\(\dfrac{1}{2}\cdot2^n+4\cdot2^n=9\cdot2^5\\ \Rightarrow2^n\left(\dfrac{1}{2}+4\right)=9\cdot2^5\\ \Rightarrow2^n\cdot\dfrac{9}{2}=9\cdot2^5\\ \Rightarrow2^{n-1}\cdot9=9\cdot2^5\\ \Rightarrow n-1=5\\ \Rightarrow n=6\)

a)\(\frac{1}{2}.2^n+4.2^n=9.2^5\)

\(\Leftrightarrow2^n\left(\frac{1}{2}+4\right)=9.32\)

\(\Leftrightarrow2^n.\frac{9}{2}=288\)

\(\Leftrightarrow2^n=288:\frac{9}{2}\)

\(\Leftrightarrow2^n=64\)

\(\Leftrightarrow2^n=2^6\) \(\Rightarrow n=6\left(TMBT\right)\)

Vậy: n=6

b) \(3^2.3^{-5}.3^n=3^{11}\)

\(\Leftrightarrow3^{2+\left(-5\right)+n}=3^{11}\)

\(\Leftrightarrow3^{\left(-3\right)+n}=3^{11}\)

\(\Rightarrow\left(-3\right)+n=11\)

\(\Leftrightarrow n=11-\left(-3\right)\) \(\Rightarrow n=11+3\Rightarrow n=14\)

Vậy n=14

c) Câu c này bạn làm giống như câu a) nha bởi vì nó cũng giống nhau thôi, bạn biến đổi \(2^{-1}=\frac{1}{2}\)rồi làm giống như trên câu a) nhé.

P/s: mình mới học lớp 6 lên 7. sai chỗ nào thông cảm cho mik với nhé. luôn  nhá..

Cảm ơn ạ!!

27^n:3^n=9

3^3n:3^n=3^2

3^(3n-n)=3^2

suy ra 3n-n=2

n(3-1)=2

n.2=2

Vậy n=2:2

n=1

còn câu sau cậu tự tính

\(n=1\)

a: \(3^n\cdot3^4\cdot\dfrac{1}{9}=3^7\)

\(\Leftrightarrow3^n\cdot3^2=3^7\)

=>n+2=7

hay n=5

b: \(\Leftrightarrow2^n\cdot\left(\dfrac{1}{2}+4\right)=9\cdot2^5\)

\(\Leftrightarrow2^n=9\cdot2^5:\dfrac{9}{2}=9\cdot\dfrac{2}{9}\cdot2^5=2^6\)

hay n=6

https://h.vn/hoi-dap/question/221389.html kham khảo ak!!! bài dài quá lười đánh máy lắm, thông cảm!!!^~

Đặt A=2.22+3.23+4.24+...+n.2nA=2.22+3.23+4.24+...+n.2n

Ta có:

A=2.22+3.23+4.24+...+n.2nA=2.22+3.23+4.24+...+n.2n

⇒2A=2(2.22+3.23+4.24+...+n.2n)⇒2A=2(2.22+3.23+4.24+...+n.2n)

⇒2A=2.23+3.24+4.25+...+n.2n+1⇒2A=2.23+3.24+4.25+...+n.2n+1

⇒2A−A=2.22+(3.23−2.23)+...+(n−n+1).2n−n.2n+1⇒2A−A=2.22+(3.23−2.23)+...+(n−n+1).2n−n.2n+1

⇒A=2.22+23+24+...+2n−n.2n+1⇒A=2.22+23+24+...+2n−n.2n+1

⇒A=22+(22+23+...+2n+1)−(n+1).2n+1⇒A=22+(22+23+...+2n+1)−(n+1).2n+1

⇒A=−22−(22+23+...+2n+1)+(n+1).2n+1⇒A=−22−(22+23+...+2n+1)+(n+1).2n+1

Đặt B=22+23+...+2n+1B=22+23+...+2n+1

⇒2B=23+24+...+2n+2⇒2B=23+24+...+2n+2

⇒2B−B=2n+2−22⇒B=2n+2−22⇒2B−B=2n+2−22⇒B=2n+2−22

⇒A=22−2n+2+22+(n+1).2n+1⇒A=22−2n+2+22+(n+1).2n+1

⇒A=(n+1).2n+1−2n+2⇒A=(n+1).2n+1−2n+2

⇒A=2n+1(n+1−2)⇒A=2n+1(n+1−2)

⇒A=(n−1).2n+1=2(n−1).2n⇒A=(n−1).2n+1=2(n−1).2n

Mà A=2(n−1).2n=2n+10A=2(n−1).2n=2n+10

⇒2(n+1)=210⇒n−1=29⇒2(n+1)=210⇒n−1=29

⇒n−1=512⇒n=513⇒n−1=512⇒n=513

Vậy n=513