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a,\(=x^{1.2.3....49.50}\)
b,\(\Rightarrow\)2Q\(=2+2^2+2^3+...+2^{50}\)
2Q-Q\(=2+2^2+2^3+...+2^{50}-1-2-2^2-...-2^{49}\)
Q\(=2^{50}-1\)
Q+1=\(2^{50}\)
Mà Q+1=\(2^n\)
\(2^{50}=2^n\Rightarrow n=50\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x1+2+3+...+50=x1275
b)Q=1+2+22+23+....+249
2Q=2+22+23+...+250
2Q-Q=250-1
Q+1=250 Mà Q+1=2n suy ra 250=2n
Vậy n=50
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x.x^2.x^3.....x^{50}\)
\(=x^{1+2+...+50}\)
b) \(Q=1+2+2^2+...+2^{49}\)
\(2Q=2+2^2+...+2^{50}\)
\(2Q-Q=2+2^2+...+2^{50}-1-2-2^2-...-2^{49}\)
\(Q=2^{50}-1\)
Thay \(Q=2^{50}-1\)vào Q + 1 = 2n ta có:
\(2^{50}-1+1=2^n\)
\(\Rightarrow2^{50}=2^n\)
\(\Rightarrow n=50\)
a)x . x2 . x3 . x4 . x5 ...x49 . x50
= x1 . x2+3+4+5+....49+50
Ta có :
Số số hạng là : ( 50 - 2 ) : 1 + 1 = 49 ( số hạng )
Tổng là : ( 50 + 2 ) . 49 : 2 = 1274
= x1 . x1274
= x1275
câu b tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(\text{a) }x.x^2.x^3.x^4.x^5.....x^{49}.x^{50}\)
\(=x^{1+2+3+4+5+...+49+50}\)
\(=x^{\frac{51.50}{2}}\)
\(=x^{1275}\)
\(\text{b) Ta có:}\)
\(4^{15}=\left(2^2\right)^{15}=2^{2.15}=2^{30}\)
\(8^{11}=\left(2^3\right)^{11}=2^{3.11}=2^{33}\)
\(\text{Vì }2^{30}< 2^{33}\text{ nên }4^{15}< 8^{11}\)
Bài 2: Tìm x
\(\left(x-1\right)^4:3^2=3^6\)
\(\Rightarrow\left(x-1\right)^4=3^6\times3^2\)
\(\Rightarrow\left(x-1\right)^4=3^8\)
\(\Rightarrow\left(x-1\right)^4=3^{2.4}\)
\(\Rightarrow\left(x-1\right)^4=\left(3^2\right)^4\)
\(\Rightarrow x-1=9\)
\(\Rightarrow x=10\)
Bài 3 và bài 4 mk làm sau
Bài 1 : a) \(x.x^2.x^3.x^4.....x^{49}.x^{50}=x^{1+2+3+...+49+50}\) (Dễ rồi tự tính)
b) \(\hept{\begin{cases}4^{15}=\left(2^2\right)^{15}=2^{30}\\8^{11}=\left(2^3\right)^{11}=2^{33}\end{cases}}\)Rồi tự so sánh đi
Bài 2 :
\(\left(x-1\right)^4\div3^2=3^6\Leftrightarrow\left(x-1\right)^4=3^8=\left(3^2\right)^4=9^4\Leftrightarrow x-1=9\Leftrightarrow x=10\)
Bài 3 :
\(\hept{\begin{cases}27^{15}=\left(3^3\right)^{15}=3^{45}\\81^{11}=\left(3^4\right)^{11}=3^{44}\end{cases}}\) nt
a ,
\(x.x^2.x^3.x^4.x^5......x^{49}.x^{50}.x=x^{24.\left(1+49\right)+51}=x^{1251}\)
a) x . x2 . x3 . ... . x50
= x(1 + 2 + 3 + ... + 50)
= x1275