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\((\frac{1}{9}-\frac{9}{17})+\frac{3}{6}-(\frac{-8}{17}-\frac{1}{2})+\frac{5}{9}=\)
=\(\frac{1}{9}-\frac{9}{17}+\frac{1}{2}+\frac{8}{17}+\frac{1}{2}+\frac{5}{9}\)
= \((\frac{1}{9}+\frac{5}{9})+(\frac{-9}{17}+\frac{8}{17})+(\frac{1}{2}+\frac{1}{2})\)
=\(\frac{2}{3}-\frac{1}{17}+1\)
=\(\frac{34}{51}-\frac{3}{51}+\frac{51}{51}\)
=\(\frac{82}{51}\)
\(\frac{24\cdot47-23}{24+47\cdot23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
\(=\frac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\frac{3\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}\)
\(=\frac{24^2+24\cdot23-23}{24+24\cdot23+23^2}\cdot\frac{3}{9}\) \(=\frac{24^2+23\cdot\left(24-1\right)}{\left(23+1\right)\cdot24\cdot23^2}\cdot\frac{1}{3}=1\cdot\frac{1}{3}=\frac{1}{3}\)
\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)=2
nha
Xét: 8.(9+m)=72+8m
9.(8+m)=72+9m
Với m E N* thì 8m<9m
=>72+8m<72+9m
=>8.(9+m)<9.(8+m)
=>8/9<8+m/9+m
Ta thấy: \(\frac{8}{9}=\frac{8.\left(9+m\right)}{9.\left(9+m\right)}=\frac{72+8m}{81+9m}\)
\(\frac{8+m}{9+m}=\frac{9.\left(8+m\right)}{9.\left(9+m\right)}=\frac{72+9m}{81+9m}\)
Vì 72+8m<72+9m (m thuộc N*) nên: \(\frac{72+8m}{81+9m}<\frac{72+9m}{81+9m}\)
hay \(\frac{8}{9}<\frac{8+m}{9+m}\)
\(\dfrac{8}{x}-8+\dfrac{11}{x}-11=\dfrac{9}{x}-9+\dfrac{10}{x}-10\)\(\Leftrightarrow\dfrac{8}{x}+\dfrac{11}{x}-\dfrac{9}{x}-\dfrac{10}{x}=8+11-9-10\)
\(\Leftrightarrow\dfrac{8+11-9-10}{x}=0\)
\(\Leftrightarrow\dfrac{0}{x}=0\)
\(\Leftrightarrow x=0\)
S=\(\left\{0\right\}\)
18
tick cho mik nha