a. 4567+(1234-4567)-4

=4567+1234-4567-4

=(4567-4567)+(1234-4)

=0+1230

=1230

2001-( 53 -1579 ) - ( - 53 )

=2001-53+1579+53

=(2001+1579)+(53-53)

=3580+0

=3580

35-17 + 2017 - 35 + ( -2017 )

=35-17+2017-35-2017

=(35-35)+(2017-2017)-17

=0+0-17

=-17

a) 4567 + ( 1234 - 4567 ) - 4

= 4567 + 1234 - 4567 - 4

= ( 4567 - 4567 ) + ( 1234 - 4 )

= 0 + 1230

= 1230

b) 2001 - ( 53 - 1579 ) - (-53)

= 2001 - 53 + 1579 + 53

= ( 2001 + 1579 ) - ( 53 - 53 )

= 3580 - 0

= 3580

c) 35 - 17 + 2017 - 35 + (-2017)

= 35 - 17 + 2017 - 35 - 2017

= ( 35 - 35 ) - 17 + ( 2017 - 2017 )

= 0 - 17 + 0

= -17 

\(a)4567+1234-4567-4\)

\(=4567-4567+1234-4\)

\(=0+1230=1230\)

\(b)2001-53-1519-\left(-53\right)\)

\(=2001-53-1519+53\)

\(=2001-1519-53+53\)

\(=482\)

A=1

B=1

a) A = 1 

b) B = 1 

A = \(\frac{1995.1996-1997}{1995.1994+1993}\)

   = \(\frac{1995.\left(1994+2\right)-1997}{1995.1994+1993}\)

   = \(\frac{1995.1994+1995.2-1997}{1995.1994+1993}\)

   = \(\frac{1995.1994+3990-1997}{1995.1994+1993}\)

   = \(\frac{1995.1994+1993}{1995.1994+1993}\)

   = \(1\)

b) B = \(\frac{12345.67890-54321}{12344.67890+13569}\)

       = \(\frac{\left(12344+1\right).67890-54321}{12344.67890+13569}\)

       = \(\frac{12344.67890+67890-54321}{12344.67890+13569}\)

       = \(\frac{12344.67890+13569}{12344.67890+13569}\)

       = \(1\)

a, = 1

b, = 1

4567 + 1234 -4567 -4

 = 4567 - 4567 + 1234 - 4

= 0+ 1230

= 1230

\(4567+\left(1234-4567\right)-4\)

\(=4567+1234-4567-4\)

\(=\left(4567-4567\right)+\left(1234-4\right)\)

\(=0+1230\)

\(=1230\)

a ) 4567 x 3 + 4 x 4567 + 4567 x 2 + 4567

= 4567 x 3 + 4567 x 4 + 4567 x 2 + 4567 x 1

= ( 3 + 4 + 2 + 1 ) x 4567

=      10    x   4567

=        45670

b ) 3896 x 20 + 40 x 3896 + 3896 x 30 + 3896

= 3896 x 20 + 3896 x 40 + 3896 x 30 + 3896 x 1

= ( 20 + 40 + 30 + 1 ) x 3896

=       91    x    3896

=         354536

Mk trả lời đầu tiên , bạn k mình nhé !

a,4567x3 + 4x4567+4567x2+4567=4567x[3+4+2+1]+4=4567x10+4=45670+4=45674

b,3896x20+40x3896+3896x30+3896=3896x[20+40+30+1]=3896x91=

con lai tu tinh nha dung quan h

Bài 1:

Tổng của 2 số là

  \(36\times2=72\) 

Số lớn là

  \(72-17=55\) 

Bài 2:

a) \(4567+y\div34=10987\) 

                 \(y\div34=10987-4567\) 

                 \(y\div34=6420\) 

                         \(y=6420\times34\) 

                         \(y=218280\) 

b) \(\dfrac{4}{3}+\dfrac{1}{2}\div y=2\) 

             \(\dfrac{1}{2}\div y=2-\dfrac{4}{3}\) 

              \(\dfrac{1}{2}\div y=\dfrac{2}{3}\) 

                      \(y=\dfrac{1}{2}\div\dfrac{2}{3}\) 

                      \(y=\dfrac{3}{4}\) 

Bài 3:

a) \(\dfrac{2}{5}\times\dfrac{2}{5}+\dfrac{9}{8}\div3=\dfrac{4}{25}+\dfrac{9}{8}\times\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}=\dfrac{107}{200}\) 

b) \(2-\left(\dfrac{1}{7}\times4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{25}{21}\)

Bài 1 : Gọi a là số lớn, b là số bé, theo đề bài ta có :

(a+b):2=36⇒a+b=72

mà b=17

Nên a=72-17=55

Bài 2 :

a) 4567+y:34=10987

⇒ y:34=10987-4567

⇒ y:34=6420

⇒ y=6420x34

⇒ y=218280

b) \(\dfrac{4}{3}+\dfrac{1}{2}:y=2\)

\(\Rightarrow\dfrac{1}{2}:y=2-\dfrac{4}{3}\)

\(\Rightarrow\dfrac{1}{2}:y=\dfrac{2}{3}\)

\(\Rightarrow y=\dfrac{1}{2}:\dfrac{2}{3}\)

\(\Rightarrow y=\dfrac{1}{2}x\dfrac{3}{2}\)

\(\Rightarrow y=\dfrac{3}{4}\)

Bài 3 :

\(\dfrac{2}{5}x\dfrac{2}{5}+\dfrac{9}{8}:3=\dfrac{4}{25}+\dfrac{9}{8}x\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}\)

= \(\dfrac{4x8}{25x8}+\dfrac{25x3}{25x8}=\dfrac{32}{200}+\dfrac{75}{200}=\dfrac{107}{200}\)

\(2-\left(\dfrac{1}{7}x4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\left(\dfrac{12}{21}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{42}{21}-\dfrac{17}{21}=\dfrac{25}{21}\)

\(A=\left(4567-97\right)+\left(-1075\right)-4567-\left(29-1075\right)\)

\(A=4567-97-1075-4567-29+1075\\ A=\left(4567-4567\right)+\left[\left(-1075\right)+1075\right]-97-29\\ A=0+0-97-29\\ A=0-97-29\\ A=-97-29\\ A=-126\)