Question

[Yên Lê Thanh]

tìm x biết: 

(x-7)^x+1-(x-7)^x+11=0

Answer 1

[Nguyễn Thanh Vân]

(x - 7)^x + 1 - (x - 7)^x + 11 = 0

(x - 7)^x . (x - 7) - (x - 7)^x . (x - 7)^11 = 0

(x - 7)^x . [(x - 7) - (x - 7)^11] = 0

=> (x - 7)^x = 0 hoặc [(x - 7) - (x - 7)^11] = 0

* TH1: (x - 7)^x = 0

=> x - 7 = 0

=> x = 0 + 7

=> x = 7

* TH2: [(x - 7) - (x - 7)^11] = 0

=> x - 7 = (x -7)^11

=> x - 7 = 1 hoặc x  - 7 = 0

x - 7 = 1

=> x = 1 + 7

     x = 8

x - 7 = 0 (TH1)

Vậy x = 7 hoặc x = 8.

Answer 2

[ngonhuminh]

\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-7\right)^x-\left(x-7\right)\left(x-7\right)^x\left(x-7\right)^{10}=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-7\right)^x\left[1-\left(x-7\right)^{10}\right]=0\)

\(\Leftrightarrow\left[\begin{matrix}x-7=0\left(1\right)\\\left(x-7\right)^x=0\left(2\right)\\1-\left(x-7\right)^{10}=0\left(3\right)\end{matrix}\right.\)

(1) và (2)=> x=7

(3)\(\Leftrightarrow\left(x-7\right)^{10}=1\Rightarrow x-7=\pm\sqrt[10]{1}=\pm1\Rightarrow\left[\begin{matrix}x=8\\x=6\end{matrix}\right.\)

Kết luận nghiệm: x={6,7,8}