Bài 2: 

Ta có: \(\frac{x+1}{x}=10\) hay \(\frac{x^1+1^1}{x^1}=10^1\)

Nên suy ra : \(\frac{x^5+1}{x^5}=10^5\)

                                 = 100000 ( do 1⁵ cũng sẽ =1 nên không viết mũ 5 cũng chả sao)

Ta có:\(A=\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)

\(\Rightarrow3A=\left(4-1\right)\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)

\(\Rightarrow3A=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)

\(\Rightarrow3A=\left(4^4-1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)

\(\Rightarrow3A=\left(4^8-1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)

\(\Rightarrow3A=\left(4^{16}-1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)

\(\Rightarrow3A=\left(4^{32}-1\right)\left(4^{32}+1\right)\)

\(\Rightarrow3A=4^{64}-1\)

\(\Rightarrow3A=B\)

xin lỗi mk làm sai kết quả là 254

1+1+2+2+4+4+8+8+16+16+32+32+64+64=254

290 nha bn.

2

4

8

16

32

64

128

256

2

4

8

16

64

128

256

          k mk nhé

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=256

tk m nhé

Tính không quy đồng mẫu:

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{32}-\frac{1}{64}\)

\(A=1-\frac{1}{64}=\frac{63}{64}\)

A = 32 + 16 + 8 + 4 + 2 + 1/64 = 63/64

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