bạn viết thế ma cũng chẳng hiểu

tớ ko biết

k cho mình nhé

ai nhanhh mk h

Bài 1: \(5^3+3^5\)

=125+243

=368

Bài 2:

a: \(\left(x-1\right)^3=125\)

=>\(\left(x-1\right)^3=5^3\)

=>x-1=5

=>x=6

b: \(\frac{720}{41-\left(2x-5\right)}=2^3\cdot5\)

=>\(\frac{720}{41-\left(2x-5\right)}=8\cdot5=40\)

=>41-(2x-5)=720/40=18

=>2x-5=23

=>2x=28

=>x=14

Bài 3:

a: \(\frac19\cdot3^4\cdot3^{n}=3^7\)

=>\(3^{n+4-2}=3^7\)

=>n+4-2=7

=>n+2=7

=>n=5

b: \(\left(2^2:4\right)\cdot2^{n}=4\)

=>\(\frac44\cdot2^{n}=4\)

=>\(2^{n}=4=2^2\)

=>n=2

(x-4):5=2⁴-3²

(x-4):5=16-9

(x-4):5=7

x-4=7.5

x-4=35

x=35+4

x=39

(2x-9):5=7

2x-9=7.5

2x-9=35

2x=35:-9

2x=-5

x=-5.2

x=-10

4(x-3)=5²-1¹⁰

4(x-3)=25-1

4(x-3)=24

x-3=24:4

x-3=6

x=6+3

x=9

(2x-9):5=7

2x-9=7x5

2x-9=35

2x=35+9 suy ra 2x=44

x=44:2 suy ra x=22

mình chỉ làm được bài này thôi mong cậu thông cảm (^-^)

\(A=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...+\frac{1}{2^{2014}}-\frac{1}{2^{2016}}\)

\(\Rightarrow2^2A=1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+\frac{1}{2^8}-...+\frac{1}{2^{2012}}-\frac{1}{2^{2014}}\)

\(\Rightarrow2^2A+A=1+\left(\frac{1}{2^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^4}-\frac{1}{2^4}\right)+...+\left(\frac{1}{2^{2014}}-\frac{1}{2^{2014}}\right)-\frac{1}{2^{2016}}\)

\(\Rightarrow5A=1-\frac{1}{2^{2016}}< 1\Rightarrow A< \frac{1}{5}=0,2\)

đây là toán lớp 2 hả?

a, S = 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰. 2S = 2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁰⁰ + 2¹⁰¹ => 2S - S = S = (2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁰⁰ + 2¹⁰¹) - (2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰) = 2¹⁰¹ - 2. Vậy S = 2¹⁰¹ - 2. b, S = 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰ = (2 + 2²) + (2³ + 2⁴) + ... + (2⁹⁹ + 2¹⁰⁰) = 2.(1 + 2) + 2³.(1 + 2) + ... + 2⁹⁹.(1 + 2) = (1 + 2).(2 + 2³ + ... + 2⁹⁹) = 3.(2 + 2³ + ... + 2⁹⁹) => S ⋮ 3. Vậy S ⋮ 3 (đpcm)

A= \(2^0+2^1+2^2+...+2^{50}\)

\(\Rightarrow\)2A =2(\(2^0+2^1+2^2+...+2^{50}\))

\(\Rightarrow\)2A= \(2+2^2+2^3+2^4+...+2^{51}\)

\(\Rightarrow\)2A-A= (\(2+2^2+2^3+2^4+...+2^{51}\))-(\(2+2^2+2^3+2^4+...+2^{50}\))

\(\Rightarrow\)A= \(2^{51}-1\)