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\(10A=\frac{10\left(10^{29}+10^{10}\right)}{10^{30}+10^{10}}=\frac{10^{30}+10^{11}}{10^{30}+10^{10}}=1+\frac{10^{11}-10^{10}}{10^{30}+10^{10}}\)
\(10B=\frac{10\left(10^{30}+10^{10}\right)}{10^{31}+10^{10}}=\frac{10^{31}+10^{11}}{10^{31}+10^{10}}=1+\frac{10^{11}-10^{10}}{10^{31}+10^{10}}\)
\(10^{30}+10^{10}< 10^{31}+10^{10}\Rightarrow\frac{10^{11}-10^{10}}{10^{30}+10^{10}}>\frac{10^{11}-10^{10}}{10^{31}+10^{10}}\)
\(\Rightarrow10A=1+\frac{10^{11}-10^{10}}{10^{30}+10^{10}}>10B=1+\frac{10^{11}-10^{10}}{10^{31}+10^{10}}\)
\(\Rightarrow A>B\)
ừa, nhưng mik còn những câu sau khó hơn nhưng mỏi tay quá nên tí gửi tiếp!
a)\(\frac{-315}{540}\)=\(\frac{7}{12}\)
b)\(\frac{25.13}{26.35}\)=\(\frac{5}{14}\)
c)\(\frac{6.9-2.17}{63.3-119}\)=\(\frac{4}{7}\)
d)\(\frac{2929-101}{2.1919+404}\)=\(\frac{2}{3}\)
\(-\frac{20+44}{56-12}=\frac{-64}{44}=\)\(\frac{-16}{11}\)
\(\frac{-120+70}{30+60}\)=\(\frac{-5}{9}\)
Ta có \(\frac{-16}{11}< \frac{-11}{11}=-1\)
\(\frac{-5}{9}>\frac{-9}{9}=-1\)
nên \(\frac{-5}{9}>-1>\frac{-16}{11}\)
Vậy \(\frac{-5}{9}>\frac{-16}{11}\)
tự kết luận nhé
\(\frac{\frac{5}{7}+\frac{5}{9}-\frac{5}{11}}{\frac{15}{7}+\frac{15}{9}-\frac{15}{11}}\)
\(=\) \(\frac{1}{3}\)
\(\frac{1}{3}\)nha
~~ tk mk đi ~~
Ai tk mk mk tk lại ~~
kb nha ~ n_n
A=\(2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\right)\)
A=\(2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\right)\)
A=\(2.\left(1-\frac{1}{13}\right)\)
A=\(2.\frac{12}{13}=\frac{24}{13}\)
A=2(2/1.3+2/3.5+2/5.7+...+2/11.13)
A=2(1/1-1/3+1/3-1/5+1/5-1/7+...+1/11-1/13)
A=2(1/1-1/13)=2.12/13=24/13
\(\frac{21^2.14^3.125^3}{35^2.36}=\frac{\left(3.7\right)^2.\left(2.7\right)^3.\left(5^3\right)^3}{\left(5.7\right)^2.2^2.3^2}=\frac{3^7.7^2.2^3.7^3.5^9}{5^2.7^2.2^2.3^2}=\frac{3^7.7^5.2^3.5^9}{5^2.7^2.2^2.3^2}=2.3^5.5^7.7^3\)