So sánh: \(\left(\frac{27}{64}\right)^{15}và\left(\frac{81}{256}\right)^{10}\)
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S
19 tháng 3 2020
a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)
\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)
\(\Leftrightarrow3x-1=3\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
b) Đề sai ! Sửa :
\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)
\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)
\(\Leftrightarrow2x+5=4\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)
\(\Leftrightarrow x+3=\frac{4}{3}\)
\(\Leftrightarrow x=-\frac{5}{3}\)
d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)
\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)
\(\Leftrightarrow x=\frac{4}{15}\)
PN
2
S
27 tháng 7 2016
\(\left(\frac{1}{27}\right)^{23}=\frac{1^{23}}{27^{23}}=\frac{1}{\left(3^3\right)^{23}}=\frac{1}{3^{69}}\)
\(\left(\frac{1}{81}\right)^{16}=\frac{1^{16}}{81^{16}}=\frac{1}{\left(3^4\right)^{16}}=\frac{1}{3^{64}}\)
Vì 369 > 364
\(\frac{1}{3^{69}}< \frac{1}{3^{64}}\)
27 tháng 7 2016
\(\left(\frac{1}{27}\right)^{23}=\frac{1^{23}}{27^{23}}=\frac{1}{\left(3^3\right)^{23}}=\frac{1}{3^{69}}\)
\(\left(\frac{1}{81}\right)^{16}=\frac{1^{16}}{81^{16}}=\frac{1}{\left(3^4\right)^{16}}=\frac{1}{3^{64}}\)
Vì 369 > 364
=> \(\frac{1}{3^{69}}< \frac{1}{3^{64}}\)
=> \(\left(\frac{1}{27}\right)^{23}< \left(\frac{1}{81}\right)^{16}\)
14 tháng 3 2020
Bài 1:
Ta có:
\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)
\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)
Lại có:
\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)
\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)
14 tháng 3 2020
Bài 2:
Ta có:
\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)
\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\Rightarrow A>B\)
27 tháng 12 2016
a) \(\left|\frac{-15}{6}\right|-\left|\frac{3}{18}\right|.\sqrt{81}+\sqrt{\frac{9}{64}}\)
\(=\frac{15}{6}-\frac{1}{6}.9+\frac{3}{8}\)
\(=\frac{15}{6}-\frac{9}{6}+\frac{3}{8}\)
\(=1+\frac{3}{8}\)
\(=\frac{11}{8}\)
b) \(\frac{6^{15}.9^{10}}{3^{34}.2^{13}}=\frac{\left(2.3\right)^{15}.\left(3^2\right)^{10}}{3^{34}.2^{13}}=\frac{2^{15}.3^{15}.3^{20}}{3^{34}.2^{13}}=2^2.3=12\)
AT
27 tháng 12 2016
a/ \(\left|\frac{-15}{6}\right|-\left|\frac{3}{18}\right|.\sqrt{81}+\sqrt{\frac{9}{64}}\)
= \(\frac{15}{6}-\frac{3}{18}.9+\frac{8}{8}\)
= \(\frac{15}{6}-\frac{3}{2}+\frac{3}{8}\)
= \(\frac{60-36+9}{24}=\frac{33}{24}=\frac{11}{8}\)
b/ \(\frac{6^{15}.9^{10}}{3^{34}.2^{13}}=\frac{\left(2.3\right)^{15}.\left(3^2\right)^{10}}{3^{34}.2^{13}}\) \(=\frac{2^{15}.3^{15}.3^{20}}{3^{34}.2^{13}}=\frac{2^2.3^{35}}{3^{34}}=\frac{4.3}{1}=12\)
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
2 tháng 8 2019
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
30 tháng 6 2015
Bạn để ý: 81 = 3^4 (34) 27 = 3^3
64 = 2^6 256 = 2^8
Vậy \(\left(\frac{27}{64}\right)^{15}=\left(\frac{3^2}{8^2}\right)^{15}=\left(\frac{3}{8}\right)^{30};\left(\frac{81}{256}\right)^{10}=\left(\frac{3^4}{4^4}\right)^{10}=\left(\frac{3}{4}\right)^{40}\)
Vì 3/8 <3/4 ; 30<40 nên \(\left(\frac{3}{8}\right)^{30}
21 tháng 6 2017
\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{15}\right)+....+\left(x+\frac{1}{575}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(13x+\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(13x+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(2x+\frac{12}{25}=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
Đặt \(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
\(3A-A=1-\frac{1}{3^5}=\frac{242}{243}=2A\)
=> \(A=\frac{121}{243}\)
=> \(2x+\frac{12}{25}=\frac{121}{243}\)
=> \(2x=\frac{121}{243}-\frac{12}{25}=\frac{109}{6075}\)
=> x = ......
\(\left(\frac{27}{64}\right)^{15}=\frac{\left(3^3\right)^{15}}{\left(2^6\right)^{15}}=\frac{3^{45}}{2^{90}}=\left(\frac{3}{2^2}\right)^{45}\)
\(\left(\frac{81}{256}\right)^{10}=\frac{\left(3^4\right)^{10}}{\left(2^8\right)^{10}}=\frac{3^{40}}{2^{80}}=\left(\frac{3}{2^2}\right)^{40}\)
Do \(\left(\frac{3}{2^2}\right)^{45}