a) \(3^5+3^4+3^3\)

\(=3^3\cdot3^2+3^3\cdot3+3^3\cdot1\)

\(=3^3\left(3^2+3+1\right)\)

\(=3^3\cdot13⋮13\)     (đpcm)

b) \(2^{10}-2^9+2^8-2^7\)

\(=2^7\cdot2^3-2^7\cdot2^2+2^7\cdot2-2^7\cdot1\)

\(=2^7\left(2^3-2^2+2-1\right)\)

\(=2^7\cdot5⋮5\)    (đpcm)

=))

\(10^{2014}=\overline{......0}\)

\(111^{100}=\overline{......1}\)

\(2^{20}=2^{4\cdot5}=\left(2^4\right)^5=\overline{....6}^5=\overline{....6}\)

\(4^{315}=4^{314}\cdot4=\left(4^2\right)^{157}\cdot4=\overline{...6}^{157}\cdot4=\overline{....6}\cdot4=\overline{....4}\)

\(7^{1995}=7^{1992}\cdot7^3=\left(7^4\right)^{498}\cdot\overline{...3}=\overline{....1}^{498}\cdot\overline{.....3}=\overline{.....3}\)

Giải : Ta có: 25 là số lẻ

=> 2x² + 7y² là số lẻ

mà 2x² luôn chẵn => 7y² là số lẻ => y² là số lẻ

=> y² = 1 (vì 7y² \(\le\)25 => y² \(\le\)25/7 \(\approx\)3}

=> y = \(\pm\)1

Với y² = 1 => 7y² = 7

=> 2x² + 7 = 25

=> 2x² = 25 - 7

=> 2x² = 18

=> x² = 9

=> x = \(\pm\)3

Vậy x = 3 hoặc x = -3 và y = 1 hoặc y = -1 (tm)

đề sai

đề đúng nha bạn

Bài 1:

\(a,\left(\frac{1}{2}\right)^{3x-1}=\frac{1}{32}\) 

\(\Rightarrow\left(\frac{1}{2}\right)^{3x-1}=\left(\frac{1}{2}\right)^5\)

\(\Rightarrow3x-1=5\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

\(b,\frac{64}{\left(-2\right)^x}=-8\)

\(\Rightarrow\left(-2\right)^x=64:\left(-8\right)\)

\(\Rightarrow\left(-2\right)^x=-8\)

\(\Rightarrow\left(-2\right)^x=\left(-2\right)^3\)

\(\Rightarrow x=3\)

\(c,\left(-2\right)^3+0,5x=1,5\)

\(\Rightarrow-8+0,5x=1,5\)

\(\Rightarrow0,5x=9,5\)

\(\Rightarrow x=19\)

Bài 2,

\(a,\frac{3}{8}.27\frac{1}{5}-51\frac{1}{5}.\frac{3}{8}+19=\frac{3}{8}\left(27\frac{1}{5}-51\frac{1}{5}\right)+19\)

\(=\frac{3}{8}\left(-24\right)+19\)

\(=-9+19=10\)

\(b,25\left(-\frac{1}{5}\right)^3+\frac{1}{5}-2.\left(-\frac{1}{2}\right)^2-\frac{1}{2}=-\frac{1}{5}+\frac{1}{5}-\left(-\frac{1}{2}\right)-\frac{1}{2}\)

\(=0\)

\(c,35\frac{1}{6}:\left(-\frac{4}{5}\right)-45\frac{1}{6}:\left(-\frac{4}{5}\right)=-\frac{4}{5}\left(35\frac{1}{6}-45\frac{1}{6}\right)\)

\(=-\frac{4}{5}.\left(-10\right)=8\)

\(d,\left(-0,25\right).8,7.2^2=\left(-0,25\right).8,7.4\)

\(=-1.8,7=-8,7\)

\(e,13\frac{1}{3}.\frac{5}{8}-23\frac{1}{3}:\frac{8}{5}=\frac{40}{3}.\frac{5}{8}-\frac{70}{3}.\frac{5}{8}\)

\(=\frac{5}{8}\left(\frac{40}{3}-\frac{70}{3}\right)\)

\(=\frac{5}{8}.-10\)

\(=-\frac{25}{4}\)

theo đề ta có :

abc=11(a+b+c)

89a=b+10c

Ta thấy \(0\le b,c\le9\)\(\Rightarrow b+10c\le99\)\(\Rightarrow a=1\)(do a khác 0)

\(\Rightarrow b=9,c=8\)

\(\Rightarrow abc=198\)

Cho a thuộc Z .Chứng tỏ 

A = a(a+1) (a+2)(a+3)(a+40) chia hết 40

\(\left|x\right|=\frac{1}{5}-\frac{1}{4}=-\frac{1}{20}\)(vô lý)

Vậy không có x thỏa mãn đề 

\(\left|x\right|=\frac{1}{5}-\frac{1}{4}\)

\(\left|x\right|=-\frac{1}{20}\)

\(\Rightarrow x\in\varnothing\)