Bai 1: Rut gon bieu thuc sau
B=(2/3)^3×(-3/4)^4×(-1)^5/(2/5)^2 × (-5/12)^3
Bai 2: Tim ia tri bieu thuc sau
M= 512- 512/2- 512/2^2- 512/2^2 - 512/2^3...512/2^10
Những câu hỏi liên quan
512 -512/2 -512/2^2 -512/2^3 -512/2^4-512/2^5-512/2^6-512/2^7-512/2^8-512/2^9-512/2^10
tính B=512-512/2-512/2^3-512/2^4-...-512/2^10
tính: A=1/6-1/39+1/51 / 1/8-1/52+1/68 B=512-512/2-512/2^2-512/2^3-...-512/2^10
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Tính :B=512 - 512/2 - 512/2^2 - 512/2^3 -.....- 512/2^10 .
\(B=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\\ =512\cdot\left(1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\right)\\ =512\cdot\left[1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\right]\)
Đặt \(H=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\Leftrightarrow B=512\cdot\left(1-H\right)\)
\(\Leftrightarrow2H=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\\ \Leftrightarrow2H-H=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\\ \Leftrightarrow H=1-\frac{1}{2^{10}}\\ \Leftrightarrow B=512\cdot\left[1-\left(1-\frac{1}{2^{10}}\right)\right]\\ \Leftrightarrow B=512\cdot\frac{1}{2^{10}}\\ \Rightarrow B=2^9\cdot\frac{1}{2^{10}}\\ \Rightarrow B=\frac{1}{2}\)
Bài 1 ;Tính :
B = \(512-\dfrac{512}{2}-\dfrac{512}{2^2}-\dfrac{512}{2^3}-...-\dfrac{512}{2^{10}}\)
B=512(1-1/2-1/2^2-1/2^3-...-1/2^10
B=512*1/1024
B=1/2
B=0.5
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tính:
\(512-\dfrac{512}{2}-\dfrac{512}{2^2}-\dfrac{512}{2^3}-...-\dfrac{512}{2^{10}}\)
Tham khảo:Câu hỏi của Nguyễn Thị Thanh Bình - Toán lớp 7 - Học trực tuyến OLM
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\(B=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\)
\(\text{M = 512 - 512/2 - .... - 512/2^10
= 2^9 - 2^9 / 2 - 2^9/2^2 - ...2^9/2^10
= 2^9 - 2^8 - 2^7 - 2^6 -.... - 1/2
2M = 2^10 - 2^9 - 2^8 - .... - 1
2M - M = 2^10 - 2^9 - 2^8 -... -1 - 2^9 + 2^8 + 2^7 +... + 1 + 1/2
M = 2^10 - 2.2^9 + 1/2
M = 2^10 - 2^10 + 1/2}\)
\(\text{ M =}\) \(\frac{1}{2}\)
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512 - 512/2 - 512/2^2 - 512/2^3 - ... - 512/2^10
tính nha
Tính : P=512-512/2-512/2^2-512/2^3-.....-512/2^10 .
Ta có:
\(P=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\)
\(\Rightarrow P=512-\left(\frac{512}{2}+\frac{512}{2^2}+\frac{512}{2^3}+...+\frac{512}{2^{10}}\right)\)
Đặt \(A=\frac{512}{2}+\frac{512}{2^2}+\frac{512}{2^3}+...+\frac{512}{2^{10}}\)
\(\Rightarrow2A=512+\frac{512}{2}+\frac{512}{2^2}+...+\frac{512}{2^9}\)
\(\Rightarrow2A-A=512-\frac{512}{2^{10}}\)
\(\Rightarrow A=512-\frac{512}{2^{10}}\)
\(\Rightarrow P=512-A=512-\left(512-\frac{512}{2^{10}}\right)=\frac{512}{2^{10}}=\frac{1}{2}\)
\(P=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\)
\(\Rightarrow P=2^9-\frac{2^9}{2}-\frac{2^9}{2^2}-\frac{2^9}{2^3}-...-\frac{2^9}{2^{10}}\)
\(\Rightarrow P=2^9-2^8-2^7-2^6-...-\frac{1}{2}\)
\(\Rightarrow2P=2^{10}-2^9-2^8-2^7-...-1\)
\(\Rightarrow2P-P=2^{10}-2^9-2^8-2^7-...-1-\left(2^9-2^8-2^7-2^6-...-\frac{1}{2}\right)\)
\(\Rightarrow2P-P=2^{10}-2^9-2^8-2^7-...-1-2^9+2^8+2^7+2^6+...+\frac{1}{2}\)
\(\Rightarrow P=2^{10}-2^9-2^9+\frac{1}{2}\)
\(\Rightarrow P=2^{10}-2.2^9+\frac{1}{2}\)
\(\Rightarrow P=2^{10}-2^{10}+\frac{1}{2}\)
\(\Rightarrow P=0+\frac{1}{2}\)
\(\Rightarrow P=\frac{1}{2}.\)
Chúc bạn học tốt!