\(\left(x-1\right)^{x+1}-\left(x-1\right)^{x+12}=0\\ \Leftrightarrow\left(x-1\right)^{x+1}\left[1-\left(x-1\right)^{11}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+11}=0\\\left(x-1\right)^{11}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

vì sao ở chỗ x-1=0 lại bỏ dấu mũ thế bạn giải thíc giúp mình với ^^

( 26 - 3x ) : 5 = 75 - 71

( 26 - 3x ) : 5 = 4

26 - 3x = 4 x 5

26 - 3x = 20

3x = 26 - 20

3x = 6

x = 6 : 3

x = 2

71 + (26-3x) :5 = 75

=> (26-3x):5 = 75-71

=> (26-3x) : 5 = 4

=> 26-3x=4.5

=> 26-3x=20

=> 3x=26-20

=> 3x=6

=> x=6:3

=> x=2

a)5-2x=3x+20

   5=3x+20+2x

   5=5x+20

=>5x+20=5

    5x=5-20

    5x=-15

   x=(-15):5

  x=-3

Bùi Ngọc Truongf Sơn làm nốt cho mình 2 bài còn lại đi

a)(3x+2):8=4

3x+2=4.8

3x+2=32

3x=32-2

3x=30

x=30:3

x=10

b)x+2x+3x+4x=100

x(1+2+3+4)=100

x.10=100

x=100:10

x=10

b) x + 2x + 3x + 4x = 100

x( 1 + 2 + 3 + 4 ) = 100

x . 10 = 100

x = 100 : 10

x = 10

\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)

\(=>\frac{3x}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right]=\frac{1}{21}\)

\(=>x\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right]=\frac{1}{21}\)

\(=>x\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{14}\right]=\frac{1}{21}\)

\(=>x\left[\frac{1}{2}-\frac{1}{14}\right]=\frac{1}{21}\)

\(=>x\cdot\frac{3}{7}=\frac{1}{21}\Leftrightarrow x=\frac{1}{9}\)

1 là lộn đề

2 là bó tay

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