S=1+5^2+5^4+...+5^200

 =>25S=5^2+5^4+5^6+...+5^202

 =>25S-S=(5^2+5^4+5^6+...+5^202)-(1+5^2+5^4+...+5^200)

 =>24S=5^202-1

 ⇒ S =(5^202−1)÷24

tổng đó bằng 5^202-1

a)  5 4 : 5 2 = 5 4 - 2 = 5 2

b)  11 4 : 11 2 = 11 2

c)  10 7 : 10 2 : 10 3 = 10 2

d)  a 11 : a 7 : a = a 3

52 = 5.5 = 25;

53 = 52.5 = 25.5 = 125;

54 = 53.5 = = 125.5 = 625.

trên đầu bài là giấu phẩy hay giấu nhân thế

\(a,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64,2^7=128,2^8=256,2^9=512,2^{10}=1024\)

\(b,3^2=9,3^3=27,3^4=81,3^5=243\)

\(c,4^2=16,4^3=64,4^4=256\)

\(d,5^2=25,5^3=125,5^4=625\)

a: \(2^2=4\)

\(2^3=8\)

\(2^4=16\)

\(2^5=32\)

\(2^6=64\)

\(2^7=128\)

\(2^8=256\)

\(2^9=512\)

\(2^{10}=1024\)

b: \(3^2=9\)

\(3^3=27\)

\(3^4=81\)

\(3^5=243\)

c: \(4^2=64\)

\(4^3=256\)

\(4^4=1024\)

d: \(5^2=25\)

\(5^3=125\)

\(5^4=625\)

a: \(3\cdot3\cdot3\cdot3\cdot3=3^5\)

b: \(y\cdot y\cdot y\cdot y=y^4\)

c: \(5\cdot p\cdot5\cdot p\cdot2\cdot q\cdot4\cdot q=25\cdot2\cdot4\cdot p^2q^2=2\cdot\left(10qp\right)^2\)

d: \(a\cdot a+b\cdot b+c\cdot c+d\cdot d\cdot d\cdot d=a^2+b^2+c^2+d^4\)

a) 3.3.3.3.3 = 3\(^5\)

b) y.y.y.y = y\(^4\)

c) 5.p.5.p.2.q.4.q = 5\(^2\).p\(^2\).q\(^2\).2\(^3\) 

(2\(^3\) ở đây là vì 2.4 = 2.2.2 = 2\(^3\))

d) a.a + b.b + c.c.c + d.d.d.d = a\(^2\)+b\(^2\)+c\(^3\)+d\(^4\)

bài 1 : Tính

2048 : 2³ = 2048 : 8 = 256

bài 1.1 : Tính

625 : 5² . 40 = 625 : 25 . 40 = 25 . 40 = 1000

Bài 2: 

a: \(=3^{11}\)

giúp mk vs nha , mk đang rất cần

Không có câu nào đúng

\(32\cdot\left(36-17\right)-\left(2\cdot50+52\right)\)

\(=32\cdot19-100-52\)

\(=456\)

Không có câu nào đúng hết bạn ơi

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam