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\(\frac{512^4.128^6.64^2.2^{13}.4^7}{1024^2.64^5.128^3.8^4.512^4}=\frac{2^{117}}{2^{119}}=\frac{2^{117}}{2^{117}.2^2}=\frac{1}{2^2}=\frac{1}{4}\)
\(=\frac{5.2^{30}.3^{18}-3^{20}.2^2.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}\cdot3^{18}}=\frac{5.2^{30}.3^{18}-3^{20}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}\left(5.3-7.2\right)}=\frac{2\left(10-9\right)}{\left(15-14\right)}==\frac{2.1}{1}=2\)
b./ \(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)(b)
Mà \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)
(b) \(\Leftrightarrow x+2010=0\Leftrightarrow x=-2010\)
a./
\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0.\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)(a)
Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)
(a) \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
a)A= \(\frac{17.(145-236-111)}{202.(34-178+127)}\)
=\(\frac{17.(-202)}{202.(-17)}\)
=\(1\)
\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{18}-2^2.2^{27}.3^{20}}{5.2^9.2^{19}.3^{19}-7.2^{29}-3^{18}}=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}\left(5.3-7.2\right)}=\frac{2.1}{1}=2\)
\(y=\frac{7.9+14.27+21.36}{21.27+42.81+63.108}=\frac{7.9+2.7.3.9-3.7.4.9}{3.7.3.9+2.3.7.9.9+7.9.4.3.9}=\frac{7.9\left(1+6-12\right)}{3.7.9\left(3+18+36\right)}=\frac{-5}{3.57}=\frac{-5}{171}\)
\(\frac{2048^2.2^{11}.8^6.64^7.128^3}{1024^4.4^8.16^7.2^{30}}=\frac{2^{22}.2^{11}.2^{18}.2^{42}.2^{21}}{2^{40}.2^{16}.2^{28}.2^{30}}=\frac{2^{114}}{2^{114}}=1\)