Bài 1:
a. $x:(\frac{-5}{9})^8=(\frac{-9}{5})^8$
$x=(\frac{-9}{5})^8.(\frac{-5}{9})^8=(\frac{-9}{5}.\frac{-5}{9})^8=1^8$
$x=1$
b. $(x+5)^3=-27=(-3)^3$
$x+5=-3$
$x=-8$
c.
$(2x+5)^4=4096=8^4=(-8)^4$
$\Rightarrow 2x+5=8$ hoặc $2x+5=-8$
$\Leftrightarrow x=\frac{3}{2}$ hoặc $x=-\frac{13}{2}$
d. $3^{x+1}=243=3^5$
$\Leftrightarrow x+1=5$
$\Leftrightarrow x=4$
e.
$\frac{-32}{(-2)^x}=4$
$(-2)^x=-8=(-2)^3$
$\Leftrightarrow x=3$
f.
$7^{x+2}+2.7^{x-1}=345$
$7^{x-1}(7^3+2)=345$
$7^{x-1}.345=345$
$7^{x-1}=1=7^0$
$\Rightarrow x-1=0\Leftrightarrow x=1$
Bài 2:
Ta thấy:
$2^{30}=(2^3)^{10}=8^{10}< 9^{10}=(3^2)^{10}=3^{20}$
Vậy $2^{30}< 3^{20}$
-------------------------
$5^{202}$ và $2^{505}$
$5^{202}=(5^2)^{101}=25^{101}< 32^{101}=(2^5)^{101}=2^{505}$
Vậy $5^{202}< 2^{505}$
Bài 2:
a) Ta có: \(2^{30}=\left(2^3\right)^{10}=8^{10}\)
\(3^{20}=\left(3^2\right)^{10}=9^{10}\)
mà 8<9
nên \(2^{30}< 3^{20}\)
b) Ta có: \(5^{202}=\left(5^2\right)^{101}=25^{101}\)
\(2^{505}=\left(2^5\right)^{101}=32^{101}\)
mà 25<32
nên \(5^{202}< 2^{505}\)
Bài 1:
a) Ta có: \(x:\left(\dfrac{-5}{9}\right)^8=\left(\dfrac{-9}{5}\right)^8\)
nên \(x=\dfrac{9^8}{5^8}\cdot\dfrac{5^8}{9^8}=1\)
b) Ta có: \(\left(x+5\right)^3=-27\)
nên x+5=-3
hay x=-8
c) Ta có: \(\left(2x+5\right)^4=4096\)
nên \(\left[{}\begin{matrix}2x+5=8\\2x+5=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=-13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{-13}{2}\end{matrix}\right.\)
