left(\dfrac{1}{5}+\dfrac{4}{5}\right)+\left(\dfrac{2}{8}+\dfrac{7}{8}-\dfrac{1}{8}\right)\)

\(=1+1=2\)

c)\(\dfrac{24}{36}\cdot\dfrac{10}{12}\cdot36\)

\(=\dfrac{24\cdot10\cdot36}{36\cdot12}=\dfrac{12\cdot2\cdot10\cdot36}{12\cdot36}\)

\(=2\cdot10=20\)

   \(\dfrac{2}{5}+\dfrac{2}{3}+\dfrac{2}{4}\)

= \(\dfrac{24}{60}\) + \(\dfrac{40}{60}\) + \(\dfrac{30}{60}\)

= \(\dfrac{64}{60}\) + \(\dfrac{30}{60}\)

= \(\dfrac{47}{30}\)

\(\dfrac{2}{6}+\dfrac{3}{12}\)

= \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\)

= \(\dfrac{7}{12}\)

\(\dfrac{5}{6}\) + \(\dfrac{1}{3}\)

= \(\dfrac{5}{6}\) + \(\dfrac{2}{6}\)

= \(\dfrac{7}{6}\)

   \(\dfrac{1}{3}\) + \(\dfrac{5}{12}\) + \(\dfrac{5}{6}\)

= \(\dfrac{4}{12}\) + \(\dfrac{5}{12}\) + \(\dfrac{10}{12}\)

= \(\dfrac{9}{12}\) + \(\dfrac{10}{12}\)

= \(\dfrac{19}{12}\)

    \(\dfrac{5}{8}\) + \(\dfrac{4}{7}\)

= \(\dfrac{35}{56}\) + \(\dfrac{32}{56}\)

=  \(\dfrac{67}{56}\) 

\(\dfrac{7}{3}\) + \(\dfrac{8}{7}\)

= \(\dfrac{49}{21}\) + \(\dfrac{24}{21}\)

= \(\dfrac{73}{21}\)

   \(\dfrac{1}{5}+\dfrac{5}{35}\)

= \(\dfrac{7}{35}\) + \(\dfrac{5}{35}\)

= \(\dfrac{12}{35}\) 

a: x=-5/11+2/11=-3/11

b: =>x=-3/24+20/24+1/24=18/24=3/4

c: =>5/8-x=1/9+5/4=4/36+45/36=49/36

=>x=5/8-49/36=-53/72

d: =>2/3-x=1/3

=>x=1/3

e: =>1/5:x=12/35

=>x=7/12

\(\text{a) }\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\\ =\dfrac{3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ \\ =\dfrac{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{\left(2^{16}-1\right)\left(2^{16}+1\right)}{3}\\ =\dfrac{2^{32}-1}{3}\\ \)

\(\text{b) }24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right) \\ =\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\\ =\left(5^{16}-1\right)\left(5^{16}+1\right)\\ =5^{32}-1\\ \)

\(\text{c) }48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^8-1\right)\left(7^8+1\right)\left(7^{16}+1\right)\\ =\left(7^{16}-1\right)\left(7^{16}+1\right)\\ =7^{32}-1\)

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