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a, ta có :
\(\hept{\begin{cases}\frac{6}{7}< 1\\\frac{11}{10}>1\end{cases}}\Rightarrow\frac{6}{7}< \frac{11}{10}\)
b, ta có :
\(\hept{\begin{cases}\frac{-5}{7}< 0\\\frac{2}{7}>0\end{cases}}\Rightarrow\frac{-5}{7}>\frac{2}{7}\)
c, ta có :
\(\hept{\begin{cases}\frac{419}{-723}< 0\\\frac{-697}{-313}>0\end{cases}}\Rightarrow\frac{419}{-723}< \frac{-679}{-313}\)
\(1)\)
\(a)(-50).37=-1850\)
\(b)(-20).(-14).25=280.25=7000\)
\(c)40.(-30).12=-1200.12=-14400\)
\(2)\)
\(a)\)
\(-40.[(-15)+30)]\)
\(=-40.15\)
\(=-600\)
\(b)\)
\(-5 . [(-27) - (-7)]\)
\(=-5 . [(-27)+7]\)
\(=-5 . (-20)\)
\(=100\)
\(c)\)
\(-10 . [(-28) + (-22) - 10]\)
\(=-10 . [-50 - 10]\)
\(=-10.(-60)\)
\(=600\)
a) \(27^{64}:81^{20}=3^{192}:3^{80}=3^{112}\)
b) \(\left(\dfrac{1}{8}\right)^{20}:\left(\dfrac{1}{16}\right)^9=\left(\dfrac{1}{2}\right)^{60}:\left(\dfrac{1}{2}\right)^{36}=\left(\dfrac{1}{2}\right)^{24}\)
c) \(\dfrac{1}{3}:\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{5}{3}-\dfrac{1}{6}=\dfrac{10}{6}-\dfrac{1}{6}=\dfrac{9}{6}=\dfrac{3}{2}\)
A=7*(1/3*13+1/13*23+1/23*33+1/33*43+1/43*53+1/53*63)
A=7/10(1/3-1/13+1/13-1/23+1/23-1/33+1/33-1/43+1/43-1/53+1/53-1/63)
A=7/10*(1/3-1/63)
A=7/10*20/63
A=2/9
a. \(\frac{6}{7}=\frac{60}{70};\frac{11}{10}=\frac{77}{70}\)
\(\frac{77}{70}>\frac{60}{70}\rightarrow\frac{11}{10}>\frac{6}{7}\)
b. \(\frac{-5}{17}< 0;\frac{2}{7}>0\rightarrow\frac{2}{7}>\frac{-5}{17}\)
c. \(\frac{419}{-723}< 0;\frac{-697}{-313}=\frac{697}{313}>0\Rightarrow\frac{-697}{-313}>\frac{419}{-723}\)
a)Ta có : 6/7 < 1
11/10 >1
Suy ra : 6/7 < 11/10
B) -5/17 < 0 và 2/7 >0
Suy ra -5/17 < 2/7
c) câu c mk ko bt xl bn nha
Ta có :
\(\frac{1}{51}\)> \(\frac{1}{100}\)
\(\frac{1}{52}\)> \(\frac{1}{100}\)
...
\(\frac{1}{99}\)> \(\frac{1}{100}\)
\(\frac{1}{100}\)= \(\frac{1}{100}\)
=> S > 50 x \(\frac{1}{100}\)
=> S > \(\frac{50}{100}\)= \(\frac{1}{2}\)
Vậy S > \(\frac{1}{2}\)
\(S=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
Ta có \(\frac{1}{51}>\frac{1}{100}\)
\(\frac{1}{52}>\frac{1}{100}\)
...
\(\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)
( có 50 phân số)
\(\Rightarrow S>50.\frac{1}{100}\)
\(\Rightarrow S>\frac{1}{2}\)
Vậy...
