a) ta có : \(5^5-5^4+5^3=5^3.\left(5^2-5+1\right)=5^3.\left(25-5+1\right)\)

\(5^3.21=5^3.3.7⋮7\) (đpcm)

b) ta có : \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.\left(49+7-1\right)\)

\(=7^4.55=7^4.5.11⋮11\) (đpcm)

c) ta có : \(3^{x+2}-2^{x+3}+3^x-2^{x+1}=3^{x+2}+3^x-2^{x+3}-2^{x+1}\)

\(=3^x\left(3^2+1\right)-2^x\left(2^3+2\right)=3^x.\left(9+1\right)-2^x.\left(8+2\right)\)

\(=3^x.10-2^x.10=10\left(3^x-2^x\right)⋮10\) (đpcm)

d) \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}=3^x.\left(3^3+3\right)+2^x.\left(2^3+2^2\right)\)

\(=3^x.\left(27+3\right)+2^x\left(8+4\right)=3^x.30+2^x.12=6.\left(3^x.5+2^x.2\right)⋮6\) (đpcm)

a)Ta có:\(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21\)(vì 21 chia hết cho 7)

\(\)\(\RightarrowĐPCM\)

b)Ta có: \(7^6+7^5-7^4⋮11=7^4\left(7^2+7-1\right)=7^4.55⋮11\)

\(\Rightarrowđpcm\)

a.\(12,5.\left(-\dfrac{5}{7}\right)+1,5.\left(-\dfrac{5}{7}\right)\)

\(=\left(-\dfrac{5}{7}\right).\left(12,5+1,5\right)\)

\(=-10\)

b,\(\left(-\dfrac{2}{5}-\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)

\(=\left(-\dfrac{2}{5}-\dfrac{3}{7}-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)

\(=-\dfrac{3}{5}:\dfrac{4}{5}\)

\(=-\dfrac{3}{4}\)

c,\(12.\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\)

\(=12.\dfrac{4}{9}+\dfrac{4}{3}\)

\(=\dfrac{16}{3}+\dfrac{4}{3}\)

\(=\dfrac{20}{3}\)

d,\(1:\left(\dfrac{2}{3}-\dfrac{3}{4}\right)^2\)

\(=\dfrac{1}{1}:\dfrac{1}{144}\)

\(=144\)

e,\(15.\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)

\(=15.\dfrac{4}{9}-\dfrac{7}{3}\)

\(=\dfrac{20}{3}-\dfrac{7}{3}\)

\(=\dfrac{13}{3}\)

a) = ( 12,5 +1,5 ). \(\left(-\dfrac{5}{7}\right)\)

= 14 . \(\left(-\dfrac{5}{7}\right)\)

= -10

b) = (\(-\dfrac{2}{5}+-\dfrac{1}{5}\)) + \(\left(\dfrac{3}{7}-\dfrac{3}{7}\right)\): \(\dfrac{4}{5}\)

= \(\left(-\dfrac{3}{5}+0\right)\): \(\dfrac{4}{5}\)

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

c) = \(\left(12.-\dfrac{2}{9}\right)\) + \(\dfrac{4}{3}\)

= \(\dfrac{8}{3}\) + \(\dfrac{4}{3}\)

= \(-\dfrac{4}{3}\)

d) = 1: \(\dfrac{23}{48}\)

=\(\dfrac{48}{23}\)

e) =\(\left(15.-\dfrac{2}{9}\right)-\dfrac{7}{3}\)

= \(\left(-\dfrac{10}{3}\right)-\dfrac{7}{3}\)

=\(-\dfrac{17}{3}\)

f) = 10 485.76

Ta có \(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21=5^3.3.7\)

Vì 5³.3 là số nguyên nên \(5^3.3.7⋮7\)

Vậy  \(5^5-5^4+5^3⋮7\)

c) \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}\)

\(=\left(3^{x+3}+3^{x+1}\right)+\left(2^{x+3}+2^{x+2}\right)\)

\(=3^x\left(3^2+3\right)+2^x\left(2^2+2\right)\)

\(=3^x.12+2^x.6\)

\(=6\left(2.3^x+2^x\right)\)

Vì \(2.3^x+2^x\in Z\)

Nên : \(6\left(2.3^x+2^x\right)⋮6\)

Vậy \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}⋮6\)

a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)

= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)

= \(\frac{17}{9}-\frac{2}{3}\)

= \(\frac{11}{9}\)

b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)

= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)

= \(\frac{2}{5}.\frac{7}{12}\)

= \(\frac{7}{30}\)

Mình lười làm quá, hay mình nói kết quả cho bn thôi nha

c) -6

d) 3

e) 3

g) 12

h) \(\frac{23}{18}\)

i) \(\frac{-69}{20}\)

k) \(\frac{-1}{2}\)

l) \(\frac{49}{5}\)

1: \(=\dfrac{3}{4}+\dfrac{5}{4}\cdot\dfrac{8}{3}-\dfrac{1}{4}\cdot\dfrac{5}{6}=\dfrac{3}{4}+\dfrac{10}{3}-\dfrac{5}{24}\)

\(=\dfrac{18}{24}+\dfrac{80}{24}-\dfrac{5}{24}=\dfrac{93}{24}=\dfrac{31}{8}\)

2: \(=\left(7+\dfrac{23}{27}-\dfrac{23}{27}\right)+\left(\dfrac{11}{25}+\dfrac{14}{25}\right)+3.25\)

\(=7+1+3.25=8+3.25=11.25\)

3: \(=\left(\dfrac{1}{9}\cdot9\right)^{2005}-4^2=1-16=-15\)

4: \(=2\cdot\dfrac{9}{4}-\dfrac{7}{2}=\dfrac{9}{2}-\dfrac{7}{2}=1\)

5: \(=\dfrac{15}{2}\cdot\dfrac{-3}{5}+\dfrac{5}{2}\cdot\dfrac{-3}{5}=\dfrac{-3}{5}\cdot\left(\dfrac{15}{2}+\dfrac{5}{2}\right)=\dfrac{-3}{5}\cdot10=-6\)

6: \(=\left(\dfrac{6}{10}+\dfrac{5}{10}\right)^2=\left(\dfrac{11}{10}\right)^2=\dfrac{121}{100}\)

7: \(=\dfrac{1}{2}\cdot\dfrac{-7}{2}=\dfrac{-7}{4}\)

\(a)\dfrac{4}{5}-\left(\dfrac{-2}{7}\right)-\dfrac{7}{10}\)

\(=\dfrac{56}{70}+\dfrac{20}{70}+\dfrac{-49}{70}\)

\(=\dfrac{27}{70}\)

\(b)\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

\(=\left(\dfrac{36}{6}+\dfrac{-4}{6}+\dfrac{3}{6}\right)-\left(\dfrac{30}{6}+\dfrac{10}{6}+\dfrac{-9}{6}\right)-\left(\dfrac{18}{6}+\dfrac{-14}{6}+\dfrac{15}{6}\right)\)

\(=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)

\(=\dfrac{35+\left(-31\right)+\left(-19\right)}{6}\)

\(=-\dfrac{15}{6}=-2\dfrac{1}{2}\)

bài dễ thấy móa lun

a,

\(\dfrac{2}{3}:\dfrac{-5}{3}+\dfrac{7}{5}=\dfrac{2}{3}\cdot\dfrac{-3}{5}+\dfrac{7}{5}=-\dfrac{2}{5}+\dfrac{7}{5}=\dfrac{5}{5}=1\)

b,

\(3^7:3^5+\left|-2\right|\cdot5^2=3^2+2\cdot25=9+50=59\)

a) \(\dfrac{2}{3}:\dfrac{-5}{3}+\dfrac{7}{5}\)

\(=\dfrac{2}{3}.\dfrac{3}{-5}+\dfrac{7}{5}\)

\(=\dfrac{-2}{5}+\dfrac{7}{5}\)

\(=\dfrac{5}{5}=1\)

b) \(3^7:3^5+\left|-2\right|.5^2\)

\(=3^{7-5}+2.25\)

\(=3^2+50\)

\(=59\)

1)\(\frac{1}{7}\).(\(\frac{1}{3}\)+\(\frac{1}{2}\)-1)

=\(\frac{1}{7}\).\(\frac{-1}{6}\)

=\(\frac{-1}{42}\)

2)\(\frac{3}{5}\).(\(\frac{7}{9}\)+\(\frac{2}{9}\)+1)

=\(\frac{3}{5}\).2=\(\frac{6}{5}\)

3)=21.\(\frac{1}{7}\)-21.\(\frac{1}{5}\)+21.\(\frac{19}{21}\)

=3-\(\frac{21}{5}\)+19

=\(\frac{89}{5}\)

cảm ơn bạn nhé

a) \(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55⋮55\)

d) \(81^7-27^9-9^{13}\)

\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)

\(=3^{28}-3^{27}-3^{26}\)

\(=3^{26}\left(3^2-3-1\right)=3^{226}.5=3^{222}.405⋮405\)