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a) Ta có:\(\frac{19}{33}\) =\(\frac{38}{66}\); \(\frac{16}{11}\)=\(\frac{96}{66}\); \(\frac{13}{22}\)=\(\frac{39}{66}\)
\(\frac{38}{66}\)<\(\frac{39}{66}\)<\(\frac{96}{66}\)hay \(\frac{19}{33}\)<\(\frac{13}{22}\)<\(\frac{16}{11}\)
Vậy các số hữu tỉ sắp xếp theo thứ tự tăng dần là :\(\frac{19}{33}\);\(\frac{13}{22}\);\(\frac{16}{11}\).
b)Ta có: \(\frac{-18}{12}\)=\(\frac{-630}{420}\); \(\frac{-10}{7}\)=\(\frac{-600}{420}\);\(\frac{-8}{5}\)=\(\frac{-672}{420}\)
\(\frac{-672}{420}\)<\(\frac{-630}{420}\)<\(\frac{-600}{420}\)hay \(\frac{-8}{5}\)<\(\frac{-18}{12}\)<\(\frac{-10}{7}\)
Vậy các số hữu tỉ sắp xếp theo thứ tự tăng dần là: \(\frac{-8}{5}\);\(\frac{-18}{12}\);\(\frac{-10}{7}\).
\(\left|5x-4\right|=\left|x-2\right|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-4=2-x\\5x-4=x-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x=6\\4x=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
Vậy \(x\in\left\{1;\frac{1}{2}\right\}\)
\(\left|5x-4\right|=\left|x-2\right|\)
\(\Rightarrow\orbr{\begin{cases}5x-4=x-2\\5x-4=-\left(x-2\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}5x-x=-2+4\\5x+x=2+4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x=2\\6x=6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)
Vậy,...........
\(\frac{x-3}{5}=\frac{x+4}{-2}\)
=> (x - 3). (-2) = 5(x + 4)
=> -2x + 6 = 5x + 20
=> -2x - 5x = 20 - 6
=> -7x = 14
=> x = 14 : (-7)
=> x = -2
x-3/5=x+4/-2
=> ﴾x ‐ 3﴿. ﴾‐2﴿ = 5﴾x + 4﴿
=> ‐2x + 6 = 5x + 20
=> ‐2x ‐ 5x = 20 ‐ 6 => ‐7x = 14 => x = 14 : ﴾‐7﴿
=> x = ‐2
> =<
\(\frac{2}{13}< ...< ...< ...< ...< ...< \frac{5}{17}\)
\(\frac{10}{65}< ...< ...< ...< ...< ...< \frac{10}{34}\)
Vậy : ta có 5/32;10/63;5/31;10/61;1/6
M=\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{17}{8^2.9^2}+\dfrac{19}{9^2.10^2}\)
=\(\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{17}{64.81}+\dfrac{19}{81.100}\)
=\(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{64}-\dfrac{1}{81}+\dfrac{1}{81}-\dfrac{1}{100}\)
=1-\(\dfrac{1}{100}\)=\(\dfrac{99}{100}\)<\(\dfrac{100}{100}=1\)
A. \(32^9=\left(2^5\right)^9=2^{45}\)
\(^{16^{10}=\left(2^4\right)^{10}=2^{40}}\)
Vì \(^{2^{45}>2^{40}}\)nên \(32^9>16^{10}\)
B. \(5^{300}=5^{3\times100}=\left(5^3\right)^{100}=125^{100}\)
\(3^{500}=3^{5.100}=\left(3^5\right)^{100}=243^{100}\)
Vì \(125^{100}< 243^{100}\) nên \(5^{300}< 3^{500}\)