Lời giải:

$2^n+34=2.2^2+3.2^3+....+n.2^n$

$2^{n+1}+68=2.2^3+3.2^4+....+n.2^{n+1}$

Trừ theo vế:

$2^n+34=n.2^{n+1}-(8+2^3+2^4+...+2^n)$

$n.2^{n+1}-2^n-42=2^3+2^4+...+2^n$

$n.2^{n+2}-2^{n+1}-84=2^4+....+2^{n+1}$

Trừ theo vế:

$n.2^{n+1}-2^n-42=2^{n+1}-8$

$2^n(2n-3)=34=17.2$

$\Rightarrow 2^n=2$ và $2n-3=17$ (vô lý)

Vậy không tìm được $n$.

\(A=2.2^2+3.2^3+...+n.2^n\)

\(2A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}\)

\(2A-A=\left(2.2^3+3.2^4+...+n.2^{n+1}\right)-\left(2.2^2+3.2^3+...+n.2^n\right)\)

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

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

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

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

Từ đây phương trình ban đầu tương đương với: 

\(\left(2n-2\right).2^n=2^{n+34}\)

\(\Leftrightarrow\left(2n-2\right).2^n=2^n.2^{34}\)

\(\Leftrightarrow n-1=2^{33}\)

\(\Leftrightarrow n=2^{33}+1\)

`#3107`

b)

`2.3^x = 162`

`\Rightarrow 3^x = 162 \div 2`

`\Rightarrow 3^x = 81`

`\Rightarrow 3^x = 3^4`

`\Rightarrow x = 4`

Vậy, `x = 4`

c)

`(2x - 15)^5 = (2 - 15)^3`

\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`

\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)

Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)

`d)`

\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!

`e)`

\(7\cdot4^{x-1}+4^{x-1}=23?\)

\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)

Bạn xem lại đề!

`f)`

\(2\cdot2^{2x}+4^3\cdot4^x=1056\)

\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)

Vậy, `x = 2`

_____

\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)

\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)

\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)

\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)

\(\Rightarrow\left(x\div3+17\right)\div10=2\)

\(\Rightarrow x\div3+17=20\)

\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)

Vậy, `x = 9.`

Đặt S = 2.2² + 3.2³ + 4.2⁴ + ... + (n - 1).2^{n - 1} + n.2ⁿ 

<=> S = 2S - S = (2.2³ + 3.2⁴ +  4.2⁵ + .... + (n - 1).2ⁿ + n. 2^{n + 1}) - (2.2² + 3.2³ + 4.2⁴ + ... + (n - 1).2^{n - 1} + n.2ⁿ)

                S = (2.2³ - 3.2³) + (3.2⁴ - 4.2⁴) + (4.2⁵ - 5.2⁵) + .... + [(n - 1).2ⁿ - n.2ⁿ] + n.2^{n + 1} - 2.2²

                   = -(2³ + 2⁴ + 2⁵ + ... + 2ⁿ) + n.2^{n + 1} - 8

Đặt A = 2³ + 2⁴ + 2⁵ + ... + 2ⁿ

  <=> 2A - A = (2⁴ + 2⁵ + 2⁶ + ... + 2^{n + 1}) - (2³ + 2⁴ + 2⁵ + ... + 2ⁿ) 

  <=> A = 2^{n + 1} - 2³ 

Khi đó S = - 2^{n - 1} + 2³ + n.2^{n - 1} - 8

              = 2^{n - 1}.(n - 1) = 2^{n + 34}

         => n - 1 = 2^{n + 34} : 2^{n - 1}

          => n - 1 = 2^{n + 34 - n + 1}

          => n - 1 = 2³⁵

          => n = 2³⁵ + 1

N=34359738369 nha

Cường xo tính lại kết quả của Xyz mà cũng làm