b: \(=\dfrac{3a-9-2a-6-6}{\left(a+3\right)\left(a-3\right)}=\dfrac{a-15}{a^2-9}\)

x=-18

x =18

x=-18

\(A=\left(1-\frac{1}{15}\right)\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)...\left(1-\frac{1}{79800}\right)\)

\(A=\frac{14}{15}.\frac{20}{21}.\frac{27}{28}...\frac{209}{210}\)

\(A=\frac{28}{30}.\frac{40}{42}.\frac{54}{56}...\frac{418}{240}\)

\(A=\frac{4.7}{5.6}.\frac{5.8}{6.7}.\frac{6.9}{7.8}...\frac{19.22}{20.21}\)

\(A=\frac{4.5.6...19}{5.6.7...20}.\frac{7.8.9...22}{6.7.8...21}\)

\(A=\frac{4}{20}.\frac{22}{6}\)

\(A=\frac{11}{15}\)

Ta có: \(\left|y+3\right|\ge0\forall y\)

\(\left|2x+y\right|\ge0\forall x,y\)

Do đó: \(\left|y+3\right|+\left|2x+y\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}y+3=0\\2x+y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-3\end{matrix}\right.\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`16^3 = (4^2)^3 = 4^6` 

`b)`

`25^6 = (5^2)^6 = 5^12`

`c)`

`81^5 = (9^2)^5 = 9^10`

`d)`

`27^5 = (3^3)^5 = 3^15`

`e)`

`64^3*16^3`

`= (4^3)^3*(4^2)^3`

`= 4^9*4^6`

`= 4^15`

_____

`@` Nâng lên lũy thừa

CT: `(a^m)^n=a^m*a^n = a^(m*n)`

\(\frac{-216}{-217}=\frac{216}{217}>\frac{-15}{16}\)

\(\frac{-216}{-217}=\frac{216}{217}>0\left(1\right)\)

\(\frac{-15}{16}< 0\left(2\right)\)

\(\text{Từ (1) và (2) }\Rightarrow\frac{-216}{-217}>\frac{-15}{16}\)