\(0,25^4.1024\\ =\left(\dfrac{1}{4}\right)^4.1024\\ =\dfrac{1}{256}.1024\\ =4\)

a) \(\left(\dfrac{1}{5}\right)^5.5^5=1\)

b) \(\left(0,125\right)^3.512=1\)

c) \(\left(0,25\right)^4.1024=4\)

a) (1/5)^5 . 5^5 = (1/5. 5)^5 = 1^5= 1

b) (0,125)^3. 512= (0,125)^3 . 8^3 = (0,125. 8)^3 = 1^3= 1

c) (0,25)^4. 1024= [(0,25)^2]^2. 32^2= (1/6)^2. 32^2=(1/6.32)^2= (32/6)^2 =2^2= 4

a, \(\left(\dfrac{1}{5}\right)^5.5^5\) =0,00032. 3125= 1

b, \(\left(0,125\right)^3.512\)= 0.001953125. 512=1

c, \(\left(0,25\right)^4.1024\) =0,00390625.1024=4

a/ \(\left(\dfrac{1}{5}\right)^5.5^5=\left(\dfrac{1}{5}.5\right)^5=1^5=1\)

b/ \(\left(0,125\right)^3.512=\left(0,125\right).8^3=\left(0,125.8\right)^3=1^3=1\)

c/ \(\left(0,25\right)^4.1024=\left(0,25^2\right)^2.32^2=\left(\dfrac{1}{6}\right)^2.32^2=\left(\dfrac{1}{6}.32\right)^2=16^2\)

\(a,\left(\dfrac{1}{5}\right)^5.5^5=\left(\dfrac{1}{5}.5\right)^5=1^5=1\)

\(b,\left(0,125\right)^3.512=\left(0,125\right)^3.8^3=\left(0,125.8\right)^3=1^3=1\)

\(c,\left(0,25\right)^4.1024=\left(0,25\right)^4.4^4.4=\left(0,25.4\right)^4.4=1^4.4=1.4=4\)

a) \(\left(\dfrac{1}{5}\right)^5.5^5\)

\(=\dfrac{1}{3125}.3125\)

= 1

b) \(\left(0,125\right)^3.512\)

\(=\dfrac{1}{512}.512\)

= 1

c) \(\left(0,25\right)^4.1024\)

= \(\dfrac{1}{256}.1024\)

= 4

50.

a) \(\left(\dfrac{1}{5}\right)^5.5^5\)

\(=\left(5^{-1}\right)^5.5^5=5^{-5}.5^5=5^{-5+5}=5^0=1\)

(0,25)4.1024 = (0,25)4.256.4 = (0,25)4.44.4 = (0,25.4)4.4 = 1.4 = 4

1) So sánh

Ta có : 2²⁴ = 2^3.8 = (2³)⁸ = 8⁸

           3¹⁶ = 3^2.8 = (3²)⁸ = 9⁸

Vì 8⁸ < 9⁸

=>  2²⁴ < 3¹⁶ 

2) Tính

\(\left(0,25\right)^4.1024=\left(\frac{1}{4}\right)^4.1024=\frac{1}{4^4}.2^{10}=\frac{1}{\left(2^2\right)^4}.2^{10}=\frac{1}{2^8}.2^{10}=\frac{2^{10}}{2^8}=2^2=4\)

3) Tìm x nguyên

(x - 1)^{x + 2} = (x - 1)^{x + 6}

=> (x - 1)^{x + 6} - (x - 1)^{x + 2} = 0

=> (x - 1)^{x + 2}.[(x - 1)⁴ - 1] = 0

=> \(\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^4-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1^4\end{cases}\Rightarrow}\orbr{\begin{cases}x-1=0\\x-1=\pm1\end{cases}}}\)

Nếu x - 1 = 0 => x = 1(tm)

Nếu x - 1 = - 1 => x = 0(tm)

Nếu x - 1 = 1 => x = 2(tm)

Vậy \(x\in\left\{1;0;2\right\}\)

Bài 1:Ta có:

2^24=2^(6.4)=64^4

3^16=3^(4.4)=81^4

Bài 2.Ta có:

(0.25)^4=1/4.1/4.1/4.1/4=1/256

=>1/256.1024=4

Bài 3:

Ta có:(x-1)^(x+2)=(x-1)^(x+6)

Chia hai vế cho (x-1)^(x+2),do đó:

1=(x-1)^(x+4)

<=>x-1=1

<=>x=2

Hoặc chia hai vế cho (x-1)^(x+6)

(x-1)^(x-4)=1

<=>x-1=1

<=>x=2

\(a,\left(\dfrac{1}{7}\right)^7.7^7=\left(\dfrac{1}{7}.7\right)^7=1\)

\(b,\left(0,125\right)^3.512=\left(0,125\right)^3.8^3=\left(0,125.8\right)^3=1^3=1\)\(c,\left(0,25\right)^4.1024=\left(\left(0,25\right)^2\right)^2.32^2=\left(\dfrac{1}{6}\right)^2.32^2=\left(\dfrac{1}{6}.32\right)^2=2^2=4\)

\(d,\dfrac{90^3}{15^3}=\left(\dfrac{90}{15}\right)^3=6^3=216\)