Tham khảo:Câu hỏi của Nguyễn Thị Thanh Bình - Toán lớp 7 - Học trực tuyến OLM

\(B=512-\dfrac{512}{2}-\dfrac{512}{2^2}-\dfrac{512}{2^3}-...-\dfrac{512}{2^{10}}\)

\(B=512-512\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+..+\dfrac{1}{2^{10}}\right)\)

Đặt: \(L=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(2L=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(2L-L=1-\dfrac{1}{2^{10}}\Leftrightarrow L=1-\dfrac{1}{2^{10}}\)

Thay Vào B

\(B=512-512\left(1-\dfrac{1}{2^{10}}\right)=512-512+\dfrac{512}{2^{10}}=\dfrac{1}{2}\)

B=512(1-1/2-1/2^2-1/2^3-...-1/2^10

B=512*1/1024

B=1/2

B=0.5

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}+\dfrac{1}{512}\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}\)

\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}+\dfrac{1}{256}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-...-\dfrac{1}{256}-\dfrac{1}{512}\)

\(\Rightarrow A=1-\dfrac{1}{512}=\dfrac{511}{512}\)

Đặt ⇒2A=1+12+14+...+1128+1256⇒2A=1+12+14+...+1128+1256

⇒A=1−1512=511512

Lời giải:

Sửa lại đề:

\(A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-...-\frac{1}{512}\)

\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)

\(2A=2-\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)

Trừ theo vế:

\(A=2A-A=\frac{1}{2^9}< 0,002\) (đpcm)

Đặt A=1/2+1/4+1/8+..+1/1024

Ax2=1+1/2+1/4+1/8+..+1/512( Nhân cả 2 vế với 2)

Ax2-A=(1+1/2+1/4+1/8+..+1/512)-(1/2+1/4+1/8+..+1/1024)

<=>A=1-1/1024

<=>A=1023/1024

Vậy biểu thức đã cho = 1023/1024

B=(2/5)⁷.5⁷+(9/4)³:(3/16)³/2⁷.5²+512

B=(2/5.5)⁷+(9/4:3/16)³/2⁷.25+512

B=2⁷+12³/2⁷.25+512

B=1856/3200+512

B=1856/3712

B=1/2

Sửa đề: (2/7)^7*7^7

\(A=\dfrac{\left(2\right)^7+\left(\dfrac{9}{3}:\dfrac{3}{16}\right)^3}{2^7\left(5^2+2^2\right)}\)

\(=\dfrac{\left(2\right)^7+\left(16\right)^3}{2^7\cdot29}\)

\(=\dfrac{2^7+2^7\cdot2^5}{2^7\cdot29}=\dfrac{1+2^5}{29}=\dfrac{33}{29}\)