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TL:
\(\frac{4.5+4.11}{4.3-8.7}=\frac{20+44}{12-56}=\frac{64}{-44}=-\frac{16}{1}=-16\)
\(\frac{4^6.3^4.9^5}{6^{12}.15}=\frac{2^{12}.3^{14}}{2^{12}.3^{13.5}}=\frac{3}{5}\)
Học 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
\(\frac{3.7.8.13}{14\cdot15\cdot26}=\frac{3\cdot7\cdot8\cdot13}{2\cdot7\cdot3\cdot5\cdot2\cdot13}=\frac{8}{5\cdot2\cdot2}=\frac{8}{20}=\frac{2}{5}\)
\(\frac{3.7.8.13}{14.15.26}=\frac{3.7.2^3.13}{2.7.3.5.13.2}=\frac{1.7.2.1}{1.1.1.5.1.1}=\frac{14}{5}\)
\(\frac{11.8-11.3}{17-6}=\frac{11.\left(8-3\right)}{11}=\frac{11.5}{11}=\frac{1.5}{1}=\frac{5}{1}=5\)
\(\frac{-17.13+17.2}{11.2-11.9}=\frac{-17.13+\left(-17\right).2}{11.\left(2-9\right)}=\frac{-17.\left(-13+2\right)}{11.\left(-7\right)}=\frac{-17.\left(-11\right)}{11.\left(-7\right)}=\frac{-17.\left(-1\right)}{-1.\left(-7\right)}=\frac{17}{7}\)
\(\frac{7}{9.10^2-2.10^2}=\frac{7}{100\left(9-2\right)}=\frac{7}{100.7}=\frac{7}{700}=\frac{1}{100}\)
mình làm được tới đó thôi, hihi.
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{100}\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A< 1-\frac{1}{10}=\frac{9}{10}\)
\(=>A>\frac{65}{132}\)
\(\frac{x-12}{3}=\frac{x+1}{4}\)
=>(x-12).4=(x+1)*3
4x-48=3x+3
4x-3x=48+3
x=51
(x-12)/3=(x+1)/4
(x-12)*4=(x+1)*3
x*4-12*4=x*3+1*3
4x-48=3x+3
4x-3x=3+48
x=51