246 x 2005 - 2005 x 148 + 2005 x 102

= 2005 x (246 - 148 + 102)

= 2005 x 200

= 401000

169 : 3 + 562 : 3 + 169 : 3

= (169 + 562 + 169) : 3

= 900 : 3

= 300

201 x 2 + 201 + 201 x 3 + 201 x 4

= 201 x 2 + 201 x 1 + 201 x 3 + 201 x 4

= 201 x (2 + 1 + 3 + 4)

= 201 x 10

= 2010

x+1/2*4/5+x+x+6/10=148

x=49

= 230

= 196490

= 36000
 

23 + 23 x 2 + 23 x 3 + 23 x 4

= 23 x (1 + 2 + 3 + 4)

= 23 x 10

= 230

246 x 2005 - 2005 x 148

= (246 - 148) x 2005

= 98 x 2005

= 196490

25 x 12 x 30 x 4

= (25 x 4) x 12 x 30

= 100 x 360

= 36000

  233 + 23 x 2 + 23 x 3 + 23 x 4

= 23 x ( 1 + 2 + 3 + 4 )

= 23 x 10

= 230

  246 x 2005 - 2005 x 148

= ( 246 - 148 ) x 2005 

= 98 x 2005

= 196490

  25 x 12 x 30 x 4 

= ( 25 x 4 ) x ( 12 x 30 )

= 100 x 360

= 36000

ai đổi ko zậy

____________
____________
ik

\(...\Rightarrow x+x+\dfrac{x}{43}+\dfrac{x}{8}=14+148+\dfrac{10}{30}+\dfrac{5}{95}\)

\(\Rightarrow\left(1+1+\dfrac{1}{43}+\dfrac{1}{8}\right)x=162+\dfrac{1}{3}+\dfrac{1}{19}\)

\(\Rightarrow\left(\dfrac{2.43.8}{43.8}+\dfrac{1.8}{43.8}+\dfrac{1.43}{43.8}\right)x=\dfrac{162.3.19}{3.19}+\dfrac{1.19}{3.19}+\dfrac{1.3}{19.3}\)

\(\Rightarrow\left(\dfrac{688}{344}+\dfrac{8}{344}+\dfrac{43}{344}\right)x=\dfrac{9234}{57}+\dfrac{19}{57}+\dfrac{3}{57}\)

\(\Rightarrow\dfrac{739}{344}x=\dfrac{9256}{57}\)

\(\Rightarrow x=\dfrac{9256}{57}:\dfrac{739}{344}=\dfrac{9256}{57}.\dfrac{344}{739}=\dfrac{\text{3184064}}{\text{42123}}\)

a. 246 x 2005 - 2005 x 148

= (246 - 148) x 2005

= 98 x 2005

= 196490

b. 25 x 12 x 30 x 4

= (25 x 4) x 12 x 30

= 100 x 360

= 36000

c. 23 + 23 x 2 + 23 x 3 + 23 x 4

= 23 x 1 + 23 x 2 + 23 x 3 + 23 x 4

= 23 x (1 + 2 + 3 + 4)

= 23 x 10

= 230

230 nhé

a, 246x2005-2005x148

=(246-148)x2005

=98x2005

=196490

b, 25x12x30x4

=(25x4)x(12x30)

=100x360

=36000

c, 23+23x2+23x3+23x4

=23x(1+2+3+4)

=23x10

=230

148 : x thì dư 20  => 148-20 = 128 chia hết cho x ;  ( x > 20)

còn 108 : x thì dư 12  => 108 - 12 = 96 chia hết cho x

=> x thuộc UC(128;96)

UCLN(128;96) =2⁵ =32 vì  96=2⁵.3 ; 128=2⁷

=> x \(\in\)U(32) ={1;2;4;8;16;32} mà x > 20

=> x = 32