a: \(4^{10}\cdot2^{30}=2^{20}\cdot2^{30}=2^{50}\)

b: \(9^{25}\cdot27^4\cdot81^3=3^{50}\cdot3^{12}\cdot3^{12}=3^{74}\)

c: \(25^{50}\cdot125^5=\left(5^2\right)^{50}\cdot\left(5^3\right)^5=5^{115}\)

d: \(64^3\cdot4^8\cdot16^4=\left(4^3\right)^3\cdot4^8\cdot\left(4^2\right)^4=4^9\cdot4^8\cdot4^8=4^{25}\)

e: \(3^8:3^6=3^{8-6}=3^2\)

f: \(2^{10}:8^3=2^{10}:2^9=2\)

g: \(12^7:6^7=\left(\dfrac{12}{6}\right)^7=2^7\)

h: \(21^5:81^3=\dfrac{7^5\cdot3^5}{3^3\cdot27^3}=\dfrac{7^5}{27^3}\)

i: \(4^9:64^2=4^9:\left(4^3\right)^2=4^9:4^6=4^3\)

j: \(2^{25}:32^4=2^{25}:2^{20}=2^5\)

k: \(125^3:25^4=\left(5^3\right)^3:\left(5^2\right)^4=5^9:5^8=5\)

Gọi A là biến cố "Thẻ lấy ra ghi số là ước của 21"

=>A={1;3;7}

=>n(A)=3

\(\Omega=\left\{1;2;3;...;20\right\}\)

=>\(n\left(\Omega\right)=20\)

\(P_A=\dfrac{3}{20}\)

\(A=1+2+2^2+...+2^{99}\\ 2A=2+2^2+2^3+...+2^{100}\\ 2A-A=\left(2+2^2+2^3+...+2^{100}\right)-\left(1+2+2^2+...+2^{99}\right)\\ A=2^{100}-1\)

\(=>A+1=2^{100}-1+1=2^{100}\)

Mà: \(A+1=2^n=>2^n=2^{100}\)

\(=>n=100\)

Có :

9 - x = 15

     x = 9 - 15

     x = - 6

Mà -6 \(\notin\) N ⇒ D = { }

9 - x = 15

=> x = 9 - 15 

=> x = -6

Mà x ∈ N => K có x thỏa mãn 

=> D = ∅

minhf cũng đang hỏi bài này luôn nè.

\(A=x\left(x^2-y\right)-x^2\left(x+y\right)+y\left(x^2-x\right)\\ =x^3-xy-x^3-x^2y+x^2y-xy\\ =-2xy\)

Thay `x=1/2;y=-100` vào A ta có:

\(A=-2\cdot\dfrac{1}{2}\cdot\left(-100\right)=100\)

\(B=\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\\=x^3+3x^2-5x-15-x^3+4x+x^2-4x^2\\ =\left(x^3-x^3\right)+\left(3x^2-4x^2+x^2\right)+\left(-5x+4x\right)-15\\ =-x-15\)

\(\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+8\dfrac{1}{5}=1,2\\\Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=1,2-8\dfrac{1}{5}\\ \Rightarrow \left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=-7\)

Nhận xét:

\(\left\{{}\begin{matrix}\left|2\dfrac{1}{5}-x\right|\ge0,\forall x\\\left|x-\dfrac{1}{5}\right|\ge0,\forall x\end{matrix}\right.\\ \Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|\ge0,\forall x\)
Mà \(-7< 0\) nên:
Không tìm được giá trị \(x\) thỏa mãn đề bài
Vậy...