`@` `\text {Ans}`

`\downarrow`

\(\dfrac{2^8-2^3}{2^5-1}=\dfrac{2^3\left(2^5-1\right)}{2^5-1}=\dfrac{2^3}{1}=2^3=8\)

_____

\(\dfrac{4^8\cdot9^4}{6^6\cdot8^3}\)

`=`\(\dfrac{\left(2^2\right)^8\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}\)

`=`\(\dfrac{2^{16}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)

`=`\(\dfrac{2^{16}\cdot3^8}{2^{15}\cdot3^6}\)

`=`\(\dfrac{3^2}{2}\) `=`\(\dfrac{9}{2}\)

______

\(\dfrac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}\)

`=`\(\dfrac{\left(3^3\right)^4\cdot2^3-3^{10}\cdot\left(2^2\right)^3}{2^4\cdot3^4\cdot\left(3^2\right)^3}\)

`=`\(\dfrac{3^{12}\cdot2^3-3^{10}\cdot2^6}{2^4\cdot3^4\cdot3^6}\)

`=`\(\dfrac{3^{10}\cdot\left(3^2\cdot2^3-2^6\right)}{3^{10}\cdot2^4}\)

`=`\(\dfrac{72-2^6}{2^4}=\dfrac{8}{16}=\dfrac{1}{2}\)

\(\dfrac{2^8-2^3}{2^5-1}=\dfrac{2^3\left(2^5-1\right)}{2^5-1}=2^3=8\)

\(\dfrac{4^8.9^4}{6^6.8^3}=\dfrac{2^{16}.3^8}{2^6.3^6.2^9}=2.3^2=18\)

\(\dfrac{27^4.2^3-3^{10}.4^3}{6^4.9^3}=\dfrac{3^{12}.2^3-3^{10}.2^6}{2^4.3^4.3^6}=\dfrac{2^3.3^{10}.\left(3^2-2^3\right)}{2^4.3^{10}}=\dfrac{9-8}{2}=\dfrac{1}{2}\)

\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{32768.6581}{46656.512}=\frac{215646208}{23887872}\)

                                             \(=\frac{6581}{729}\)

\(nha^{ }\)

\(chucbnhocgioi!\)

giúp mk ik, năn nỉ đó.

\(=\dfrac{2^{19}\cdot3^9-3\cdot3^8\cdot2^{18}\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}=\dfrac{-3^{10}\cdot2^{18}}{2^{19}\cdot3^9\cdot7}=-\dfrac{3}{14}\)

m.n ơi giúp mk 1 hoặc 2 câu đc ko ạ mk cần gấp lắm mà mk ko bt cách lm

câu 2a) xét (x-1)²> hoặc = 0

(x-1)²+(y+1)²> hoặc bằng 0

(x-1)²+(y+1)²+3> hoặc =3

=> GTNN của biểu thức trên là 3

GIÚP minh vs mai mình nộp rui!!!!!!!!!!!!!!!!!!!!@@@@@@@@@@

a) ĐKXĐ: 

\(\left\{{}\begin{matrix}x^2-9\ne0\\x+3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne-3\end{matrix}\right.\Leftrightarrow x\ne\pm3\) 

b) \(A=\dfrac{x+15}{x^2-9}-\dfrac{2}{x+3}\)

\(A=\dfrac{x+15}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{x+15-2x+6}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{21-x}{\left(x+3\right)\left(x-3\right)}\)

c) Thay x = - 1 vào A ta có: 

\(A=\dfrac{21-\left(-1\right)}{\left(-1+3\right)\left(-1-3\right)}=\dfrac{21+1}{2\cdot-4}=\dfrac{22}{-8}=-\dfrac{11}{4}\)