Tìm n sao cho \(\left(0,25\right)^n=\dfrac{1}{256}\)
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Tính m , n ,p biết :
a) \(\left(\dfrac{1}{3}\right)^m=\dfrac{1}{81}\) b) \(\left(\dfrac{3}{5}\right)^n=\left(\dfrac{9}{25}\right)^5\)
c) \(\left(-0,25\right)^p=\dfrac{1}{256}\)
a) \(\left(\dfrac{1}{3}\right)^m=\dfrac{1}{81}\)
\(\Rightarrow\dfrac{1^m}{3^m}=\dfrac{1}{81}\)
\(\Rightarrow\dfrac{1}{3^m}=\dfrac{1}{3^4}\)
\(\Rightarrow m=4\)
b) \(\left(\dfrac{3}{5}\right)^n=\left(\dfrac{9}{25}\right)^5\)
\(\Rightarrow\left(\dfrac{3}{5}\right)^n=\left[\left(\dfrac{3}{5}\right)^2\right]^5\)
\(\Rightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^{10}\)
\(\Rightarrow n=10\)
c) \(\left(-0,25\right)^p=\dfrac{1}{256}\)
\(\Rightarrow\left(\dfrac{-1}{4}\right)^p=\dfrac{1}{256}\)
\(\Rightarrow\left(\dfrac{-1}{4}\right)^p=\dfrac{1}{4^4}\)
\(\Rightarrow\left(\dfrac{-1}{4}\right)^p=\left(\dfrac{1}{4}\right)^4\)
\(\Rightarrow p=4\)
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Tính:
\(N=\left(0,25\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)\(N=\left(0,25\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)
\(N=4\cdot16\cdot\dfrac{9}{16}\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=4\cdot9\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}\)
\(=\dfrac{16}{5}\cdot\dfrac{243}{8}=\dfrac{486}{5}\)
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Tìm n, biết: \(\dfrac{\left(-4\right)^n}{16}=-64\)
A. n = 6 B. n = 5 C. n = 3 D. n = 256
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cho I = \(\dfrac{1.3+2}{4}.\dfrac{3.5+2}{16}.\dfrac{15.17+2}{256}.\dfrac{255.257+2}{65536}.....\dfrac{\left(2^{2^n}-1\right)\left(2^{2^n}+1\right)+2}{2^{2^n}}\)với n ϵ N. Chứng minh: I < \(\dfrac{4}{3}\)
Tính m, n, p biết
a) \(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)
b) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
c) \(\left(-0,25\right)^p=\frac{1}{256}\)
a ) \(\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow m=4\)
b ) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Leftrightarrow\left(\frac{3}{5}^2\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Leftrightarrow\left(\frac{9}{25}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Leftrightarrow n=5\)
c ) \(\left(-0,25\right)^p=\frac{1}{256}\)
\(\Leftrightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}\)
\(\Leftrightarrow\left(-\frac{1}{4}\right)^p=\left(-\frac{1}{4}\right)^4\)
\(\Leftrightarrow p=4\)
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\(a.\)
\(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow m=4\)
Vậy : \(m=4\)
\(b.\)
\(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Rightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{15}\)
\(\Rightarrow n=5\)
Vậy : \(n=5\)
\(c.\)
\(\left(-0,25\right)^p=\frac{1}{256}\)
\(\Rightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}\)
\(\Rightarrow\left(-\frac{1}{4}\right)^p=\left(\frac{1}{4}\right)^4\)
\(\Rightarrow p=4\)
Vậy : \(p=4\)
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\(\left(\dfrac{1}{2}\right)^x+\left(\dfrac{1}{2}\right)^{x+3}=\dfrac{9}{256}\)
tìm x thõa mãn
Lời giải:
\(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+3}=\frac{9}{256}\)
\(\Leftrightarrow \left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^x.\left(\frac{1}{2}\right)^3=\frac{9}{256}\)
\(\Leftrightarrow \left(\frac{1}{2}\right)^x.(1+\frac{1}{8})=\frac{9}{256}\Rightarrow \left(\frac{1}{2}\right)^x=\frac{1}{32}=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow x=5\)
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Với mọi số tự nhiên \(n>1\) giải thích tại sao \(\dfrac{2}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\)
Ta có: \(\dfrac{2}{\left(n-1\right)n\left(n+1\right)}=\dfrac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\)
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Tìm x biết:
a) \(\left(3\dfrac{1}{2}+2x\right).3\dfrac{2}{3}=5\dfrac{1}{3}\)
b) \(\left(\dfrac{x}{7}+0,25\right)=\dfrac{-1}{28}\)
a: \(\Leftrightarrow2x+\dfrac{7}{2}=\dfrac{16}{3}:\dfrac{11}{3}=\dfrac{16}{11}\)
=>2x=-45/22
hay x=-45/44
b: =>x/7=-1/28:1/4=-1/7
=>x=-1
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a)(7/2+2x).11/3=16/3
7/2+2x=16/3:11/3
7/2+2x=16/3.3/11
7/2+2x=16/11
2x=16/11-7/2
2x= -45/22
x= -45/22:2
x= -45/44
Vậy x= -45/44
b)x/7 +1/4= -1/28
x/7= -1/28-1/4
x/7= -2/7
=>x= -2
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Tìm số tự nhiên n sao cho:
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}=\dfrac{637}{2550}\)
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}=\dfrac{637}{2550}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{637}{2550}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{637}{2550}\)
\(\Leftrightarrow\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{637}{1275}\)
\(\Leftrightarrow\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{1}{2}-\dfrac{637}{1275}=\dfrac{1}{2550}\)
\(\Leftrightarrow\left(n+1\right)\left(n+2\right)=2550\)
\(\Leftrightarrow n^2+3n-2548=0\)
\(\Rightarrow n=49\)
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