\(S=a+a^3+...+a^{2n+1}\)

\(S.a^2=a^3+a^5+...+a^{2n+1}+a^{2n+3}\)

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

\(\Rightarrow S=\dfrac{a^{2n+3}-a}{a^2-1}\)

\(S_1=1+a^2+...+a^{2n}\)

\(S_1.a^2=a^2+a^4+...+a^{2n}+a^{2n+2}\)

\(\Rightarrow S_1\left(a^2-1\right)=a^{2n+2}-1\)

\(\Rightarrow S_1=\dfrac{a^{2n+2}-1}{a^2-1}\)

a/ \(I=lim\dfrac{5^n+2^n}{3^n+4^n}=lim\dfrac{1+\left(\dfrac{2}{5}\right)^n}{\left(\dfrac{3}{5}\right)^n+\left(\dfrac{4}{5}\right)^n}=\dfrac{1}{0}=+\infty\)

b/ \(I=lim\dfrac{\sqrt{n^3+2n}+3n}{n+\sqrt{n^2+1}}=lim\dfrac{\sqrt{\dfrac{n^3}{n^3}+\dfrac{2n}{n^3}}+\dfrac{3n}{n^{\dfrac{3}{2}}}}{\dfrac{n}{n^{\dfrac{3}{2}}}+\sqrt{\dfrac{n^2}{n^3}+\dfrac{1}{n^3}}}=\dfrac{1}{0}=+\infty\)

c/ \(I=lim\left[n\left(\sqrt{2+\dfrac{n}{n^2}}-\sqrt{1+\dfrac{2n}{n^2}+\dfrac{3}{n^2}}\right)\right]=+\infty.\left(\sqrt{2}-1\right)=+\infty\)

a) bằng 9 nha bạn

b) thì mik ko bik làm.

Đúng thì bạn tim giúp mik nha bạn. Thx bạn

\(a=\lim\dfrac{\dfrac{1}{n}+\dfrac{1}{n^2}}{1+\dfrac{2}{n}}=\dfrac{0}{1}=0\)

\(b=\lim n^3\left(-2+\dfrac{1}{n}+\dfrac{2}{n^3}\right)=+\infty.\left(-2\right)=-\infty\)

\(c=\lim\dfrac{\sqrt{9-\dfrac{1}{n}-\dfrac{1}{n^2}}}{4-\dfrac{2}{n}}=\dfrac{\sqrt{9}}{4}=\dfrac{3}{4}\)

\(d=\lim\dfrac{\left(\dfrac{3}{4}\right)^n+5}{1+\left(\dfrac{2}{4}\right)^n}=\dfrac{5}{1}=5\)

a) \(3{x^5}.5{x^8} = 3.5.{x^5}.{x^8} = 15.{x^{5 + 8}} = 15.{x^{13}}\).

b) \( - 2{x^{m + 2}}.4{x^{n - 2}} =  - 2.4.{x^{m + 2}}.{x^{n - 2}} =  - 8.{x^{m + 2 + n - 2}} =  - 8.{x^{m + n}}\) (m, n \(\in\) N; n > 2).

a) x-y

b) 8x²+x

c) 1

a) \({x^5}:{x^3} = {x^{5 - 3}} = {x^2}\);

b) \((4{x^3}):{x^2} = (4:1).({x^3}:{x^2}) = 4x\);

c) \((a{x^m}):(b{x^n}) = (a:b).({x^m}:{x^n}) = (a:b).{x^{m - n}}\)(a ≠ 0; b ≠ 0; m, n \(\in\) N, m ≥ n).

a)5!=1.2.3.4.5=120

4!-3!=3!(4-1)=6.3=18

a) 5! = 1.2.3.4.5= 120

b) 4! - 3!= 3! (4-1) = 6.3 = 18

Đáp án là 120 , 18

Bài 2:

a) \(\left(x+5\right)^2=x^2+10x+25\)

b) \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)

c) \(\left(2u+3v\right)^2=4u^2+12uv+9v^2\)

d) \(\left(-\dfrac{1}{8}a+\dfrac{2}{3}bc\right)^2=\dfrac{1}{64}a^2-\dfrac{1}{6}abc+\dfrac{4}{9}b^2c^2\)

e) \(\left(\dfrac{x}{y}-\dfrac{1}{z}\right)^2=\dfrac{x^2}{y^2}-\dfrac{2x}{yz}+\dfrac{1}{z^2}\)

f) \(\left(\dfrac{mn}{4}-\dfrac{x}{6}\right)\left(\dfrac{mn}{4}+\dfrac{x}{6}\right)=\dfrac{m^2n^2}{16}-\dfrac{x^2}{36}\)

Bài 1:

$M=(2a+b)^2-(b-2a)^2=[(2a+b)-(b-2a)][(2a+b)+(b-2a)]$

$=4a.2b=8ab$

$N=(3a+1)^2+2a(1-2b)+(2b-1)^2$

$=(9a^2+6a+1)+2a-4ab+(4b^2-4b+1)$
$=9a^2+8a+4b^2-4b-4ab+2$

$A=(m-n)^2+4mn=m^2-2mn+n^2+4mn$

$=m^2+2mn+n^2=(m+n)^2$

Bài 1: 

a: Ta có: \(M=\left(2a+b\right)^2-\left(b-2a\right)^2\)

\(=4a^2+4ab+b^2-b^2+4ab-4a^2\)

\(=8ab\)

b: Ta có: \(N=\left(3a+2\right)^2+2a\left(1-2b\right)+\left(2b-1\right)^2\)

\(=\left(3a+2+1-2b\right)^2\)

\(=\left(3a-2b+3\right)^2\)

\(=9a^2+4b^2+9-12ab+18a-12b\)

c: Ta có: \(A=\left(m-n\right)^2+4nm\)

\(=m^2-2mn+n^2+4mn\)

\(=m^2+2mn+n^2\)

\(=\left(m+n\right)^2\)

2: 

a: \(\left(x+5\right)^2=x^2+10x+25\)

b: \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)