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a: \(\Leftrightarrow3^n:27^n=\dfrac{1}{9}\)
\(\Leftrightarrow\left(\dfrac{1}{9}\right)^n=\dfrac{1}{9}\)
hay n=1
b: \(\Leftrightarrow3^n\cdot3^2=3^8\)
=>n+2=8
hay n=6
c: \(\Leftrightarrow2^n\cdot\dfrac{9}{2}=9\cdot2^5\)
\(\Leftrightarrow2^n=2^6\)
hay n=6
d: \(\Leftrightarrow8^n=512\)
hay n=3
\(\frac{1}{2}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(=>\left(\frac{1}{2}+4\right)\cdot2^n=\frac{9}{2}\cdot2^6\)
\(=>\frac{9}{2}\cdot2^n=\frac{9}{2}\cdot2^6\)
\(=>2^n=2^6\)
\(=>n=6\)
\(B=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)\)
\(=\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)...\left(\frac{n.\left(n+2\right)+1}{n.\left(n+2\right)}\right)\)
\(=\left(\frac{2^2}{1.3}\right).\left(\frac{3^2}{2.4}\right).\left(\frac{4^2}{3.5}\right)...\left(\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\right)\)
\(=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)
\(=\frac{\left(n+1\right)}{1}.\frac{2}{\left(n+2\right)}\)
\(=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}=2.\frac{n+1}{n+2}< 2\)(vì \(\frac{n+1}{n+2}< 1\))
Vậy B < 2
Ta có:
\(1+\frac{1}{1.3}=\frac{4}{1.3}=\frac{2^2}{1.3}\)
\(1+\frac{1}{2.4}=\frac{9}{2.4}=\frac{3^2}{2.4}\)
\(1+\frac{1}{3.5}=\frac{16}{3.5}=\frac{4^2}{3.5}\)
...
\(1+\frac{1}{n\left(n+2\right)}=\frac{n^2+2n+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
=>
\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2^2.3^2.4^2...\left(n+1\right)^2}{1.2.3^2.4^2...\left(n+1\right)\left(n+2\right)}=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}\)
\(=\frac{2\left(n+2\right)-2}{n+2}=2-\frac{2}{n+2}< 2\)
Vậy B < 2
1,
a. \(\left(-3,8\right)+\left[\left(-5,7\right)+\left(+3,8\right)\right]\)
= \(\left(-3,8\right)+\left[-1,9\right]\)
= \(-5,7\)
b. \(\left(31,4\right)+\left[\left(6,4\right)+\left(-2,8\right)\right]\)
= \(\left(31,4\right)+\left[3,6\right]\)
= \(35\)
2.
a.\(\left|2x-3\right|=5\)
\(\Rightarrow2x-3=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=8:2=4\\x=-2:2=-1\end{matrix}\right.\)
Vậy \(x=4\) hoặc \(x=-1\)
Cậu có thể vào đây tham khảo : http://h.vn/hoi-dap/question/119685.html