1không thuc hiên phép tính hay ss :\(\frac{221}{222};\frac{443}{445};\frac{665}{668}\)
2 rut gon \(A=\frac{2+\sqrt{3+\sqrt{5-\sqrt{48}}}}{\sqrt{6}+\sqrt{2}}\)
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\(\frac{221}{222};\frac{443}{445};\frac{665}{668}\)
\(\frac{221}{222}< \frac{443}{445}< \frac{665}{668}\)
.....
\(\frac{221}{222};\frac{443}{445};\frac{668}{665}\)
\(\frac{221}{222}< \frac{443}{445}< \frac{668}{665}\)
.....
= \(\frac{5^6.5^5}{6.5^{10}}\) = \(\frac{5^{11}}{6.5^{10}}\)= \(\frac{5}{6}\)
\(\frac{2^5.6^3}{8^2.9^2}\) = \(\frac{2^5.2^3.3^3}{2^6.3^4}\) = \(\frac{2^8.3^3}{2^6.3^4}\) = \(\frac{2^2}{3}\) = \(\frac{4}{3}\)
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 \(A=25\)