221/222>443/445>665/668

chuc ban hoc tot ^-^

\(\frac{221}{222};\frac{443}{445};\frac{665}{668}\)

\(\frac{221}{222}< \frac{443}{445}< \frac{665}{668}\)

.....

giải ra đc ko

\(\frac{221}{222};\frac{443}{445};\frac{668}{665}\)

\(\frac{221}{222}< \frac{443}{445}< \frac{668}{665}\)

.....

= \(\frac{5^6.5^5}{6.5^{10}}\) = \(\frac{5^{11}}{6.5^{10}}\)= \(\frac{5}{6}\)

\(\frac{25^3.5^5}{6.5^{10}}\)=\(\frac{5^{11}}{6.5^{10}}\)=\(\frac{5}{6}\)

ok 100% chính xác

\(\frac{25^3\cdot5^5}{6\cdot5^{10}}=\frac{25^3}{6\cdot5^5}\)

\(\frac{\left(5^2\right)^3}{6\cdot5^5}=\frac{5^6}{6\cdot5^5}=\frac{5}{6}\)

\(\frac{2^5.6^3}{8^2.9^2}\) = \(\frac{2^5.2^3.3^3}{2^6.3^4}\) = \(\frac{2^8.3^3}{2^6.3^4}\) = \(\frac{2^2}{3}\) = \(\frac{4}{3}\)

bâm máy tính đi

\(=\frac{6912}{5184}=\frac{4}{3}\)

Đặt \(A=\frac{1}{4.9}+\frac{1}{9.14}++\frac{1}{14.19}+......+\frac{1}{44.49}\)

\(A=\frac{1}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+.....+\frac{5}{44.49}\right)\)

\(A=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+.....+\frac{1}{44}-\frac{1}{49}\right)\)

\(A=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{49}\right)=\frac{1}{5}.\frac{45}{196}=\frac{9}{196}\)

Đặt \(B=\frac{1-3-5-7-.......47-49}{89}\)

\(B=\frac{1-\left(3+5+7+......+47+49\right)}{89}\)

Từ 3 -> 49 có: (49-3):2+1=24(số hạng)

=>\(3+5+7+....+47+49=\frac{\left(49+3\right).24}{2}=624\)

=>\(B=\frac{1-624}{89}=\frac{-623}{89}=-7\)

Vậy \(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right).\frac{1-3-5-,,,,,-49}{89}=A.B=\frac{9}{196}.\left(-7\right)=-\frac{9}{28}\)

\(A=\frac{125^{100}}{5^{298}}\cdot\frac{2^{160}}{4^{80}}=>A=\frac{\left(5^3\right)^{100}}{5^{298}}\cdot\frac{2^{160}}{\left(2^2\right)^{80}}\)

\(=>A=\frac{5^{300}}{5^{298}}\cdot\frac{2^{160}}{2^{160}}=>A=5^2\cdot1=>A=25\)

\(A=\frac{125^{100}}{5^{298}}.\frac{2^{160}}{4^{80}}\)

    \(=\frac{\left(5^3\right)^{100}}{5^{298}}.\frac{2^{160}}{\left(2^2\right)^{80}}\)

    \(=\frac{5^{300}}{5^{298}}.\frac{2^{160}}{2^{160}}\)

    \(=5^2.1=25\)

Vậy \(A=25\)

\(A=\frac{125^{100}}{5^{298}}.\frac{2^{160}}{4^{80}}=\frac{\left(5^3\right)^{100}}{5^{298}}.\frac{2^{160}}{\left(2^2\right)^{80}}\)

                                          \(=\frac{5^{300}}{5^{298}}.\frac{2^{160}}{2^{160}}\)

                                           \(=5^2.1\) 

                                             \(=25\)