Tính: \(\left(0,25\right)^4.1024\)
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\(0,25^4.1024\\ =\left(\dfrac{1}{4}\right)^4.1024\\ =\dfrac{1}{256}.1024\\ =4\)
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/ \(\left(\dfrac{1}{5}\right)^5.5^5=\left(\dfrac{1}{5}.5\right)^5=1^5=1\)
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
(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 = 23.8 = (23)8 = 88
316 = 32.8 = (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
\(a,\left(\dfrac{1}{7}\right)^7.7^7=\left(\dfrac{1}{7}.7\right)^7=1\)
\(b,\left(0,125\right)^3.512=\left(0,125\right)^3.8^3=\left(0,125.8\right)^3=1^3=1\)\(c,\left(0,25\right)^4.1024=\left(\left(0,25\right)^2\right)^2.32^2=\left(\dfrac{1}{6}\right)^2.32^2=\left(\dfrac{1}{6}.32\right)^2=2^2=4\)
\(d,\dfrac{90^3}{15^3}=\left(\dfrac{90}{15}\right)^3=6^3=216\)
\(\left(0,25\right)^4.1024\)
\(=\left(\frac{1}{4}\right)^4.32^2\)
\(=\left[\left(\frac{1}{4}\right)^2\right]^2.32^2\)
\(=\left(\frac{1}{16}\right)^2.32^2\)
\(=\left(\frac{1}{16}.32\right)^2\)
\(=2^2\)
\(=4.\)
Chúc bạn học tốt!
Ta có: \(\left(0,25\right)^4.1024\)
\(=\left(\frac{1}{4}\right)^4.32^2\)
\(=\left[\left(\frac{1}{4}\right)^2\right]^2.32^2\)
\(=\left(\frac{1}{16}\right)^2.32^2\)
\(=\left(\frac{1}{16}.32\right)^2\)
\(=2^2\)
\(=4\)
Chúc bạn hok tốt!!! Đặng Quốc Huy