a) 2x + (-61) - (21 - 61) = 2x - 21 + (61 - 61) = 2x - 21.

b) (- 3 - x + 5) + 3 = (- 3 + 3) + 5 - x = 5 - x.

c) 11- (13 - x) + (13 - 11) = (11- 11) + (13- 13) + x = x

d) 25 - ( 15 - x + 303) + 303 = 25 - 15 + (303 - 303) + x = x + 10

\(\frac{202}{303}=\frac{202:101}{303:101}=\frac{2}{3}\)

202/303=2/3

202/303=2/3

Chúc bạn học giỏi nha!!!

K cho mik vs nhé

= -15 - x + 25 +x = 10

(-15)+ (-x)+ (-25)+ (-x)

Không thể rút gọn được nha I can rest your love on me

1 Tìm số nguyên x bik:

a.-( 2x+2)+25=-19           

-2x - 2 + 25 = - 19

-2x = -19+2-25

-2x = -42

   x = 21   

 b.1-(12+3x)=7

1 - 12 - 3x = 7

- 11 - 3x = 7

 -3x = 7 + 11

- 3x = 18

    x = -6

c.-(2x+2)+25=-19

 giống câu a nhé

2.Rút gọn biểu thức 

a.2x+(-61)-(21-61)       

= 2x - 61 - 21 + 61

= 2x - 21  

b. (-3-x+5)+3

= 2 - x + 3

= -1 -x 

c.11-(13-x)+(13-11)         

= 11 - 13 + x + 13 - 11

= x

d.25-(15-x+303)+303

= 25 - 15 + x - 303 + 303

= 10 + x

e.x+(-81)-(11-8)                     

= x - 81 - 11 + 8

= x - 84

f. (-1-x+2)+1

= -1 - x + 2 + 1

= - x + 2

g.15-(11-x)+(11-15)             

= 15 - 11 + x + 11 - 15

= x  

i) 15-(15-x+202)+202

= 15 - 15 + x - 202 + 202

= x

3.Chứng minh đẳng thức

a.-(59-3x)+39=3x-20       

- 59 + 3x + 39 = 3x - 20

- 59 + 39 + 20 = 3x - 3x

           0          = 0

b.-(a+b+c)+(b-c)-(a-c-1)=1+c-29

b - a - b - c + b - c - a + c + 1 = 1 + c - 29

b - a - b - c + b - c - a + c - c = 1 - 29 - 1

                   - 2a + b - 2c          = - 29

chịu luôn  -__- 

c.-(19-2x)+39=2x+20

- 19 + 2x + 39 = 2x + 20

2x - 2x = 20 + 19 - 29

0          = 0

d.-(a+b+c)+b-c-( a-c+1)=c-2a-1

- a - b - c + b - c - a + c - 1 = c - 2a - 1

- a - b - c + b - c - a + c + 2a - c = -1 + 1

                     - 2c  =  0

                         c  = 0

Vậy c = 0

\(log_{a^3}b.log_ba=\dfrac{1}{3}.log_ab.log_ba=\dfrac{1}{3}\)

\(log_{a^{10}}b^5.log_{b^3}a^9=\dfrac{1}{10}.5.log_ab.\dfrac{1}{3}.9.log_ba=\dfrac{3}{2}\)

\(log_{a^{107}}b^{101}.log_{b^{303}}a^{428}=\dfrac{1}{107}.101.log_ab.\dfrac{1}{303}.428.log_ba=\dfrac{4}{3}.log_ab.log_ba=\dfrac{4}{3}\)

a: \(log_{a^3}b\cdot log_ba=\dfrac{1}{3}\cdot log_ab\cdot log_ba=\dfrac{1}{3}\)

b: \(log_{a^{10}}b^5\cdot log_{b^3}a^9\)

\(=\dfrac{1}{10}\cdot log_ab^5\cdot\dfrac{1}{3}\cdot log_ba^9\)

\(=\dfrac{1}{30}\cdot5\cdot log_ab\cdot9\cdot log_ba=\dfrac{45}{30}=\dfrac{3}{2}\)

c: \(log_{a^{107}}b^{101}\cdot log_{b^{303}}a^{428}\)

\(=\dfrac{1}{107}\cdot log_ab^{101}\cdot\dfrac{1}{303}\cdot log_ba^{428}\)

\(=\dfrac{1}{107}\cdot101\cdot log_ab\cdot\dfrac{1}{303}\cdot428\cdot log_ba\)

\(=4\cdot\dfrac{1}{3}=\dfrac{4}{3}\)

C = (-15 + |x|) + (25 - |-x|)

= -15 + |x| + 25 - |-x|

= (25 - 15) + |x| - |-x|

= 10 + |x| - |-x|

Ta có: |x| = |-x| => |x| - |-x| = 0

=> C = 10 + |x|

C = (-15 + IxI ) + ( 25 - I-xI)

C = (-15 + x ) + ( 25 - x )

C = -15 + x + 25 - x

C = ( -15 + 25 ) + ( x - x )

C = 10 + 0

C = 10

Vậy C = 10

P = ( - 15 + | x | ) + ( 25 - |  - x | )

P = ( - 15 + x ) + ( 25 - x )

P = - 15 + x + 25 - x

P = - 15  + 25 + x - x

P = 25 - 15 + 0

P = 10

P=(-15)+x+25-x

P=[(-15)+25]+(x-x)

P=10+0

P=10

a) P = ( - 15 + | x | ) + ( 25 - | -x | )

       = ( - 15 + x ) + ( 25 - x )

       = - 15 + x + 25 - x

       = ( -15 ) + 25 + x - x

       = 25 - 15 + 0

       = 10