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Ta có:
\(\left(\frac{1}{2}\right)^{225}=\left[\left(\frac{1}{2}\right)^9\right]^{25}=\left(\frac{1}{516}\right)^{25}\)
\(\left(\frac{1}{3}\right)^{100}=\left[\left(\frac{1}{3}\right)^4\right]^{25}=\left(\frac{1}{81}\right)^{25}\)
\(\frac{1}{516}< \frac{1}{81}\Rightarrow\left(\frac{1}{516}\right)^{25}< \left(\frac{1}{81}\right)^{25}\Rightarrow\left(\frac{1}{2}\right)^{225}< \left(\frac{1}{3}\right)^{100}\)
\(\frac{2008}{2009};\frac{20}{19}\)
\(1-\frac{2008}{2009}=\frac{1}{2009}\)
\(1-\frac{20}{19}=\frac{-1}{19}=\frac{1}{19}\)
Vì 19 < 2009 Nên \(\frac{1}{2009}< \frac{1}{19}\)
Vậy \(\frac{2008}{2009}>\frac{20}{19}\)
Xét \(\frac{1}{\sqrt{13}}>\frac{1}{\sqrt{14}}\Rightarrow\frac{1}{\sqrt{13}}-1< \frac{1}{\sqrt{14}}+1\)
Mà \(\sqrt{225}< \sqrt{289}\)
\(\Rightarrow\sqrt{225}-\left(\frac{1}{\sqrt{13}}-1\right)< \sqrt{289}-\left(\frac{1}{\sqrt{14}}+1\right)\)
Vậy....................
a) \(A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{2}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)
\(=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)
\(=\frac{1}{5}+\frac{2}{7}\)
\(=\frac{7}{35}+\frac{10}{35}\)
\(=\frac{17}{35}\)
Vậy \(A=\frac{17}{35}\)
b) \(B=\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...+\frac{5^2}{56.61}\)
\(=5.\left(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{56.61}\right)\)
\(=5.\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{56}-\frac{1}{61}\right)\)
\(=5.\left(\frac{1}{11}-\frac{1}{61}\right)\)
\(=5.\left(\frac{61}{671}-\frac{11}{671}\right)\)
\(=5.\frac{50}{671}\)
\(=\frac{250}{671}\)
Vậy \(B=\frac{250}{671}\)
\(\frac{1}{9}\cdot3^4\cdot3^n=3^8\)
\(=>3^n=3^8:3^4:\frac{1}{9}\)
\(=>3^n=3^8:3^4\cdot9\)
\(=>3^n=3^8:3^4\cdot3^2\)
\(=>3^n=3^6\)
\(=>n=6\)
b) \(\frac{1}{9}.3^4.3^n=3^8\)
\(\Rightarrow\left(\frac{1}{3}\right)^2.3^4.3^n=3^8\)
\(\Rightarrow\frac{1}{3^2}.3^4.3^n=3^8\)
\(\Rightarrow3^2.3^n=3^8\)
\(\Rightarrow3^n=3^8:3^2\)
\(\Rightarrow3^n=3^6\)
\(\Rightarrow n=6\)
Vậy n = 6
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
\(\frac{\left(x+1\right)3}{111\cdot3}=\frac{3x+3}{333}\)
\(\frac{\left(y+2\right)2}{222\cdot2}=\frac{2y+4}{444}\)
Ta có: \(\frac{3x+3}{333}=\frac{2y+4}{444}=\frac{z+3}{333}\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{3x+3}{333}=\frac{2y+4}{444}=\frac{z+3}{333}=\frac{3x+3+2y+4+z+3}{333+444+333}=\frac{\left(3x+2y+z\right)+\left(3+4+3\right)}{1110}=\frac{989+10}{1110}=\frac{999}{1110}=\frac{9}{10}\)
\(\frac{3x+3}{333}=\frac{9}{10}\Rightarrow3x+3=\frac{2997}{10}\Rightarrow3x=\frac{2967}{10}\Rightarrow x=\frac{989}{10}=98,9\)
Tìm y và z tương tự nhé! Ko hiểu chỗ nào thì nói tớ!
b)\(\frac{1}{330}< \frac{1}{225}\)vi day la truong hop cung tu
c)\(\frac{1}{3^{11}}=\frac{1}{177147}\)
\(\frac{1}{7^{14}}=1,474441139_{X10}^{12}\)
nen \(\frac{1}{3^{11}}< \frac{1}{7^{14}}\)vi day cung la truong hop cung tu
\(nha^{ }\)