\(\Rightarrow\left(7n-11\right)^3=32\times25+200\)

\(\Rightarrow\left(7n-11\right)^3=1000\)

\(\Rightarrow7n-11=10\)

\(\Rightarrow7n=10+11\)

\(\Rightarrow n=21:7=3\)

\(\left(7n-11\right)^3=2^5.5^2+200\)

\(\left(7n-11\right)^3=32.25+200\)

\(\left(7n-11\right)^3=800+200\)

\(\left(7n-11\right)^3=1000\)

=>TH1:(7n-11)³=10³

=>7n-11=10

=>7n=10+11=21

=>n=21:7=3

TH2: (7n-11)³=-10³

=>7n-11=-10

=>7n=-10+11

=>7n=1

=>n=1:7=1/7

Mà n thuộc N nên n=3

Kết luận n=3

Trà My~ Cần j lm 2 TH; you lm sai roay đs. Vì nếu (7n-11)³=-10 thì lm sao -10³=1000 cho được

câu a đề sai

2n-15=17

=> 2n=17+15

=>2n=32

=> n=5

c,\(2.3^n=162=>3^n=162:2=>3^n=81\)

=>\(n=4\)

d, (7n-11)^3=2^5.5^2+200 = (7n-11)^3=32.25+200

= (7n-11)^3=800 +200 = 1000

câu này đè sai 

(7x−11)³ = 2⁵ . 5² + 200

(7x−11)³ = 32 . 25 + 200

(7x−11)³ = 1000

(7x−11)³ = 10³

7x−11 = 10

7x       = 10 + 11

7x       = 21

  x      = 21 : 7

  x = 3

giúp mik ik mik tic cho

(7x−11)³ = 2⁵ . 5² + 200

(7x−11)³ = 32 . 25 + 200

(7x−11)³ = 1000

(7x−11)³ = 10³

7x−11 = 10

7x       = 10 + 11

7x       = 21

  x      = 21 : 7

  x = 3

\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)

\(=\left(\frac{12}{8}+\frac{3}{8}\right)+\left(\frac{12}{128}+\frac{3}{128}\right)+\frac{3}{512}\)

\(=\frac{15}{8}+\frac{15}{128}+\frac{3}{512}\)

\(=\frac{240}{128}+\frac{15}{128}+\frac{3}{512}\)

\(=\frac{255}{128}+\frac{3}{512}\)

\(=\frac{1020}{512}+\frac{3}{512}\)

\(=\frac{1023}{512}\)

\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}=\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
                                                        \(=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{8}-\frac{3}{8}+\frac{3}{16}-\frac{3}{16}+\frac{3}{32}\)
                                                         \(=3+\frac{3}{32}=\frac{3.32}{32}+\frac{3}{32}=\frac{96+3}{32}=\frac{99}{32}\)

Ta có \(\left(n^2+7n+9\right)⋮\left(n+3\right)\)

\(\Leftrightarrow\left[\left(n^2+3n\right)+\left(4n+12\right)-3\right]⋮\left(n+3\right)\)

\(\Leftrightarrow\left[n\left(n+3\right)+4\left(n+3\right)-3\right]⋮\left(n+3\right)\)

\(\Rightarrow-3⋮\left(n+3\right)\)Hay \(n+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\)

Vậy \(n\in\left\{-6;-4;-2;0\right\}\)

Ta có: \(\frac{n^2+7n+9}{n+3}=\frac{n^2+3n+3n+9}{n+3}+\frac{n}{n+3}\)

= \(\frac{\left(n+3\right)^2}{n+3}+\frac{n+3-3}{n+3}=n+3+1-\frac{3}{n+3}\)=> x + 4 - 3/n+3

Do n thuộc N => n+ 4 thuộc N; Để \(n^2+7n+9⋮n+3=>3⋮n+3\)

Hay n+3 thuộc Ư(3)

=> n+ 3 thuộc { -3;-1;1;3}

=> n thuộc { -6; -4; -2;0}

Mà n thuộc N nên n =0

bạn làm chi tiết hộ mình 3 câu C,D,E. Mình đang gấp! tung nguyen viet

ko hieu bang cach hop la the nao

4C=4^2+4^3+..+4^(n+1)

4C-C=4^(n+1)-4=4(4^n-1)

C=4.(4^n-1)/3

5^2.D=5^2+5^4+..+5^202

5^2D-D=5^202-1

D=(5^202-1)/24

E=(2^202-1)/3

trả lời nhanh cho mk nha,hôm nay cần.tính hợp lí

A=2^0+Ơ2^1+2^2+...+2^2006

B=1+3+3^2+...+3^100

a) \(2^x.4=128\Rightarrow2^x=32=2^5\Rightarrow x=5\)

b) \(x^{17}=x\Rightarrow x^{17}-x=0\Rightarrow x\left(x^{16}-1\right)=0\Rightarrow x=0\) hay \(x=1\)

c) \(\left(2x-2\right)^3=8\Rightarrow\left(2x-2\right)^3=2^3\Rightarrow2x-2=2\Rightarrow2x=4\Rightarrow x=2\)

d) \(\left(x-6\right)^3=\left(x-6\right)^2\Rightarrow\left(x-6\right)^3-\left(x-6\right)^2=0\)

\(\Rightarrow\left(x-6\right)^2\left(x-6-1\right)=0\Rightarrow\Rightarrow\left(x-6\right)^2\left(x-7\right)=0\)

\(\Rightarrow x-6=0\) hay \(x-7=0\Rightarrow x=6\) hay \(x=7\)

e) \(\left(7x-11\right)^3=2^5.5^2+200\Rightarrow\left(7x-11\right)^3=32.25+200\)

\(\Rightarrow\left(7x-11\right)^3=1000=10^3\Rightarrow7x-11=10\Rightarrow7x=21\Rightarrow x=3\)

f) \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\Rightarrow2^{x-1}=24-\left[16-3\right]-3\)

\(\Rightarrow2^{x-1}=24-13-3\Rightarrow2^{x-1}=8=2^3\Rightarrow2x-1=3\Rightarrow2x=4\Rightarrow x=2\)

Cảm ơn bạn Nguyễn Đức Trí rất nhìu nha!