* \(2n=2560\Leftrightarrow n=\dfrac{2560}{2}=1280\) vậy \(n=1280\)

* \(3n=729\Leftrightarrow n=\dfrac{729}{3}=243\) vậy \(n=243\)

* \(4n=256\Leftrightarrow n=\dfrac{256}{4}=64\) vậy \(n=64\)

* \(2.2n=256\Leftrightarrow n=\dfrac{256}{2.2}=\dfrac{256}{4}=64\) vậy \(n=64\)

\(2n=2560\Rightarrow n=1280\)

\(3n=729\Rightarrow n=243\)

\(4n=256\Rightarrow n=64\)

\(2.2n=256\Rightarrow n=64\)

a: \(=\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot x^{n-1+2n+1+1}\cdot y^{2n+1+n+1}=\dfrac{1}{2}x^{3n+1}y^{3n+2}\)

Hệ số: 1/2

Bậc: 6n+3

b: \(=\dfrac{6}{5}\cdot\dfrac{4}{2}\cdot\dfrac{2}{6}\cdot x^{3-n+4-n}\cdot y^{5-n+6-n}=\dfrac{4}{5}x^{7-2n}y^{11-2n}\)

Hệ số: 4/5

bậc: 18-4n

c: \(=\dfrac{4}{7}x^{2-n+2n-3+1}y^{1+n-1+1}=\dfrac{4}{7}x^{n-1}y^{n+1}\)

Hệ số: 4/7

Bậc: 2n

d: =4/7x^(2n+2)*y^(2n+2)

Hệ số: 4/7

Bậc: 4n+4

Ta có : 

4 . 10²ⁿ + 4 . 10ⁿ + 1

= 4 . 10ⁿ . 10² + 4 . 10ⁿ + 1

= 10ⁿ . (4 . 100 + 4) + 1

= 10ⁿ . 404 + 1

Ủng hộ mk nha !!! ^_^

Ta có : 

4 . 10²ⁿ + 4 . 10ⁿ + 1

= 4 . 10ⁿ . 10² + 4 . 10ⁿ + 1

= 10ⁿ . (4 . 100 + 4) + 1

= 10ⁿ . 404 + 1

câu a: 14a + 6b = 84 + ab

<=> 14a + 6b - 84 - ab =0

<=> (14a -84) + (6b -ab)=0

<=> 14( a- 6) - b(a-6)=0

<=> (a - 6)(14-b) = 0

Vậy a=6, b=14

Đặt \(A=\dfrac{n}{4+n^4}\)

\(=\dfrac{n}{n^4+4n^2+4-4n^2}\)

\(=\dfrac{n}{\left(n^2+2\right)^2-\left(2n\right)^2}\)

\(=\dfrac{n}{\left(n^2+2-2n\right)\left(n^2+2+2n\right)}\)

\(\Rightarrow4A=\dfrac{4n}{\left(n^2-2n+2\right)\left(n^2+2n+2\right)}\)

\(=\dfrac{1}{n^2-2n+2}-\dfrac{1}{n^2+2n+2}\)

Đặt \(P=\dfrac{1}{4+1^4}+\dfrac{3}{4+3^4}+...+\dfrac{2n-1}{4+\left(2n-1\right)^4}\)

\(\Rightarrow4P=\dfrac{4}{4+1^4}+\dfrac{12}{4+3^4}+...+\dfrac{4\left(2n-1\right)}{4+\left(2n-1\right)^4}\)

\(=\dfrac{1}{1^2-2\times1+2}-\dfrac{1}{1^2+2\times1+2}\)

\(+\dfrac{1}{3^2-2\times3+2}-\dfrac{1}{3^2+2\times3+2}+...+\)

\(\dfrac{1}{\left(2n-1\right)^2-2\left(2n-1\right)+2}-\dfrac{1}{\left(2n-1\right)^2+2\left(2n-1\right)+2}\)

\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{17}+...+\)

\(\dfrac{1}{\left(2n-1\right)^2-2\left(2n-1\right)+2}-\dfrac{1}{4n^2-4n+1+4n-2+2}\)

\(=1-\dfrac{1}{4n^2+1}\)

\(\Rightarrow P=\dfrac{1}{4}-\dfrac{1}{4\left(4n^2+1\right)}\)

a) \(A=2^{n-1}+2.2^{n+3}-8.2^{n-4}-16.2^n\)

\(=2^{n-1}+2^{n+3+1}-2^{n-4+3}-2^{n+4}\)

\(=2^{n-1}+2^{n+4}-2^{n-1}-2^{n+4}\)

\(=0\)

b) \(B=\left(3^{n+1}-2.2^n\right)\left(3^{n+1}+2.2^n\right)-3^{2n+2}+\left(8.2^{n-2}\right)^2\)

\(=\left(3^{n+1}-2^{n+1}\right)\left(3^{n+1}-2^{n+1}\right)-3^{2n+2}+2^{2n+2}\)

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

\(=0\)

mẫu các phân số này có dạng a⁴ + 4 = a⁴ + 4a² + 4 - 4a² = (a² - 2a + 2)(a² + 2a + 2)

do đó các phân số sẽ biến đổi như sau:

\(\frac{a}{4+a^4}=\frac{a}{\left(a^2-2a+2\right)\left(a^2+2a+2\right)}=\frac{1}{4}\frac{4a}{\left(a^2-2a+2\right)\left(a^2+2a+2\right)}\)

\(=\frac{1}{4}\left(\frac{1}{a^2-2a+2}-\frac{1}{a^2+2a+2}\right)\)

do đó biểu thức M = \(\frac{1}{4}\left(\frac{1}{1}-\frac{1}{\left(2n-1\right)^2+2\left(2n-1\right)+2}\right)=\frac{n^2}{4n^2+1}\)

Mẫu của các phân số có dạng : \(a^4+4=a^4+4a^2+4-4a^2=\left(a^2+2\right)^2-\left(2a\right)^2=\left(a^2+2-2a\right)\left(a^2+2+2a\right)\)

Do đó các phân số biến dổi như sau:

\(\dfrac{a}{a^4+4}=\dfrac{a}{\left(a^2+2-2a\right)\left(a^2+2+2a\right)}=\dfrac{1}{4}.\dfrac{4a}{\left(a^2+2-2a\right)\left(a^2+2+2a\right)}\)

Đặt biểu thức trên là M nhé!!!

Vậy M=\(M=\dfrac{1}{4}\left(\dfrac{1}{1}-\dfrac{1}{\left(2n-1\right)^2+2\left(2n-1\right)+2}\right)=\dfrac{n^2}{4n^2+1}\)

Bài này của lớp 9 á nha bạn!!! Em mới học lớp 6 à năm nay lên 7. Do thầy dạy trước nên có gì sai sót thì bỏ qua nhé!!!

a,

\(A=2^{n-1}+2.2^{n+3}-8.2^{n-4}-16.2^n\)

\(=2^{n-1}+2^{n+3+1}-2^{n-4+3}-2^{n+4}\)

\(=2.2^{n-1}+2.2^{n+4}=2^n+2^{n+5}\)

b,

\(B=\left(3^{n+1}-2.2^n\right)\left(3^{n+1}+2.2^n\right)-3^{2n+2}+\left(8.2^{n-2}\right)^2\)

\(=\left(3^{n+1}\right)^2-\left(2.2^n\right)^2-\left(3^{n+1}\right)^2+\left(2^{n-2+3}\right)^2\)

\(=-2^{n+1}+2^{n+1}=0\)