1/2 x 1/4 x 1/8 x 1/16 x 1/32 =?
Những câu hỏi liên quan
1/1-x +1/1+x +2/1+x^2 +4/1+x^4 +8/1+x^8 +16/1+x^16 = 32/1-x^32 c/m
\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)
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Tính:\(\frac{1}{x}+\frac{1}{x+1}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
(x^2+x+1)(x^4+x^2+1)(x^8+x^4+1)(x^16+x^8+1)(x^32+x^16+1) rút gọn zùm mình với
https://www.youtube.com/watch?v=cFZDEMTQQCs
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Thực hiện phép tính :
\(\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}+\frac{32}{1+x^{32}}\)
khó quá làm sao mà trả lời đc
tự đầu mình vắt óc mà suy nghĩ
Xem thêm câu trả lời
viết gọn biểu thức :
(x^2-x+1)(x^4-x^2+1)(x^8-x^4+1)(x^16-x^8+1)(x^32-x^16+1)
1. tinh nhanh: 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
2. tim x: ( x + 1/2 ) + ( x +1/4 ) + ( x + 1/8 ) + ( x + 1/16 ) = 1
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(2\times A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2\times A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\right)\)
\(A=1-\frac{1}{128}\)
\(A=\frac{127}{128}\)
\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(2\times B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)
\(B=1-\frac{1}{16}=\frac{15}{16}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\Leftrightarrow4\times x+\frac{15}{16}=1\)
\(\Leftrightarrow4\times x=\frac{1}{16}\)
\(\Leftrightarrow x=\frac{1}{64}\)
(x^2+x+1)(x^4+x^2+1)(x^8+x^4+1)(x^16+x^8+1)(x^32+x^16+1) rút gọn với mình tạo nick max t i c k cho
Tìm x, biết:(x+1/2)+(x+1/4)+(x+1/8)+(x+1/16)+(x+1/32)=2
Ta có:
1/2+1/4+1/8+1/16+1/32=31/32
Vì có 5 tổng=> ta gọi phần còn lại là 5X
=>5X+31/32=2
=>5X+31/32=64/32
=>5X=64-32-31/32
=>5X=33/32
=>X=33/32:5
=>X=33/160
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64/32 chứ không phải là 64-32 đâu nha
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Chứng minh đẳng thức
\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}=\dfrac{32}{1-x^{32}}\)
Các bạn giúp mk nha
Đề sai nha bạn 
mình sửa luôn
\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2\left(1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4\left(1+x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\dfrac{4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8\left(1+x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16\left(1+x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}+\dfrac{16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\dfrac{16+16x^{16}+16-16x^{16}}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\dfrac{32}{1-x^{32}}=VP\left(đpcm\right)\)
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Rút gọn biểu thức:
A) x – ( x/2 + 1/2 )
B) (x/2 – 1/2) : 2 + 1/2
C) (x/2 – 1/2) – (x/4 + 1/4)
D) (x/4 – 3/4) : 2 + 1/2
E) (x/4 – 3/4) – (x/8 + 1/8)
F) (x/8 – 7/8) : 2 + 1/2
G) (x/8 – 7/8) – (x/16 + 1/16)
H) (x/16 – 15/16) : 2 + 1/2
I) (x/16 – 15/16) – (x/32 + 1/32)
