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Ta có:
\(2n:\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+.....+\frac{1}{1+2+...+n}\right)=2020\)
<=> \(2n:\left(\frac{2}{2}+\frac{2}{3.2}+\frac{2}{4.3}+...+\frac{2}{\left(n+1\right).n}\right)=2020\)
<=> \(n:\left(1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=2020\)
<=> \(n:\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=2020\)
<=> \(n:\left(1-\frac{1}{n+1}\right)=2020\)
<=> \(n:\frac{n}{n+1}=2020\)
<=> n + 1 = 2020
<=> n = 2019
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\(\left(8x-3\right)^{2n}=5^{2n}\)
Do 2n chẵn
\(\Rightarrow\hept{\begin{cases}8x-3=5\\8x-3=-5\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=1\\x=-\frac{1}{4}\end{cases}}\)
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Ta co n^2 chia 5 du 1 hoac du 4
=>n^4 chia 5 du 1 hoac du 4
\(\orbr{\begin{cases}n^4\equiv1\left(mod5\right)\\n^4\equiv4\left(mod5\right)\end{cases}}=>\orbr{\begin{cases}n^5\equiv n\left(mod5\right)\\n^4-4+5⋮5\end{cases}}\)\(=>\orbr{\begin{cases}n^5-n⋮5\\n^4\equiv1\left(mod5\right)\left(#\right)\end{cases}}\)
Theo (#) ta co:\(n^5\equiv n\left(mod5\right)\Rightarrow n^5-n⋮5\)
Vay n^5-n chia het cho 5
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\(n\in N\)(n>0)\(\Rightarrow\left(x_1a-y_1b\right)^{2n}\ge0,...,\left(x_ma-y_mb\right)^{2n}\ge0\)\(\Rightarrow VT\ge0\)
Dấu "=" xra khi \(x_1a-y_1b=0;...;x_ma-y_mb=0\left(a,b>0\right)\Rightarrow\frac{x_1}{y_1}=\frac{x_2}{y_2}=...=\frac{x_m}{y_m}=\frac{b}{a}\)
Theo t/c dãy tỉ số bằng nhau:
\(\Rightarrow\frac{b}{a}=\frac{x_1+x_2+...+x_m}{y_1+y_2+...+y_m}\)(đpcm)
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\(81^{2n}.27^n=9^5\)
\(3^{8n}.3^{3n}=3^{10}\)
\(3^{11n}=3^{10}\)
\(11n=10\)
\(n=\frac{10}{11}\)
81^2n=3^8n
27^n=3^3n
->81^2n*27^n=3^12n mà 9^5=3^10->12n=10->n=10/12
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a. \(\left(\frac{-1}{5}\right)^n=\frac{-1}{125}\)
<=> \(\left(\frac{-1}{5}\right)^n=\left(\frac{-1}{5}\right)^3\)
<=> n = 3
b. \(\left(\frac{-2}{11}\right)^m=\frac{4}{121}\)
<=> \(\left(\frac{-2}{11}\right)^m=\left(\frac{2}{11}\right)^2\)
<=> m = 2
c. 72n + 72n+2 = 2450
<=> 72n + 72n . 72 = 2450
<=> 72n.(1+72) = 2450
<=> 72n = 72
<=> 2n = 2
<=> n = 1
\(\frac{3^{2n}.11^{2n}}{11^{2n}}=81\Leftrightarrow3^{2n}=3^4\)
\(\Leftrightarrow2n=4\Leftrightarrow n=2\)