(3/5)^5 và (5/243)^3
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\(\left(\frac{3}{8}\right)^5=\frac{243}{8^5}>\frac{125}{8^5}>\frac{125}{9^5}=\frac{125}{\left(3^2\right)^5}=\frac{125}{3^{10}}>\frac{125}{3^{15}}=\frac{5^3}{\left(3^5\right)^3}=\left(\frac{5}{243}\right)^3\)
=>\(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)
a/
+ \(\frac{1}{243^6}=\frac{1}{3^6.81^6}=\frac{1}{3^2.3^4.81^6}=\frac{1}{9.81^7}\) (1)
+ \(80< 81\Rightarrow80^7< 81^7\Rightarrow\frac{1}{80^7}>\frac{1}{81^7}\) (2)
+ \(81^7< 9.81^7\Rightarrow\frac{1}{81^7}>\frac{1}{9.81^7}\) (3)
Từ (1) (2) (3) \(\Rightarrow\frac{1}{80^7}>\frac{1}{243^6}\)
b/ Xem lại đề bài
a) Ta có: \(\left(\dfrac{1}{243}\right)^6=\left(\dfrac{1}{3}\right)^{5\cdot6}=\left(\dfrac{1}{3}\right)^{30}\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{28}>\left(\dfrac{1}{243}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{3^4}\right)^7>\left(\dfrac{1}{243}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{81}\right)^7>\left(\dfrac{1}{243}\right)^6\)
mà \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{81}\right)^7\)
nên \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)
\(\left(\dfrac{3}{8}\right)^5\&\left(\dfrac{5}{243}\right)^3\)
\(\left(\dfrac{3}{8}\right)^5=\left(\dfrac{90}{240}\right)^5=\dfrac{90^5}{240^5}\)
\(\left(\dfrac{5}{243}\right)^3=\dfrac{5^3}{243^3}\)
\(=>\dfrac{90^5}{240^5}>\dfrac{5^3}{243^3}\)
\(=>\left(\dfrac{3}{8}\right)^5>\left(\dfrac{5}{243}\right)^3\)
1: 243^5=(3^5)^5=3^25
3*27^8=3*3^24=3^25=243^5
3: 3^300=27^100
2^200=4^100
mà 27>4
nên 3^300>2^200
4: 15^2=3^2*5^2
81^3*125^3=3^12*5^9
=>15^2<81^3*125^3
6: 125^5=5^15
25^7=5^14
mà 15>14
nên 125^5>25^7
3 và 5/81
nhớ k nha