Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có
\(\left(\frac{1}{2}\right)^{225}\)=\(\left(\frac{1}{2}\right)^{9.25}\)=\(\left(\frac{1}{512}\right)^{25}\)
\(\left(\frac{1}{3}\right)^{100}\)=\(\left(\frac{1}{3}\right)^{4.25}\)=\(\left(\frac{1}{81}\right)^{25}\)
Vì \(\frac{1}{512}\)<\(\frac{1}{81}\) => \(\left(\frac{1}{512}\right)^{25}\)<\(\left(\frac{1}{81}\right)^{25}\)
Hay \(\left(\frac{1}{2}\right)^{225}\)<\(\left(\frac{1}{3}\right)^{100}\)
Mong bạn tích cho mình nhé
\(\left(\frac{1}{2}\right)^{225}=\left[\left(\frac{1}{2}\right)^9\right]^{25}=\left(\frac{1}{81}\right)^{25}\)\(\left(\frac{1}{2}\right)^{225}=\left[\left(\frac{1}{2}\right)^9\right]^{25}=\left(\frac{1}{81}\right)^{25}\)
\(\left(\frac{1}{3}\right)^{100}=\left[\left(\frac{1}{3}\right)^4\right]^{25}=\left(\frac{1}{81}\right)^{25}\)
vì \(\left(\frac{1}{81}\right)^{25}=\left(\frac{1}{81}\right)^{25}\Rightarrow\left(\frac{1}{2}\right)^{225}=\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrowđpcm\)
Ta có \(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2014^2}\right)\)
\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)...\left(\frac{2014^2-1}{2014^2}\right)\)
\(=\frac{\left(2-1\right)\left(2+1\right)}{2^2}.\frac{\left(3-1\right)\left(3+1\right)}{3^2}...\frac{\left(2014-1\right)\left(2014+1\right)}{2014^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{2013.2015}{2014.2014}\)
\(=\frac{1.2...2013}{2.3...2014}.\frac{3.4...2015}{2.3...2014}\)
\(=\frac{1}{2014}.\frac{2015}{2}\)
\(=\frac{2015}{2014.2}>\frac{1}{2}\)hay -A>1/2
=>\(A< \frac{-1}{2}\)hay A<B
a)\(\left(\frac{1}{5}\right)^5\).\(5^5\)=\(\frac{1}{3125}\).3125=1
=> 6x - 3 - 5 - 15x = 44
=> -9x - 8 = 44
=> -9x = 52
=> x = \(\frac{-52}{9}\)
nhớ
3(2x-1)-5(1+3x)=44
\(\Leftrightarrow\)6x-3-5-15x=44
\(\Leftrightarrow\)-11x=52
\(\Leftrightarrow\)x=-52/11
a) \(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=2x+3\\x+\frac{1}{2}=-\left(2x+3\right)\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x-x=\frac{1}{2}-3\\x+\frac{1}{2}=-2x-3\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x+2x=-3-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\3x=\frac{-7}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x=\frac{-7}{6}\end{array}\right.\)
Vậy \(x\in\left\{\frac{-5}{2};\frac{-7}{6}\right\}\)
\(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)
\(Ta\) \(có\): \(x+\frac{1}{2}=2x+3\)
\(x+\frac{1}{2}=x+x+3\\\)
\(x+\frac{1}{2}=x+\left(x+3\right)\)
\(\Rightarrow\frac{1}{2}=x+3\)
\(\Rightarrow x=\frac{1}{2}-3\)
\(\Rightarrow x=-\frac{5}{2}\)
Vậy \(x=-\frac{5}{2}\)
b, \(\left|x+\frac{1}{5}\right|+\left|x+\frac{2}{5}\right|+\left|x+1\frac{2}{5}\right|=4x\)
\(Ta\) \(có\)
\(x+\frac{1}{5}+x+\frac{2}{5}+x+1\frac{2}{5}\)\(=4x\)
\(3x+\left(\frac{1}{5}+\frac{2}{5}+1\frac{2}{5}\right)=4x\)
\(3x+2=4x\)
\(3x+2=3x+x\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
Ta có:
\(\left(\frac{1}{2}\right)^{225}=\left[\left(\frac{1}{2}\right)^9\right]^{25}=\left(\frac{1}{516}\right)^{25}\)
\(\left(\frac{1}{3}\right)^{100}=\left[\left(\frac{1}{3}\right)^4\right]^{25}=\left(\frac{1}{81}\right)^{25}\)
\(\frac{1}{516}< \frac{1}{81}\Rightarrow\left(\frac{1}{516}\right)^{25}< \left(\frac{1}{81}\right)^{25}\Rightarrow\left(\frac{1}{2}\right)^{225}< \left(\frac{1}{3}\right)^{100}\)