Đặt N=2^49+2^48+...+2^3+2^2+2

=>2N=2^50+2^49+...+2^4+2^3+2^2

=>2N-N=2^50-2

=>N=2^50-2

=>M=2^50-2^50+2=2

a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.\left(5^2\right)^4}=7.\frac{5^8}{5^8}=7\)

b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3.3^8.5^2.5^3}{3.5.5^4.3^8}=\frac{5^5}{5^5}=1\)

c) Đề hơi sai roi bạn oi

d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2=\frac{1}{100}+\frac{121}{100}=\frac{61}{50}\)

Khanh Nguyễn Ngọc  :câu d ko sai bạn nha dấu "/" là trên nhó

a:

Sửa đề: \(B=\frac{5^{99}+1}{5^{100}+1}\)

Ta có: \(5A=\frac{5^{50}+5}{5^{50}+1}=\frac{5^{50}+1+4}{5^{50}+1}=1+\frac{4}{5^{50}+1}\)

\(5B=\frac{5^{100}+5}{5^{100}+1}=\frac{5^{100}+1+4}{5^{100}+1}=1+\frac{4}{5^{100}+1}\)

Ta có: \(5^{50}+1<5^{100}+1\)

=>\(\frac{4}{5^{50}+1}>\frac{4}{5^{100}+1}\)

=>\(\frac{4}{5^{50}+1}+1>\frac{4}{5^{100}+1}+1\)

=>5A>5B

=>A>B

b: \(\frac{A}{3}=\frac{3^{49}-5}{3^{49}-15}=\frac{3^{49}-15+10}{3^{49}-15}=1+\frac{10}{3^{49}-15}\)

\(\frac{B}{3}=\frac{3^{50}-5}{3^{50}-15}=\frac{3^{50}-15+10}{3^{50}-15}=1+\frac{10}{3^{50}-15}\)

Ta có: \(3^{49}-15<3^{50}-15\)

=>\(\frac{10}{3^{49}-15}>\frac{10}{3^{50}-15}\)

=>\(\frac{10}{3^{49}-15}+1>\frac{10}{3^{50}-15}+1\)

=>\(\frac{A}{3}>\frac{B}{3}\)

=>A>B

d, D = 40² - 28² + 32² +80.32

    D = (40² + 2.40.32 + 32²) - 28²

    D = (40 + 32)² - 28²

    D = (40 + 32 - 28)(40 + 32 + 28)

    D = 44.100

   D = 4400

  e, E = 10.80,5 + 10.19,5 - 8.20,5 - 8. 79,5

      E =  10.(80,5 + 19,5) - 8.( 20,5 + 79,5)

      E = 10.100 - 8.100

     E = 100.(10-8)

     E = 200

F = 50² - 18² + 32² + 100.32

F = (50² - 18²) + 32.( 32 + 100)

F = (50 -18)(50+18) + 32. 132

F = 32.68 + 32.132

F = 32.( 68 + 132)

F = 32. 200

F = 6400

Đặt \(A=5+5^2+5^3+5^4+...+5^{49}+5^{50}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{49}+5^{50}\right)\)

\(=5.\left(1+5\right)+5^3.\left(1+5\right)+...+5^{49}.\left(1+5\right)\)

\(=5.6+5^3.6+...+5^{49}.6\)

\(=6.\left(5+5^3+...+5^{49}\right)⋮6\)

Vậy \(A⋮6\)

Bài 1:

a: \(10^{10}=\left(2\cdot5\right)^{10}=2^{10}\cdot5^{10}=2^9\cdot5^{10}\cdot2\)

\(48\cdot50^5=2^4\cdot3\cdot\left(2\cdot5^2\right)^5=2^4\cdot3\cdot2^5\cdot5^{10}=2^9\cdot5^{10}\cdot3\)

mà 2<3

nên \(10^{10}<48\cdot50^5\)

b: \(1990^{10}+1990^9=1990^9\left(1990+1\right)=1990^9\cdot1991\)

\(1991^{10}=1991^9\cdot1991\)

mà 1990<1991

nên \(1990^{10}+1990^9<1991^{10}\)

c: \(107^{50}<108^{50}=\left(2^2\cdot3^3\right)^{50}=2^{100}\cdot3^{150}\)

\(73^{75}>72^{75}=\left(2^3\cdot3^2\right)^{75}=2^{225}\cdot3^{150}\)

mà \(2^{225}\cdot3^{150}>2^{100}\cdot3^{150}=108^{50}>107^{50}\)

nên \(73^{75}>107^{50}\)

d: \(2^{91}=\left(2^{13}\right)^7=8192^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\)

mà 8192>3125

nên \(2^{91}>5^{35}\)

e: \(A=72^{45}-72^{44}=72^{44}\left(72-1\right)=72^{44}\cdot71\)

\(B=72^{44}-72^{43}=72^{43}\left(72-1\right)=72^{43}\cdot71\)

mà 44>43

nên A>B

Bài 2:

a:

ĐKXĐ: x<>2023

\(\frac{x-2023}{4}=\frac{1}{x-2023}\)

=>\(\left(x-2023\right)\left(x-2023\right)=4\cdot1\)

=>\(\left(x-2023\right)^2=4\)

=>\(\left[\begin{array}{l}x-2023=2\\ x-2023=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2+2023=2025\left(nhận\right)\\ x=-2+2023=2021\left(nhận\right)\end{array}\right.\)

b: \(\left(2x+1\right)^4=\left(2x+1\right)^6\)

=>\(\left(2x+1\right)^6-\left(2x+1\right)^4=0\)

=>\(\left(2x+1\right)^4\cdot\left\lbrack\left(2x+1\right)^2-1\right\rbrack=0\)

=>\(\left(2x+1\right)^4\cdot\left(2x+1-1\right)\left(2x+1+1\right)=0\)

=>\(2x\left(2x+1\right)^4\cdot\left(2x+2\right)=0\)

=>\(\left[\begin{array}{l}2x=0\\ 2x+1=0\\ 2x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-\frac12\\ x=-1\end{array}\right.\)

c: \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)

=>\(\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\)

=>\(\left(3x-1\right)^{10}\cdot\left\lbrack\left(3x-1\right)^{10}-1\right\rbrack=0\)

=>\(\left[\begin{array}{l}\left(3x-1\right)^{10}=0\\ \left(3x-1\right)^{10}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x-1=0\\ \left(3x-1\right)^{10}=1\end{array}\right.\)

=>\(\left[\begin{array}{l}3x-1=0\\ 3x-1=1\\ 3x-1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\\ x=\frac23\\ x=0\end{array}\right.\)

d: Sửa đề \(2^{x+1}\cdot3^{y}=12^{x}\)

=>\(2^{x+1}\cdot3^{y}=\left(2^2\cdot3\right)^{x}=2^{2x}\cdot3^{x}\)

=>\(\begin{cases}2x=x+1\\ y=x\end{cases}\Rightarrow\begin{cases}x=1\\ y=x=1\end{cases}\)

A=(1+3^2)+(3^4+3^6)+...+(3^48+3^50)

A=1(1+3^2)+3^4(1+3^2)+...+3^48(1+3^2)

A=1.10+3^4.10+...+3^48.10

A=10(1+3^4+...+3^48)

A=2.5(1+3^4+...+3^48)

=>A chia hết cho 2 và 5 nên 8.A cũng chia hết cho 2 và 5