a) Ta có :  2 x : 2 2 = 2 5  nên x = 7.

b) Ta có:  3 x : 3 2 = 3 5  nên x = 7.

c) Ta có :  4 4 : 4 x = 4 2  nên x = 2.

d) Ta có :  5 x : 5 2 = 5 2  nên x = 4,

e) Ta có:  5 x + 1 : 5 = 5 4  nên x = 4.

f) Ta có :  4 2 x - 1 : 4 = 4 2  nên x = 2

64\(^x\) : 16\(^x\) = 256

(64: 16)^\(x\)  = 256

        4\(^x\)  = 4⁴

          \(x\) = 4

Vậy \(x\) = 4

128 x 68 + 16 x 256 

= 128 x 68 + 16 x 2 x 128

= 128 x 68 + 32 x 128

= 128 x (68+32)

= 128 x 100 = 12 800

Chọn (C) 16.

2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 

= 2 x 1 + 2 x 2 + 2 x 4 + 2 x 8 + 2 x 16 + 2 x 32 + 2 x 64 + 2 x 128 + 2 x 256 

= 2 x ( 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 ) 

= 2 x 511 

= 1022 

2 + 4 + 8 + 16 + 1022 + 256 + 512

2 + 4 + 8 + 16 ________ + 256 + 512

Đáp án: 1022

1 + 1 = 2

2 + 2 = 4

4 + 4 = 8

8 + 8 = 16

16 + 16 = 32

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

512 + 512 = 1024

1024 + 1024 = 2048

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=265

256+256=512

512+512=1024

1024+1024=2048

Học tốt ^_^

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=256

256+256=512

512+512=1024

1024+1024=2048

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=256

256+256=512

512+512=1024

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=156

256+256=512

512+512=1024

Lần lượt nhe:2

                    4

                    8

                    16

                    32

                    128

                    256

                    512

                    1024

các tổng này đều = các số hạng của phép tính trên đúng ko

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128 = 256

256+256=512

512+512= 1024

1024+1024 = 2048 

2048 + 2048 = 4096

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=256

256+256=512

512+512=1024

1024+1024=2048

2048+2048=4096

1+1=2

2+2=4

4+4=8

8+8=16

16+16=32

32+32=64

64+64=128

128+128=256

256+256=512

512+512=1024

1024+1024=2048

2048+2048=4096

a, \(\frac{7}{4x}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=22\)

\(\frac{7}{4x}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=22\)

\(\frac{7}{4x}\left[33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]=22\)

\(\frac{7}{4x}\left[33.\left(\frac{35}{420}+\frac{21}{420}+\frac{14}{420}+\frac{10}{420}\right)\right]=22\)

\(\frac{7}{4x}\left[33.\frac{4}{21}\right]=22\)

\(\frac{7}{4x}.\frac{44}{7}\)=22

\(\frac{11}{x}=22\)

x=11:22

x=\(\frac{1}{2}\)

b,\(\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right).x=1\)

Đặt A\(=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

Ta có :\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(\Rightarrow4A=4.\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right)\)

\(\Rightarrow4A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{64}+\frac{1}{64}\)

\(\Rightarrow4A=\frac{32+16+8+4+2+1}{64}=\frac{63}{64}\)

\(\Rightarrow A=\frac{63}{64}:4=\frac{63}{256}\)

\(\Rightarrow\frac{63}{256}.x=1\)

\(\Leftrightarrow x=1:\frac{63}{256}=\frac{256}{63}\)