\(\frac{16}{2^n}=2\)=> \(2^n=16:2=8\)=> \(n=3\)

b) \(\frac{\left(-3\right)^n}{81}=-27\)=> \(\left(-3\right)^n=-27.81=-2187\)=> n=7

c) \(8^n.2^n=16^n\)

a) \(\frac{16}{2^n}=2\)

\(2^n=\frac{16}{2}=8\)

\(2^n=2^3\)

\(\Rightarrow n=3\)

b) \(\frac{\left(-3\right)^n}{81}=-27\)

\(\left(-3\right)^n=\left(-27\right).81=-2187\)

\(\left(-3\right)^n=\left(-3\right)^7\)

\(\Rightarrow n=7\)

c) Bạn viết thiếu đề nhé !

\(8^n.2^n=?\)

Cảm ơn hai bạn nha

2ⁿ : 2⁴ = 1

=> 2^n-4 = 1 = 2⁰

=> n - 4 = 0

=> n = 0 + 4

=> n = 4

2ⁿ:2⁴=1

=>2ⁿ=1.2⁴

=>2ⁿ=2⁴

=>n=4

a) Đặt A = \(6^5.5-3^5\)

\(=\left(2.3\right)^5.5-3^5\)

\(=2^5.3^5.5-3^5\)

\(=3^5.\left(2^5.5-1\right)\)

\(=3^5.\left(32.5-1\right)\)

\(=3^5.159\)

\(=3^5.3.53⋮53\)

Vậy \(A⋮53\)

b) Đặt \(B=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\)

\(=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{119}.\left(1+2\right)\)

\(=2.3+2^3.3+...+2^{119}.3\)

\(=3.\left(2+2^3+...+2^{59}\right)⋮3\)

Vậy \(B⋮3\)

\(B=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2\right)+3^4.\left(1+2+2^2\right)+...+2^{118}.\left(1+2+2^2\right)\)

\(=2.7+2^4.7+...+2^{118}.7\)

\(=7.\left(2+2^4+...+2^{118}\right)⋮7\)

Vậy \(B⋮7\)

\(B=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)

\(+...+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2+2^3+2^4\right)+2^6.\left(1+2+2^2+2^3+2^4\right)\)

\(+2^{116}.\left(1+2+2^2+2^3+2^4\right)\)

\(=2.31+2^6.31+...+2^{116}.31\)

\(=31.\left(2+2^6+...+2^{116}\right)⋮31\)

Vậy \(B⋮31\)

\(B=\left(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}+2^{13}+2^{14}+2^{15}+2^{16}\right)\)

\(+...+\left(2^{113}+2^{114}+2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)+2^9.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)

\(+...+2^{113}.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)

\(=2.255+2^9.255+...+2^{113}.255\)

\(=255.\left(2+2^9+...+2^{113}\right)\)

\(=17.15.\left(2+2^9+...+2^{113}\right)⋮17\)

Vậy \(B⋮17\)

c) Đặt C = \(3^{4n+1}+2^{4n+1}\)

Ta có:

\(3^{4n+1}=\left(3^4\right)^n.3\)

\(2^{4n}=\left(2^4\right)^n.2\)

\(3^4\equiv1\left(mod10\right)\)

\(\Rightarrow\left(3^4\right)^n\equiv1^n\left(mod10\right)\equiv1\left(mod10\right)\)

\(\Rightarrow3^{4n+1}\equiv\left(3^4\right)^n.3\left(mod10\right)\equiv1.3\left(mod10\right)\equiv3\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(3^{4n+1}\) là \(3\)

\(2^4\equiv6\left(mod10\right)\)

\(\Rightarrow\left(2^4\right)^n\equiv6^n\left(mod10\right)\equiv6\left(mod10\right)\)

\(\Rightarrow2^{4n+1}\equiv\left(2^4\right)^n.2\left(mod10\right)\equiv6.2\left(mod10\right)\equiv2\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(2^{4n+1}\) là \(2\)

\(\Rightarrow\) Chữ số tận cùng của C là 5

\(\Rightarrow C⋮5\)

d) Đặt \(D=75+\left(4^{2006}+4^{2005}+4^{2004}+...+1\right).25\)

Đặt \(E=4^{2006}+4^{2005}+4^{2004}+...+1\)

\(\Rightarrow4E=4^{2007}+4^{2006}+4^{2005}+...+4\)

\(\Rightarrow3E=4E-E\)

\(=\left(4^{2007}+4^{2006}+4^{2005}+...+4\right)-\left(4^{2006}+4^{2005}+4^{2004}+...+1\right)\)

\(=4^{2007}-1\)

\(\Rightarrow E=\dfrac{\left(4^{2007}-1\right)}{3}\)

\(\Rightarrow D=75+\dfrac{4^{2007}-1}{3}.25\)

Ta có:

\(4^{2007}=\left(4^2\right)^{1003}.4\)

\(4^2\equiv6\left(mod10\right)\)

\(\left(4^2\right)^{1003}\equiv6^{1003}\left(mod10\right)\equiv6\left(mod10\right)\)

\(\Rightarrow4^{2007}\equiv\left(4^2\right)^{1003}.4\left(mod10\right)\equiv6.4\left(mod10\right)\equiv4\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(4^{2007}\) là 4

2⁴ : (x + 1) + 2 = 6

16 : (x + 1) = 6 - 2

16 : (x + 1) = 4

x + 1 = 16 : 4

x + 1 = 4

x = 4 - 1

x = 3

a.    x=(x-1)^2

b.    câu hỏi chưa xong kìa

(x-1)³\(=\)(x-5)³

\(\Leftrightarrow\)(x-1)³-(x-1)⁵\(=\)0

\(\Leftrightarrow\)(x-1)³\([\)1-(x-1)²\(]\)\(=\)0

\(\Leftrightarrow\)(x-1)³\(=\)0 hoặc 1-(x-1)²\(=\)0

\(\Leftrightarrow\)x-1\(=\)0 hoặc x-1\(=\pm\)1

\(\Leftrightarrow\)x\(=\)1 hoặc x\(=\)2; x\(=\)0

Vậy x\(\in\){1;2;0}

b) (x-1)ⁿ\(=\)(x-1)ⁿ⁺²

\(\Leftrightarrow\)(x-1)ⁿ-(x-1)ⁿ⁺²\(=\)0

\(\Leftrightarrow\)(x-1)ⁿ\([\)1-(x-1)²\(]\)\(=\)0

\(\Leftrightarrow\)(x-1)ⁿ\(=\)0 hoặc (x-1)²\(=\)1

\(\Leftrightarrow\)x\(=\)1 hoặc x\(=\)2; x\(=\)0

Vậy x\(\in\){1;2;0}

a, đề k rõ :v

b, (x + 2)⁴ = 625

=> (x + 2) = (+5)⁴  

=> x + 2 = + 5

=> x = 3 hoặc x = -7

vậy_

a, \(4^{n+2}\) = 64

\(4^{n+2}\) = \(4^3\)

n + 2 = 3

n = 3 -2

n = 1

a, \(\left(x+2\right)^4\) = 625

\(\left(x+2\right)^4\) = \(5^4\)

x +2 = 5

x= 5 - 2

x= 3

Tk mk nha

(x+2)⁴=625

(x+2)⁴=5⁴

x+2=5(vì  số mũ 4>0)

x=3

kb nha

a) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}.\)

\(\Rightarrow\frac{a}{c}=\frac{a-b}{c-d}\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)

b) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a^n}{c^n}=\frac{b^n}{d^n}=\frac{\left(a-b\right)^n}{\left(c-d\right)^n}\)(*)

mà \(\frac{a^n}{c^n}=\frac{b^n}{d^n}=\frac{a^n-b^n}{c^n-d^n}\)

Từ (*) \(\Rightarrow\frac{a^n-b^n}{c^n-d^n}=\frac{\left(a-b\right)^n}{\left(c-d\right)^n}\)

c) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}=\frac{3a-b}{3c-d}\)

\(\Rightarrow\frac{a}{c}=\frac{3a+b}{3c+d}=\frac{3a-b}{3c-d}\)

...

phần c mk ko bk xl bn nha! nom giùm mk đề