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a)\(\frac{1}{95}>\frac{1}{96}\Rightarrow\frac{91}{95}>\frac{91}{96}\Rightarrow\frac{93}{95}>\frac{91}{95}>\frac{91}{96}\Rightarrow\frac{93}{95}>\frac{91}{96}\)
b)\(\frac{37}{40}=1-\frac{3}{40}\)\(\frac{43}{46}=1-\frac{3}{46}\)
\(\frac{1}{40}>\frac{1}{46}\Rightarrow\frac{-3}{40}< -\frac{3}{46}\Rightarrow1-\frac{3}{40}< 1-\frac{3}{46}\Rightarrow\frac{37}{40}< \frac{43}{46}\)
c)
\(\frac{27}{55}=\frac{54}{110}< \frac{55}{110}=\frac{1}{2}\)
\(\frac{12}{23}=\frac{24}{46}>\frac{23}{46}=\frac{1}{2}\)
Vậy \(\frac{27}{55}< \frac{1}{2}< \frac{12}{23}\Rightarrow\frac{27}{55}< \frac{12}{23}\)
A=98-97+96-95+......+2-1
A= (98-97)+(96-95)+.....+(2-1)
A= 1+1+..............+1
A= 98
Bài làm
= \(\frac{18.25+9.45.2+3.27.6}{100-99+98-97+96-95+.....+2-1}\)
= \(\frac{18.25+9.2.45+3.6.27}{50}\)
= \(\frac{18.25+18.45+18.27}{50}\)
= \(\frac{18.\left(25+45+27\right)}{50}\)
= \(\frac{18.97}{50}\)
= \(\frac{1746}{50}\)
Đặt biểu thức trên là A ta có:
A = \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{24}\)+ \(\frac{1}{48}\)+ \(\frac{1}{96}\)
A x 3 = \(1\)+ \(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)
A x 3 = \(1\)+ \(1\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{8}\)+ \(\frac{1}{8}\)- \(\frac{1}{16}\)+ \(\frac{1}{16}\)- \(\frac{1}{32}\)
A x 3 = 2 - \(\frac{1}{32}\)= \(\frac{63}{32}\)
A = \(\frac{63}{32}\): 3 = \(\frac{63}{96}\)
Trả lời:
\(y\times\frac{15}{2}-\frac{1}{3}\times\left(\frac{1}{4}+y\right)=96\frac{2}{3}\)
\(\Leftrightarrow y\times\frac{15}{2}-\frac{1}{12}-\frac{1}{3}\times y=\frac{290}{3}\)
\(\Leftrightarrow y\times\left(\frac{15}{2}-\frac{1}{3}\right)=\frac{387}{4}\)
\(\Leftrightarrow y\times\frac{43}{6}=\frac{387}{4}\)
\(\Leftrightarrow y=\frac{27}{2}\)
Vậy \(y=\frac{27}{2}\)
Giả sử a > = b ko làm mất đi tính tổng quát của bài toán.
=> a= m+b (m>=0)
Ta có: \(\frac{a}{b}+\frac{b}{a}\)= \(\frac{b+m}{b}\)+ \(\frac{b}{b+m}\)=1 + \(\frac{m}{b}\)+\(\frac{b}{b+m}\)< 1 + \(\frac{m}{b+m}\)+\(\frac{b}{b+m}\)= 1 + \(\frac{m+b}{b+m}\)= 1+1=2
Vậy a/b + b/a < 2 (ĐPCM)
\(\frac{379-x}{12}=21,7\) Đ/S : 118,6
\(\frac{95}{96}=1-\frac{1}{96}\)
\(\frac{96}{97}=1-\frac{1}{97}\)
Vì \(\frac{1}{96}>\frac{1}{97}\)
Suy ra \(1-\frac{1}{96}< 1-\frac{1}{97}\)
\(\frac{95}{96}=\frac{9215}{9312}\) (1)
\(\frac{96}{97}=\frac{9216}{9312}\)(2)
từ (1) và (2) suy ra \(\frac{95}{96}< \frac{96}{97}\)