SO SÁNH
\(\left(\frac{1}{80}\right)^7va\left(\frac{1}{243}\right)^6\)
\(\left(\frac{3}{5}\right)^5va\left(\frac{5}{243}\right)^3\)
\(\frac{-22}{35}va\frac{-103}{177}\)
\(\frac{84}{-83}va\frac{-337}{331}\)
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a) 80<243 nên \(\frac{1}{80}>\frac{1}{243}\)
\(\Rightarrow\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^7\) mà \(\left(\frac{1}{243}\right)^7>\left(\frac{1}{243}\right)^6\)
Nên \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
b) Ta so sánh \(\frac{3}{8}\) với \(\frac{5}{243}\)
Ta có: \(\frac{3}{8}=\frac{3.243}{8.243}=\frac{729}{1944}\)
\(\frac{5}{243}=\frac{5.8}{243.8}=\frac{40}{1944}\)
Suy ra: \(\frac{3}{8}>\frac{5}{243}\)
\(\Rightarrow\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^5\) mà \(\left(\frac{5}{243}\right)^5>\left(\frac{5}{243}\right)^3\)
Nên \(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)
\(a,\left(\frac{1}{8}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\frac{1}{3^{28}}\)
\(\left(\frac{1}{234}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{3^{30}}\)
Vì \(\frac{1}{3^{28}}>\frac{1}{3^{30}}\)nên \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^8\)
\(b,\left(\frac{3}{5}\right)^5=\left(\frac{3}{2^3}\right)^5=\frac{243}{2^{15}};\)
\(\left(\frac{5}{243}\right)^3=\left(\frac{5}{3^5}\right)^3=\frac{125}{3^{15}}\)
Chọn phân số \(\frac{243}{3^{15}}\)làm phân số trung gian để so sánh hai phân số trên , ta thấy :
\(\frac{243}{2^{15}}>\frac{243}{3^{15}}>\frac{125}{3^{15}}\)
=> \(\frac{243}{2^{15}}>\frac{125}{3^{15}}\)
Từ đó => \(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)
a) \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\frac{1}{3^{4.7}}=\frac{1}{3^{28}}\)
\(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{3^{3.6}}=\frac{1}{3^{30}}\)
Vì \(\frac{1}{3^{28}}< \frac{1}{3^{30}}\left(3^{28}< 3^{30}\right)\)
Nên \(\left(\frac{1}{80}\right)^7< \left(\frac{1}{243}\right)^6\)
b) \(\left(\frac{3}{8}\right)^5=\frac{3^5}{\left(2^3\right)^5}=\frac{243}{2^{3.5}}=\frac{243}{2^{15}}>\frac{243}{3^{15}}>\frac{125}{3^{15}}\)
\(=\frac{5^3}{\left(3^5\right)^3}=\frac{5^3}{3^{5.3}}=\frac{5^3}{3^{15}}=\left(\frac{5}{243}\right)^3\)
\(\Rightarrow\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)
1. Ta có : Dùng phân số \(\frac{84}{71}\)làm phân số trung gian ta được : \(\frac{84}{81}>\frac{73}{71}\)
2. Ta có : \(\left[\frac{3}{8}\right]^5=\left[\frac{3}{2^3}\right]^5=\frac{243}{2^{15}};\left[\frac{5}{243}\right]^3=\left[\frac{5}{3^5}\right]^3=\frac{125}{3^{15}}\)
Chọn phân số \(\frac{243}{3^{15}}\)làm phân số trung gian để so sánh ta được : \(\frac{243}{2^{15}}>\frac{125}{3^{15}}\)
Vậy : \(\left[\frac{3}{8}\right]^5>\left[\frac{5}{243}\right]^3\)