Bài 6: Lũy thừa của một số hữu tỉ (tiếp theo...)

`24 xx 5^5 + 5^2 xx 5^3`

`=24 xx 5^5 + 5^5`

`=5^5 (24 +1)`

`=5^5 . 25`

`=5^5 .5^2`

`=5^7`

=24x3125 +\(5^2+5^3\)

=75000+ \(5^5\)

=75000+3125

=78125

\(24\cdot5^5+5^2\cdot5^3\)

\(=24\cdot5^5+5^5\)

\(=5^5\cdot5^2=5^7\)

`(2^39 +2^35): 2^33`

`=2^39 : 2^33 + 2^35 :2^33`

`=2^(39-33)+2^(35-33)`

`=2^6+2^2`

`=68`

Theo mình là 68, còn cách làm bạn làm giống như bạn bên trên nha!

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 \(5^{202}< 2^{505}\)

Làm bài nào vậy bạn

Bài 2: 

a) Ta có: \(2^{30}=8^{10}\)

\(3^{20}=9^{10}\)

mà 8<9

nên \(2^{30}< 3^{20}\)

b) \(5^{202}=25^{101}\)

\(2^{505}=32^{101}\)

mà 25<32

nên \(5^{202}< 2^{505}\)

Bài 6: 

Ta có: \(\left(-\dfrac{1}{9}\right)^{200}=\left(\dfrac{1}{81}\right)^{100}\)

\(\left(-\dfrac{1}{4}\right)^{300}=\left(-\dfrac{1}{64}\right)^{100}\)

mà \(\dfrac{1}{81}>-\dfrac{1}{64}\)

nên \(\left(\dfrac{-1}{9}\right)^{200}>\left(-\dfrac{1}{4}\right)^{300}\)

Bài 5:

a) Ta có: \(x-\dfrac{3}{7}=\dfrac{1}{3}\)

nên \(x=\dfrac{1}{3}+\dfrac{3}{7}\)

hay \(x=\dfrac{16}{21}\)

b) Ta có: \(\dfrac{-5}{3}+\dfrac{1}{4}x=\left(-\dfrac{2}{3}\right)^2\)

\(\Leftrightarrow\dfrac{1}{4}x-\dfrac{5}{3}=\dfrac{4}{9}\)

\(\Leftrightarrow\dfrac{1}{4}x=\dfrac{4}{9}+\dfrac{5}{3}=\dfrac{4}{9}+\dfrac{15}{9}=\dfrac{19}{9}\)

hay \(x=\dfrac{19}{9}:\dfrac{1}{4}=\dfrac{76}{9}\)

c) Ta có: \(\left|2x+\dfrac{3}{4}\right|-6=-5\)

\(\Leftrightarrow\left|2x+\dfrac{3}{4}\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{3}{4}=-1\\2x+\dfrac{3}{4}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-7}{4}\\2x=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)

Ta có:15^x:5^x=81

<=> 3^x.5^x:5^x=81

<=> 3^x=81=3⁴

<=> x=4

15^x:5^x=81

3^.5^x:5^x=81

3^x=81

3^4=81

\(\Leftrightarrow\)\(15^{x}:5^{x}=3^{4}\)

\(\Leftrightarrow\)\(x=4\)

\(\dfrac{12^3\cdot28^2}{18^4\cdot24^3}=\dfrac{2^6\cdot3^3\cdot2^4\cdot7^2}{2^4\cdot3^8\cdot2^9\cdot3^3}=\dfrac{2^{10}\cdot7^2}{2^{13}\cdot3^8}=\dfrac{1}{8}\cdot\dfrac{49}{6561}=\dfrac{49}{52488}\)

\(\dfrac{2^6.3^3.2^2.2^2.7^2}{3^4.2^4.3^4.3^3.2^9}=\dfrac{2^{10}.3^3.7^2}{3^{11}.2^{13}}=\dfrac{7^2}{3^8.2^3}=\dfrac{49}{52488}\)

Bài 3: 

a) Ta có: \(\dfrac{\left(-2\right)^x}{16}=-8\)

\(\Leftrightarrow\left(-2\right)^x=-128\)

\(\Leftrightarrow x=7\)

b) Ta có: \(\left(1-x\right)^3=-64\)

\(\Leftrightarrow1-x=-4\)

\(\Leftrightarrow x=5\)

c) Ta có: \(\dfrac{20^x}{5^x}=256\)

\(\Leftrightarrow4^x=256\)

\(\Leftrightarrow x=4\)

d) Ta có: \(25\le5^x\le5\cdot5^4\)

\(\Leftrightarrow2\le x\le5\)

\(\Leftrightarrow x\in\left\{2;3;4;5\right\}\)

e) Ta có: \(\left|x+0.5\right|-1=0\)

\(\Leftrightarrow\left|x+0.5\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=1\\x+\dfrac{1}{2}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

Bài 3: 

f) Ta có: \(\dfrac{1}{9}\cdot3^4\cdot3^x=3^7\)

\(\Leftrightarrow3^{x+2}=3^7\)

\(\Leftrightarrow x+2=7\)

hay x=5

Bài 4:

a) Ta có: \(\left(-3\right)^4=81\)

\(\left(-3\right)^3=-27\)

mà 81>-27

nên \(\left(-3\right)^4>\left(-3\right)^3\)

3^x+1+4.3^x=567

3^x.3+4.3^x=567

3^x(3+4)=567

3^x.7=567

3^x=567:7=81

3^x=3⁴

=>x=4

Vậy x=4

c) `3^(x+1)+4.3^3=567`

`3^(x+1)+108 = 567`

`3^x . 3 = 459`

`3^x=153`

`3^x = 3^2 . 17`

`=>` Không có `x` thỏa mãn.

.

`P=(x-2)^2+11/5`

Vì `(x-2)^2 >=0 forall x `

`=> (x-2)^2 + 11/5 >= 11/5 forall x`

`<=> P >=11/5`

`=> P_(min)=11/5 <=> x-2=0 <=>x=2`