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\(D=\frac{\frac{88}{132}-\frac{33}{132}+\frac{60}{132}}{\frac{55}{132}+\frac{132}{132}-\frac{84}{132}}\)
\(D=\frac{\frac{115}{132}}{\frac{103}{132}}\)
\(D=\frac{115}{103}\)
=\(\frac{3\left(\frac{1}{1}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{2}{4}+\frac{2}{6}+\frac{2}{8}}{5\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\right)}\)
=\(\frac{3}{5}+\frac{2\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\right)}{5\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\right)}\)=\(\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\)
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{19}{20}\)
\(=\frac{1.2.3.....19}{2.3.4.....20}\)
\(=\frac{1}{20}\)
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{20}\right)\)
\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{18}{19}.\frac{19}{20}\)
\(B=\frac{1}{20}\)
Hok tốt
\(=\frac{-\frac{1}{9}+1-\frac{2}{10}+1-\frac{3}{11}+1-...-\frac{92}{100}+1}{\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}}\)
\(=\frac{\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}}{\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}}\)
\(=\frac{8\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}{\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}}\)
= 8
a, \(\frac{24.315+3.561.8+4.124.6}{1+3+5+7+...+97+99-500}\) (1)
Đặt : S = 1 + 3 + 5 + 7 + ... + 97 + 99
SSH của S là : (99 -1) : 2 + 1 = 50(sh)
Tổng của S là : \(\frac{\left(99+1\right).50}{2}=\frac{100.50}{2}=\frac{5000}{2}=2500\)
Thay S vào biểu thức (1) Ta có :
\(\frac{24.315+3.561.8+4.124.6}{2500-500}\)
\(=\frac{3.8.315+3.561.8+4.2.124.3}{2000}\)
\(=\frac{3.8.315+3.561.8+8.124.3}{2000}\)
\(=\frac{\left(3.8\right).\left(315+561+124\right)}{2000}=\frac{24.1000}{2000}=\frac{24000}{2000}=12\)
b, \(\frac{3^9.3^{20}.2^8}{3^{24}.243.2^6}=\frac{3^{29}.2^8}{3^{24}.3^5.2^6}=\frac{3^{29}.2^6.2^2}{3^{29}.2^6}=2^2=4\)
\(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot\cdot\cdot\frac{2016^2-1}{2016^2}=\frac{1.3}{2.3}\cdot\frac{2.4}{3.3}\cdot\cdot\cdot\cdot\frac{2015.2017}{2016.2016}\)
\(=\frac{\left(1.2.3....2015\right).\left(3.4....2016.2017\right)}{\left(2.3....2016\right)\left(2.3......2015.2016\right)}=\frac{2017}{2.2016}=\frac{2017}{4032}\)
Ta có; \(\left(\frac{a}{2}-b\right)^2\ge0;\forall x\)
\(\Rightarrow\frac{a^2}{4}+b^2\ge2.\frac{a}{2}.b=ab\)
đpcm
