Đặt \(A=2^{49}+2^{48}+\cdots+2^2+2\)

=>\(2A=2^{50}+2^{49}+\cdots+2^3+2^2\)

=>2A-A=\(2^{50}+2^{49}+\cdots+2^3+2^2-2^{49}-2^{48}-\cdots-2^2-2\)

=>\(A=2^{50}-2\)

Ta có: \(2^{50}-2^{49}-2^{48}-\cdots-2^2-2\)

\(=2^{50}-\left(2^{50}-2\right)\)

\(=2^{50}-2^{50}+2=2\)

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

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 \(A⋮6\)

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

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

hbewjfewi

Câu 3 = (5 mũ 51 - 1) : 4

Lời giải:
a. $x^3=4^3\Rightarrow x=4$

b. $x^2=49=7^2=(-7)^2$

$\Rightarrow x=7$ hoặc $x=-7$

c. $x^3+1=28$

$x^3=28-1=27=3^3$

$\Rightarrow x=3$

d. $2^x=16=2^4$

$\Rightarrow x=4$

e. $2^4.2^x=2^6$

$\Rightarrow 2^{4+x}=2^6$

$\Rightarrow 4+x=6$

$\Rightarrow x=2$

g.

$5^x=25.5^3=5^2.5^3=5^5$

$\Rightarrow x=5$

Lần sau bạn lưu ý viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để đề được rõ ràng hơn nhé.