a, \(\left|x+2\right|-\left|x+7\right|=0\Rightarrow\left|x+2\right|=\left|x+7\right|\Rightarrow\orbr{\begin{cases}x+2=x+7\\x+2=-x-7\end{cases}\Rightarrow\orbr{\begin{cases}0=5\left(loại\right)\\2x=-9\end{cases}\Rightarrow}x=\frac{-9}{2}}\)

b, - Nếu \(2x-1\ge0\Rightarrow x\ge\frac{1}{2}\), ta có: 2x - 1 = 2x - 1 => 2x = 2x (thỏa mãn với mọi x)

- Nếu 2x - 1 < 0 => \(x< \frac{1}{2}\), ta có: 2x - 1 = 1 - 2x => 4x = 2 => x = \(\frac{1}{2}\) (không thỏa mãn điều kiện)

Vậy \(x\ge\frac{1}{2}\)

c,d tương tự b

e, tương tự a

a) | 2x - 1 | = 1- 3x

\(\orbr{\begin{cases}2x-1=1-3x\\2x-1=-\left(1-3x\right)\end{cases}}\)

\(\orbr{\begin{cases}2x-3x=1+1\\2x-1=-1+3x\end{cases}}\)

\(\orbr{\begin{cases}-x=2\\2x+3x=-1+1\end{cases}}\)

\(\orbr{\begin{cases}x=-2\\5x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=0\end{cases}}\)

b) | 1 - 2x | = x + 1 

\(\orbr{\begin{cases}1-2x=x+1\\1-2x=-\left(x+1\right)\end{cases}}\)

\(\orbr{\begin{cases}-2x-x=1-1\\-2x+x=-1-1\end{cases}}\)

\(\orbr{\begin{cases}-3x=0\\-x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

tương tự

 a, 7x . (x-10 ) = 0

=> \(\hept{\begin{cases}7x=0\\x-10=0\end{cases}}\) => \(\hept{\begin{cases}x=0\\x=10\end{cases}}\)

b,17 . (3x-6).(2x-8) = 0

          (3x-6).(2x-8) = 0 : 17

          (3x-6).(2x-8)  = 0

=>\(\hept{\begin{cases}3x-6=0\\2x-8=0\end{cases}}\)=>\(\hept{\begin{cases}x=2\\x=4\end{cases}}\)

Nhớ nha

a) (3x-1)³ = 27 = 3³

=> 3x - 1 = 3

3x = 4

x = 4/3

b) 3^2x-1 = 27 = 3³

=> 2x - 1 = 3

2x = 4

x = 2

c) 6^8-2x = 36 = 6²

=> 8 -2x = 2

=> 2x = 6

x = 3

d) (3x-2)¹⁰ = (3x-2)⁷

=> (3x-2)¹⁰ - (3x-2)⁷ = 0

(3x-2)⁷. [(3x-2)³ - 1] = 0

=> (3x-2)⁷ = 0 => 3x-2 = 0 => 3x = 2 => x = 2/3

(3x-2)³ - 1 = 0 => (3x-2)³ =  1 => 3x - 2 = 1 => 3x = 3 => x = 1

KL:...

\(\left(3x-1\right)^3=3^3\)

\(3x-1=3=>3x=4\)

\(=>x=\frac{4}{3}\)