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a) Ta có :
\(\frac{19}{23}=\frac{19.101}{23.101}=\frac{1919}{2323}\)
\(\Rightarrow\frac{19}{23}=\frac{1919}{2323}\)
b) Ta có
\(\frac{10}{60}=\frac{10.10101}{60.10101}=\frac{101010}{606060}\)
\(\Rightarrow\frac{10}{60}=\frac{101010}{606060}\)
c) Ta có
\(\frac{123}{124}=\frac{123.1001}{124.1001}=\frac{123123}{124124}\)
\(\Rightarrow\frac{123}{124}=\frac{123123}{124124}\)
Xét :
\(\frac{122}{123}-1=-\frac{1}{123}\)
\(\frac{123}{124}-1=-\frac{1}{124}\)
Vì \(-\frac{1}{123}< -\frac{1}{124}\)
\(\Rightarrow\frac{122}{123}< \frac{123}{124}\)
\(\Rightarrow\frac{122}{123}< \frac{123123}{124124}\)
a) i)\(\frac{7\cdot25-7\cdot7}{7\cdot24+7\cdot3}=\frac{7\left(25-7\right)}{7\left(24+3\right)}=\frac{18}{27}=\frac{2}{3}\) ii)\(\frac{2\cdot\left(-1\right)\cdot13\cdot\left(-3\right)^2\cdot\left(-2\right)\cdot\left(-5\right)}{\left(-3\right)\cdot2\cdot2\cdot\left(-5\right)\cdot13\cdot2}=\frac{-3}{2}\)
b) i)\(\frac{3}{-4}< 0;\frac{-1}{-4}>0=>\frac{3}{-4}< \frac{-1}{-4}\)
ii) ta có \(\frac{15}{17}+\frac{2}{17}=1;\frac{25}{27}+\frac{2}{27}=1\)
mà \(\frac{2}{17}>\frac{2}{27}\) =>\(\frac{15}{17}< \frac{25}{27}\)
Dựa vào tính chất :x<y và y<z thì x<z, ta có :
-12/-37<0 và 0< 13/38
=> -12/-37<13/38
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Ta có: \(\frac{13}{38}>\frac{13}{39}=\frac{1}{3}\) (1)
\(\frac{-12}{-37}=\frac{12}{37}< \frac{12}{36}=\frac{1}{3}\) (2)
Từ (1)(2) => \(\frac{13}{38}>\frac{-12}{-37}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)
\(A=1-\frac{1}{256}\)
\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3B-B=\left(1+\frac{1}{3}+...+\frac{1}{243}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}\right)\)
\(2B=1-\frac{1}{729}\)
\(B=\frac{1-\frac{1}{729}}{2}\)
\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2C=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2C-C=\left(1+\frac{1}{2}+...+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)\)
\(C=1-\frac{1}{64}\)
a)Ta có:
\(\frac{1212}{1313}=\frac{12\cdot101}{13\cdot101}=\frac{12}{13}\)
Suy ra \(\frac{12}{13}=\frac{1212}{1313}\)
câu b lạ
còn on 0
a)
Ta có
\(\frac{12}{13}=\frac{12.101}{13.101}=\frac{1212}{1313}\)
\(\Rightarrow\frac{12}{13}=\frac{1212}{1313}\)
b)
Ta có
\(1-\frac{123}{124}=\frac{1}{124}\)
\(1-\frac{124}{125}=\frac{1}{125}\)
Mà \(\frac{1}{124}>\frac{1}{125}\)
\(\Rightarrow1-\frac{123}{124}>1-\frac{124}{125}\)
\(\Rightarrow\frac{123}{124}< \frac{124}{125}\)
\(\Rightarrow\frac{123}{124}< \frac{124.1001}{125.1001}\)
\(\Rightarrow\frac{123}{124}< \frac{124124}{125125}\)
ngu quá
ông giỏi toán thế
Nguyễn Thị Mai
chuyên toán mà
\(\frac{12}{13}và\frac{1212}{1212}\)
\(\frac{1212}{1313}=\frac{12.101}{13.101}\) \(\Rightarrow\) \(\frac{12}{13}\)
Mà \(\frac{12}{13}=\frac{12}{13}\) \(\Rightarrow\) \(\frac{12}{13}=\frac{1212}{1313}\)
\(\frac{123}{124}\) và \(\frac{124124}{125125}\)
\(1-\frac{123}{124}\) = \(\frac{1}{124}\)
\(1-\frac{124124}{125125}\) = \(\frac{1}{125}\)
Mà \(\frac{1}{124}>\frac{1}{125}\)
\(\Rightarrow\) \(1-\frac{123}{124}\) > \(1-\frac{124}{125}\)
\(\Rightarrow\) \(\frac{123}{124}\) <\(\frac{124}{125}\) \(\Rightarrow\) \(\frac{123}{124}\) < \(\frac{124.1001}{125.1001}\)
\(\Rightarrow\) \(\frac{123}{124}< \frac{124124}{125125}\)
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