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a/ \(\frac{15}{-50}=-\frac{3}{10}\)
\(\frac{7}{10}\)
\(\frac{24}{-20}=-\frac{12}{10}\)
\(\Rightarrow-\frac{12}{10};-\frac{3}{10};\frac{7}{10}\)
KL: ..........................................
b/ \(\frac{17}{39}=\frac{17.20}{39.20}=\frac{340}{780}\)
\(\frac{11}{65}=\frac{11.12}{65.12}=\frac{132}{780}\)
\(\frac{9}{52}=\frac{9.15}{52.15}=\frac{135}{780}\)
\(\Rightarrow\frac{132}{780};\frac{135}{780};\frac{340}{780}\)
KL:.............................................
c/ \(\frac{17}{20}=\frac{17.9}{20.9}=\frac{153}{180}\)
\(\frac{-19}{30}=-\frac{19.6}{30.6}=-\frac{114}{180}\)
\(\frac{38}{45}=\frac{38.4}{45.4}=\frac{152}{180}\)
\(\frac{-13}{18}=-\frac{13.10}{18.10}=-\frac{130}{180}\)
\(\Rightarrow-\frac{130}{180};-\frac{114}{180};\frac{152}{180};\frac{153}{180}\)
KL:..................................
két bn vớ mk . mk bày cho chớ làm vào đây tốn thời gian lắm
\(A=\frac{\left(-2\right)^3\cdot3^3\cdot5^3\cdot7\cdot8}{3\cdot5^3\cdot2^4\cdot42}\)
\(=\frac{\left(-2\right)^3\cdot3^3\cdot6^3\cdot5^3\cdot7\cdot2^3}{3\cdot5^3\cdot2^4\cdot2\cdot3\cdot7}\)
\(=\frac{\left(-2\right)^3\cdot3^8\cdot5^3\cdot2^3\cdot7}{3^2\cdot5^3\cdot2^5\cdot7}=-2\cdot3^6\)
câu b để nghĩ...
a/ \(\frac{5}{6n}\)và \(\frac{7}{15}\)
=> MSC = \(6n\cdot15=90n\)
\(\Rightarrow\frac{5}{6n}=\frac{5\cdot15}{90n}=\frac{75}{90n}\)
\(\Rightarrow\frac{7}{15}=\frac{7\cdot6n}{90n}=\frac{42n}{90n}\)
b/ \(\frac{9x}{24}\)và \(\frac{12}{36}\)
=> MSC = 72
\(\Rightarrow\frac{9x}{24}=\frac{9x\cdot3}{72}=\frac{27x}{72}\)
\(\Rightarrow\frac{12}{36}=\frac{12\cdot2}{72}=\frac{24}{72}\)
a)MSC = 6n . 15 = 90n
5/6n = 5 . 15/60n . 15 = 75/90n
7/15 = 7 . 6n/15 . 6n =42n/90n
#Louis
Ta có :
\(A=-\frac{7}{10^{2005}}+-\frac{15}{10^{2006}}=-\frac{7}{10^{2005}}+-\frac{8}{10^{2006}}+-\frac{7}{10^{2006}}\)
\(B=-\frac{15}{10^{2005}}+-\frac{7}{10^{2006}}=-\frac{7}{10^{2005}}+-\frac{8}{10^{2005}}+-\frac{7}{10^{2006}}\)
Do \(-\frac{7}{10^{2005}}=-\frac{7}{10^{2005}};-\frac{7}{10^{2006}}=-\frac{7}{10^{2006}};-\frac{8}{10^{2006}}>-\frac{8}{10^{2005}}\)
\(\Rightarrow\frac{-7}{10^{2005}}+-\frac{7}{10^{2006}}+-\frac{8}{10^{2006}}>-\frac{7}{10^{2005}}+-\frac{7}{10^{2006}}+-\frac{8}{10^{2005}}\)
\(\Rightarrow A>B\)
Vậy \(A>B\)
Chúc bạn học tốt !!!
\(A-B=\left(-\frac{7}{10^{2005}}-\frac{-15}{10^{2005}}\right)+\left(-\frac{15}{10^{2006}}-\frac{-7}{10^{2006}}\right)=\frac{8}{10^{2005}}-\frac{8}{10^{2006}}=8\left(\frac{1}{10^{2005}}-\frac{1}{10^{2006}}\right)\)
Do \(10^{2005}< 10^{2006}\Rightarrow\frac{1}{10^{2005}}>\frac{1}{10^{2006}}\Rightarrow\frac{1}{10^{2005}}-\frac{1}{10^{2006}}>0\Leftrightarrow8\left(\frac{1}{10^{2005}}-\frac{1}{10^{2006}}\right)>0\Rightarrow A-B>0\Leftrightarrow A>B\)
\(\frac{15}{50}=\frac{15\cdot2}{50\cdot2}=\frac{30}{100}\)
\(\frac{7}{10}=\frac{7\cdot10}{10\cdot10}=\frac{70}{100}\)
\(\frac{24}{-20}=\frac{-24\cdot5}{20\cdot5}=\frac{-120}{100}\)