a) 27 × 10 = 270;    72 × 100 = 7200;    14 × 1000 = 14000

86 × 10 = 860;     103 × 10 = 10300;    452 × 1000 = 452000

358 × 10 = 3580;     1977 × 100 = 197700;    300 × 1000 = 300000

b) 80 :10 = 8;     400 : 100 = 4    ;6000 : 1000 = 6

300 : 10 = 30;     4000 : 100 = 40    ; 60000 : 1000 = 60

2000 :10 = 200;     40000 :100 = 400    ;600000 : 1000 = 600

c) 64 × 10 = 640;    32 × 100 = 3200;    95 × 1000 = 95000

640 : 10 = 64;     3200 : 100 = 32;    95000 : 1000 = 95

a) Ta có \(0,625^{200}=\left(\dfrac{5}{8}\right)^{200}\) và \(0,5^{1000}=\left(\dfrac{1}{2}\right)^{1000}=\left(\dfrac{1}{2}\right)^{5.200}\) \(=\left[\left(\dfrac{1}{2}\right)^5\right]^{200}\) \(=\left(\dfrac{1}{32}\right)^{200}\). Mà hiển nhiên \(\left(\dfrac{5}{8}\right)^{200}>\left(\dfrac{1}{32}\right)^{200}\) nên suy ra \(0,625^{200}>0,5^{1000}\)

b) Ta thấy \(\left(-32\right)^{27}< 0\) trong khi \(\left(-27\right)^{32}>0\) nên đương nhiên \(\left(-32\right)^{27}< \left(-27\right)^{32}\)

c) Ta thấy \(-\dfrac{3}{2}>-2\) nên \(\left(-\dfrac{3}{2}\right)^5>\left(-2\right)^5\)

1000 : [30 + (2^x - 6)] = 3² + 4²

= 1000 : [ 30 + (2^x-6)] = 9 + 16 = 25

= 30 + (2^x - 6) = 1000 : 25 = 40

= 2^x - 6 = 40 - 30 = 10

2^x = 10 + 6 = 16

⇒ 2^x = 2⁴

⇒ x = 4

\(\Leftrightarrow2^x+24=40\)

hay x=4

\(1000:\text{[}30+\left(2^x-6\right)\text{]}=3^2+4^2\)

\(1000:\text{[}30+\left(2^x-6\right)\text{]}=9+16\)

\(1000:\text{[}30+\left(2^x-6\right)\text{]}=25\)

\(\text{ }30+\left(2^x-6\right)\text{ }=40\)

\(2^x-6=10\)

\(2^x=16\)

\(=>2^x=2^4\)

\(=>x=4\)

\(1000:\left[30+\left(2^x-6\right)\right]=3^2+4^2\\ 1000:\left[30+\left(2^x-6\right)\right]=9+16\\ 1000:\left[30+\left(2^x-6\right)\right]=25\\ 30+\left(2^x-6\right)=1000:25\\ 30+\left(2^x-6\right)=40\\ 2^x-6=40-30\\ 2^x-6=10\\ 2^x=10+6\\ 2^x=16\\ 2^x=2^4\\ x=4\)

\(1000\left[32+\left(2x-6\right)\right]=3.2+4.2\)

\(\Leftrightarrow1000\left(32+2x-6\right)=14\)

\(\Leftrightarrow2x+26=\frac{7}{500}\)

\(\Leftrightarrow2x=\frac{-12993}{500}\)

\(\Leftrightarrow x=\frac{-12993}{1000}\)

Phải là 3 mũ 2 cộng với 4 mũ 2 

Đặt \(A=1-3+3^2-3^3+\cdots+\left(-3\right)^{x}\)

=>\(A=\left(-3\right)^0+\left(-3\right)^1+\cdots+\left(-3\right)^{x}\)

=>\(-3\cdot A=\left(-3\right)^1+\left(-3\right)^2+\cdots+\left(-3\right)^{x+1}\)

=>\(-3\cdot A-A=\left(-3\right)^1+\left(-3\right)^2+\cdots+\left(-3\right)^{x+1}-\left(-3\right)^0-\left(-3\right)^1-\cdots-\left(-3\right)^{x}\)

=>\(-4A=\left(-3\right)^{x+1}-1\)

=>\(4A=1-\left(-3\right)^{x+1}\)

=>\(1-\left(-3\right)^{x+1}=1+9^{1000}=1+3^{2000}\)

=>\(-\left(-3\right)^{x+1}=3^{2000}\)

=>\(\left(-3\right)^{x+1}=-3^{2000}\)

=>x∈∅

Công thứ nè bạn : 

(n-1) nhân khoảng cách + số hạng đầu tiên 

Làm tốt nha

Ta có 7=4+3 ; 10=4+2*3 \(\Rightarrow\)số thứ 32 là 4+3*31=97

a) 961

b) 10   

c) 256

d) 29