\(\left(\frac{4}{27}+\frac{4}{165}+\frac{4}{285}\right):\left(\frac{5}{84}+\frac{3}{180}+\frac{4}{285}\right)=\frac{4}{27}+\frac{4}{165}+\frac{4}{285}:\frac{5}{84}+\frac{3}{180}+\frac{4}{285}=\frac{1052}{5643}:\frac{12}{133}=\frac{1841}{891}\)
Tính đúng :
\(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(2013^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(2014^4+\frac{1}{4}\right)}\)
Rút gọn: \(S=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(2004^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(2005^4+\frac{1}{4}\right)}\)
Tính giá trị biểu thức \(M=\frac{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)....\left(2014^4+\frac{1}{4}\right)}{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)....\left(2013^4+\frac{1}{4}\right)}\) .
a)Tính giá trị của biểu thức : S=\(\frac{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(100^4+\frac{1}{4}\right)}{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)..\left(99^4+\frac{1}{4}\right)}\)
b) Cho x,y là các số thực dương.Tìm GTNN của biểu thức : P=\(\frac{x+y}{\sqrt{x\left(4x+5y\right)}+\sqrt{y\left(4y+5x\right)}}\)
tính B=\(\frac{\left(1^4+\frac{1}{4}\right).\left(3^4+\frac{1}{4}\right).....\left(29^4+\frac{1}{4}\right)}{\left(2^{\text{4}}+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)......\left(30^4+\frac{1}{4}\right)}\)
Tính;
R=\(\frac{\sqrt{\left(-\frac{2}{5}\right)^5.\left(-\frac{5}{8}\right)^3.5^2}}{\sqrt[3]{\left(-\frac{3}{4}\right)^3.\left(-\frac{5}{24}\right)^2.\left(-\frac{5}{3}\right)^4}}\)
Đơn giản biểu thức \(\frac{1}{\left(a+b\right)^3}\left(\frac{1}{a^4}-\frac{1}{b^4}\right)+\frac{2}{\left(a+b\right)^4}\left(\frac{1}{a^3}-\frac{1}{b^3}\right)+\frac{2}{\left(a+b\right)^5}\left(\frac{1}{a^2}-\frac{1}{b^2}\right)\)
Cho a,b,c là các số thực dương thỏa mãn a+b+c=3abc. Chứng minh rằng :
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\left[\frac{a^4}{\left(ab+1\right)\left(ac+1\right)}+\frac{b^4}{\left(bc+1\right)\left(ab+1\right)}+\frac{c^4}{\left(ca+1\right)\left(bc+1\right)}\right]\ge\frac{27}{4}\)
\(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)...\left(29^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)...\left(30^4+\frac{1}{4}\right)}=?\)
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