tìm n biết: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.....\frac{30}{64}=2^n\)
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Tìm X
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot\frac{5}{12}\cdot.......\cdot\frac{30}{62}\cdot\frac{31}{64}=4^x\)
1/4.2/6.3/8.4/10.........30/62.31/64=4x
=1/2.1/2.1/2.1/2.............1/2.1/64=4^x
=1/2^30.1/2^6=4^x
=1/2^36=4^x
=1/4^18=4^x
=>x=-18
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1. Tìm x, biết:
a) \(\left(x-\frac{3}{4}\right)^2=0\)
b) \(\left(x+\frac{1}{2}\right)^2=\frac{9}{64}\)
c) \(\frac{\left(-2\right)^x}{16}=-8\)
2. Tính:
\(\frac{6^3.12^4}{4^7.9^5}\)
3. Tìm \(x\in N\), biết:
\(3^2.81\le3^x\le27.243\)
1. Tìm x, biết :
a. ( x - \(\frac{3}{4}\)) \(^2\)= 0
=> x - \(\frac{3}{4}\)= 0
=> x = 0 + \(\frac{3}{4}\)
=> x = \(\frac{3}{4}\)
b. ( x + \(\frac{1}{2}\)) \(^2\)= \(\frac{9}{64}\)
=> ( x + \(\frac{1}{2}\)) \(^2\)= ( \(\frac{3}{8}\)) \(^2\)
=> x + \(\frac{1}{2}\)= \(\frac{3}{8}\)
=> x = \(\frac{3}{8}\)- \(\frac{1}{2}\)
=> x = \(\frac{-1}{8}\)
c. \(\frac{\left(-2\right)^x}{16}=-8\)
=> \(\frac{\left(-2\right)^x}{16}=\frac{-8}{1}=\frac{-128}{16}\)
=> ( -2)\(^x\)= -128
=> ( -2 ) \(^x\)= ( -2) \(^7\)
=> x = 7
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Bài 1.So Sánh
a,frac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{10^2}+frac{1}{12^2} và frac{1}{2}
b,frac{1}{2^2}+frac{1}{4^2}+frac{1}{6^2}+frac{1}{8^2}+frac{1}{10^2}+frac{1}{12^2}và frac{1}{2}
Bài 2: a,Tìm n để frac{3n+1}{n+1}
là 1 số nguyên
b,(n+1)^n 64 (n thuộc Z)
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Bài 1.So Sánh
a,\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}+\frac{1}{12^2} và \frac{1}{2}\)
b,\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}và \)\(\frac{1}{2}\)
Bài 2: a,Tìm n để \(\frac{3n+1}{n+1} \)là 1 số nguyên
b,\((n+1)^n\)= 64 (n thuộc Z)
\(\frac{1}{4}\times\frac{2}{6}\times\frac{3}{8}\times\frac{4}{10}\times\frac{5}{12}.....\frac{30}{62}\times\frac{31}{64}=2^x\)
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}.....\frac{30}{62}.\frac{31}{64}=2^x\)
=>\(\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}....\frac{30}{2.31}.\frac{31}{2.32}=2^x\)
=>\(\frac{1.2.3.4.5....30.31}{2.2.2.3.2.4.2.5.2.6...2.31.2.32}=2^x\)
=>\(\frac{2.3.4.5...30.31}{2^{31}.32.\left(2.3.4.5...31\right)}=2^x\)
=>\(\frac{1}{2^{31}.2^5}=2^x\)
=>\(\frac{1}{2^{36}}=2^x\)
=> x=36
Vậy x=36
Chúc bn học tốt nhé!
Bài 1: Tìm số nguyên n để phân số M=\(\frac{2n-7}{5n}\)có giá trị là số nguyên.
Bài 2: Tìm x biết :
a) / x - 3 /=2.(x + 2)
b) \(1\frac{1}{3}\div(24\frac{1}{6}-24\frac{1}{5})-\frac{1\frac{1}{2}-\frac{3}{4}}{4x-\frac{1}{2}}=-1\frac{1}{15}\div(8\frac{1}{5}-8\frac{1}{3})\)
Giúp mình với.
\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot....\cdot\frac{30}{62}\cdot\frac{31}{64}\)
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}\)
\(=\frac{1.2.3.4...30.31}{2.2.2.3.2.4.2.5...2.31.2.32}\)
\(=\frac{1.2.3.4...30.31}{2^{31}.\left(2.3.4.5...31\right).32}\)
\(=\frac{1}{2^{31}.32}\)
\(=\frac{1}{2^{31}.2^5}\)
\(=\frac{1}{2^{36}}\)
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\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}......\frac{30}{62}.\frac{31}{64}\)=2x
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}.....\frac{30}{62}.\frac{31}{64}\)
\(=\frac{1.2.3.4....30.31}{4.6.8.10.12....62.64}\)
\(=\frac{1.\left(2.3.4.5....30.31\right)}{2.\left(2.3.4....30.31\right).64}\)
\(=\frac{1}{2.64}\)
\(=\frac{1}{128}\)
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Xem thêm câu trả lời
Bài 1: Tìm số nguyên n để phân số Mfrac{2n-7}{5n}có giá trị là số nguyên.Bài 2: Tìm x biết :a) /x - 3/ 2.(x+2)b)1frac{1}{30}: (24frac{1}{6}- 24frac{1}{5})-frac{1frac{1}{2}-frac{3}{4}}{4x-frac{1}{2}} (-1frac{1}{15}) : (8frac{1}{5}- 8frac{1}{3})
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Bài 1: Tìm số nguyên n để phân số M=\(\frac{2n-7}{5n}\)có giá trị là số nguyên.
Bài 2: Tìm x biết :
a) /x - 3/ =2.(x+2)
b)1\(\frac{1}{30}\): (24\(\frac{1}{6}\)- 24\(\frac{1}{5}\))-\(\frac{1\frac{1}{2}-\frac{3}{4}}{4x-\frac{1}{2}}\)= (-1\(\frac{1}{15}\)) : (8\(\frac{1}{5}\)- 8\(\frac{1}{3}\))
Tìm n biết
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\) <=>\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+.....\frac{1}{n}-\frac{1}{n-1}\right)=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{1999}{2001}\)
<=>\(\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\)
<=>\(\frac{1}{n+1}=\frac{1}{2001}\)
<=>n+1 =2001
<=>n = 2000
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ta có:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{2001}\)
\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}\right)=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{1}{2.3}+\frac{1}{2.6}+\frac{1}{2.10}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\)
\(\frac{1}{n+1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\frac{1}{n+1}=\frac{1}{2001}\)
=>\(n+1=2001\)
=>\(n=2000\)
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Ta có :
\(A.\frac{1}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+..............+\frac{1}{n.\left(n+1\right)}=\frac{1999}{2001}\)
=> \(A.\frac{1}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..............+\frac{1}{n.\left(n+1\right)}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..............+\frac{1}{n}-\frac{1}{n+1}=\frac{1999}{4002}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\Rightarrow\frac{1}{n+1}=\frac{1}{2}-\frac{1999}{4002}=\frac{1}{2001}\)
\(\Rightarrow n+1=2001\Rightarrow n=2001-1=2000\)
Vậy n = 2000
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