\(a,\left(\frac{1}{8}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\frac{1}{3^{28}}\)

\(\left(\frac{1}{234}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{3^{30}}\)

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 => \(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)

 Theo  thứ tự nhé

a) <

b) <

c) >

Tìm x :

x - 0,27 = \(\frac{73}{100}\)

x           = \(\frac{73}{100}+0,27\)

x           = 1

Cậu P khó quá mik chưa nghĩ ra cách tính nhanh nhất !

Cậu tự giải nhé !

Hok tốt

\(x-0,27=\frac{\frac{73}{77}+\frac{73}{165}+\frac{73}{285}}{25\times\left(\frac{5}{84}+\frac{3}{180}+\frac{4}{285}\right)}.\)

\(x-0,27=\frac{\frac{146}{105}+\frac{73}{285}}{25\times\left(\frac{8}{105}+\frac{4}{285}\right)}\)

\(x-0,27=\frac{\frac{219}{133}}{25\times\frac{12}{133}}\)

\(x-0,27=\frac{\frac{219}{133}}{\frac{300}{133}}\)

\(x-0,27=0,73\)

\(x=0,73+0,27\)

\(x=1\)

Câu một \(=3^2.\frac{1}{3^5}.\left(3^4\right)^2.\frac{1}{3^3}=3^{10}.\frac{1}{3^8}=3^2=9\)

Câu hai \(=\left(2^2.2^5\right):\left(2^3.\frac{1}{2^4}\right)=\frac{2^7}{\frac{2^3}{2^4}}=2^8=256\)

Chờ chút nhá :D

Câu 3  \(=9-64-25^2=-680\)

Câu 4 \(=8+1-1+1=9\)

Câu 5 \(=4,75-0,37+0,125-1,28-2,5+3\frac{1}{12}=0,725+3\frac{1}{12}=3\frac{97}{120}\)

Sai thì mình xin lỗi :v, vội quá

Câu 1:\(3^2.\frac{1}{243}.81^2.\frac{1}{3^3}=3^2.\frac{1}{3^5}.3^8.\frac{1}{3^3}=3^2:3^5.3^8:3^3=3^{2-5+8-3}=3^2=9\)

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 \(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)