| 2x - 1 | = | 2x + 3 |

=> \(\left[{}\begin{matrix}2x-1=2x+3\\2x-1=-2x+3\end{matrix}\right.\)

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

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

-4 = 0 ( ktm )

2) 2x - 1 = -2x + 3

2x - 1 + 2x - 3 = 0

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

4x - 4 = 0

4x = 4

x = 1 ( tm )

Vậy x = 1

( a + 2 )² - ( a + 2 ) ( a - 2 )

( a + 2 - a + 2 ) ( a + 2 )

4 ( a + 2 )

a) 4x( 2x - 5 ) + 3 ( 5 - 2x ) = 0

4x( 2x - 5 ) - 3 ( 2x - 5 ) = 0

( 4x - 3 ) ( 2x - 5 ) = 0

\(\Rightarrow\left[{}\begin{matrix}4x-3=0\Rightarrow x=\dfrac{3}{4}\\2x-5=0\Rightarrow x=\dfrac{5}{2}\end{matrix}\right.\)

b) 3( 2x - 1 ) = x ( 1 - 2x )

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

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

3x ( 2x - 1 ) = 0

\(\Rightarrow\left[{}\begin{matrix}3x=0\Rightarrow x=0\\2x-1=0\Rightarrow x=\dfrac{1}{2}\end{matrix}\right.\)

c) x³ - 10x² = -16x

x³ - 10x² + 16x = 0

x³ - 2x² - 8x² + 16x = 0

x² ( x - 2 ) - 8x ( x - 2 ) = 0

( x² - 8x ) ( x - 2 ) = 0

x ( x - 8 ) ( x - 2 ) = 0

=> x = 0

hoặc x - 8 = 0 => x = 8

hoặc x - 2 = 0 => x = 2

a) \(\dfrac{5}{6}-x=\dfrac{3}{12}\)

\(\dfrac{10}{12}-x=\dfrac{3}{12}\)

\(x=\dfrac{10}{12}-\dfrac{3}{12}\)

\(x=\dfrac{7}{12}\)

b) \(\dfrac{12}{7}:x+\dfrac{9}{10}=\dfrac{7}{5}\)

\(\dfrac{12}{7}:x=\dfrac{7}{5}-\dfrac{9}{10}\)

\(\dfrac{12}{7}:x=\dfrac{14}{10}-\dfrac{9}{10}\)

\(\dfrac{12}{7}:x=\dfrac{1}{2}\)

\(x=\dfrac{12}{7}:\dfrac{1}{2}\)

\(x=\dfrac{12}{7}\cdot2\)

\(x=\dfrac{24}{7}\)

a) 67 x 99 = 67 x 100 - 67 x 1 = 6700 - 67 = 6633

b) 43 x 11 = 43 x 10 + 43 x 1 = 430 + 43 = 473

c) 67 x 101 = 67 x 100 + 67 x 1 = 6700 + 67 = 6767

d) 25 x 12 = 25 x 10 + 25 x 2 = 250 + 50 = 300

2x - ( x + 2 ) = 10

2x - x - 2 = 10

x - 2 = 10

x = 10 + 2

x = 12