\(M=4.6\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)\(=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

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

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

Vậy M <N

3: =(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)

=(5^4-1)(5^4+1)(5^8+1)(5^16+1)

=(5^8-1)(5^8+1)(5^16+1)

=(5^16-1)(5^16+1)

=5^32-1

4:

D=(4^4-1)(4^4+1)(4^8+1)*....*(4^64+1)

=(4^8-1)(4^8+1)*...*(4^64+1)

=...

=4^128-1

5: =(5^2-1)(5^2+1)(5^4+1)*...*(5^128+1)+(5^256-1)

=(5^4-1)(5^4+1)*...*(5^128+1)+5^256-1

=5^256-1+5^256-1

=2*5^256-2

3, \(C=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)

\(C=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)

\(C=\left(5^4-1\right)\left(5^4+1\right)....\left(5^{16}+1\right)\)

\(C=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(C=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(C=5^{32}-1\)

4, \(D=15\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^4-1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^8-1\right)\left(4^8+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^{16}-1\right)\left(4^{16}+1\right)...\left(4^{64}+1\right)\)

\(D=\left(4^{32}-1\right)\left(4^{32}+1\right)\left(4^{64}+1\right)\)

\(D=\left(4^{64}-1\right)\left(4^{64}+1\right)\)

\(D=4^{128}-1\)

5, \(E=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{256}+1\right)\)

\(E=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{128}+1\right)\left(5^{256}+1\right)\)

\(E=\left(5^4-1\right)\left(5^4+1\right)....\left(5^{256}+1\right)\)

....

\(E=\left(5^{128}-1\right)\left(5^{128}+1\right)\left(5^{256}+1\right)\)

\(E=\left(5^{256}-1\right)\left(5^{256}+1\right)\)

\(E=5^{512}-1\)

\(B=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

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

\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(=5^{32}-1< 5^{32}\)

Vậy \(B< A\)

\(1+2+...+n=\frac{n\left(n+1\right)}{2}\)

\(\Rightarrow E=1+\frac{1}{2}\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{200}.\frac{200.201}{2}\)

\(=1+\frac{1}{2}\left(3+4+5+...+201\right)\)

\(=1+\frac{1}{2}\left(1+2+3+...+201-1-2\right)\)

\(=1+\frac{1}{2}\left(\frac{201.202}{2}-3\right)=10150\)

\(\frac{21}{5}\left|x\right|< 2019\Rightarrow\left|x\right|< 2019\div\frac{21}{5}=\frac{3365}{7}\)

\(\Rightarrow-480\le x\le480\)

\(\Rightarrow\sum x=-480+480-479+479+...+-1+1+0=0\)

\(\frac{2^{24}\left(x-3\right)}{\frac{81}{35}.\left(6.2^{24}-2^{26}\right)}=\frac{25}{9}\)

\(\Leftrightarrow\frac{2^{24}\left(x-3\right)}{2^{24}\left(6-2^2\right)}=\frac{25}{9}.\frac{81}{35}\)

\(\Leftrightarrow\frac{x-3}{2}=\frac{45}{7}\)

\(\Leftrightarrow x-3=\frac{90}{7}\)

\(\Rightarrow x=\frac{111}{7}\)

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{15}+1\right)\)

\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(\frac{1}{2}\left(5^{32}+1\right)=\frac{5^{32}+1}{2}\)

a)

 Ta có

a chia 5 dư 4

=> a=5k+4 ( k là số tự nhiên )

\(\Rightarrow a^2=\left(5k+4\right)^2=25k^2+40k+16\)

Vì 25k^2 chia hết cho 5

    40k chia hết cho 5

    16 chia 5 dư 1

=> đpcm

2) Ta có

\(12=\frac{5^2-1}{2}\)

Thay vào biểu thức ta có

\(P=\frac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)}{2}\)

\(\Rightarrow P=\frac{\left[\left(5^2\right)^2-1^2\right]\left[\left(5^2\right)^2+1^2\right]\left(5^8+1\right)}{2}\)

\(\Rightarrow P=\frac{\left[\left(5^4\right)^2-1^2\right]\left[\left(5^4\right)^2+1^2\right]}{2}\)

\(\Rightarrow P=\frac{5^{16}-1}{2}\)

3)

\(\left(a+b+c\right)^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)

\(=a^3+b^3+c^2+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+cb+c^2\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

a)\(\left(4-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=\left(\dfrac{4}{1}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=\left(\dfrac{20}{5}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=\dfrac{8}{5}.\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=1-\dfrac{2}{5}.\dfrac{25}{-4}\)

\(=1-\dfrac{-5}{2}\)

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

\(=\dfrac{7}{2}\)

dài quá nên mik sẽ giải lần lượt mỗi câu trả lời là một câu nhá bạn!!

Giải:

a)(4-12/5).25/8-2/5:-4/25

=8/5.25/8-(-5/2)

=5+5/2

=15/2

b)(-5/24+3/4-7/12):(-5/16)

=-1/24:(-5/16)

=2/15

c)6/7+5/4:(-5)-(-1/28).(-2)²

=6/7+(-1/4)-(-1/28).4

=6/7-1/4-(-1/7)

=6/7-1/4+1/7

=(6/7+1/7)-1/4

=1-1/4

=3/4

Chúc bạn học tốt!

\(A=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(A=\dfrac{24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)

\(A=\dfrac{5^{32}-1}{2}\)

\(A=\left|2x+4\right|+\left|2x+6\right|+\left|2x+8\right|\)

\(A=\left|2x+4\right|+\left|2x+8\right|+\left|2x+6\right|\)

\(A=\left|2x+4\right|+\left|-2x-8\right|+\left|2x+6\right|\)

\(A\ge\left|2x+4-2x-8\right|+\left|2x+6\right|\)

\(A\ge4+\left|2x+6\right|\)

Vì \(\left|2x+6\right|\ge0\) nên \(A\ge4\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}2x+4\le0\\2x+6=0\\2x+8\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x\le-4\\2x=-6\\2x\ge-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-2\\x=-3\\x\ge-4\end{matrix}\right.\)

Vậy \(x=-3\)

a) (2n-1)⁴ : (2n-1) = 27

(2n-1)³ = 27  =3³

=> 2n - 1= 3

=> 2n = 4

n = 2

phần b,c làm tương tự nha bn

d) (21+n) : 9 = 9⁵:9⁴

(2n+1) : 9 = 9

2n + 1 = 81

2n = 80

n = 40

Tìm số tự nhiên n, biết :

a/ (2.n−1)4:(2.n−1)=27

\(\left(2.n-1\right)^3=27\)

\(2.n-1=3^3\Rightarrow2.n-1=3\)

2.n - 1 = 3

2.n      = 3 + 1

   n      =  4 : 2

   n      =  2

B,C tương tự nha

d) \(\left(21+n\right):9=9^5:9^4\)

   \(\left(21+n\right):9=9\)

        \(21+n=9.9\)

         \(21+n=81\)

                 \(n=81-21\)

                  \(n=60\)