a) \(A=27\cdot36+73\cdot99+27\cdot14-49\cdot73\)

\(A=27\cdot\left(36+14\right)+73\cdot\left(99-49\right)\)

\(A=27\cdot50+73\cdot50\)

\(A=50\cdot\left(27+73\right)\)

\(A=50\cdot100\)

\(A=5000\)

b) \(B=\left(4^5\cdot10\cdot5^6+25^5\cdot2^8\right):\left(2^8\cdot5^4+5^7\cdot2^5\right)\)

\(B=\dfrac{\left(2^2\right)^5\cdot2\cdot5\cdot5^6+\left(5^2\right)^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)

\(B=\dfrac{2^{11}\cdot5^7+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)

\(B-\dfrac{2^8\cdot5^7\cdot\left(2^3\cdot1+5^3\cdot1\right)}{2^5\cdot5^4\cdot\left(2^3\cdot1+5^3\cdot1\right)}\)

\(B=\dfrac{2^8\cdot5^7}{2^5\cdot5^4}\)

\(B=2^3\cdot5^3\)

\(B=10^3\)

\(B=1000\)

=113 x (54+45)

=113x 99 

= 11187

b, 24x3 - 12x5

=12 x 2x3 - 12x5

=12x6 - 12 x5

=12x(6-5)

=12

Giups mk mấy bài này với nhé!

1.Tính giá trị biểu thức bằng cách hợp lí:

a)54x113+45x113

b)24x3-12x5

a)54x113+45x113

=113*(54+45)

=113*100

=11300

b)24x3-12x5

=12*2*3-12*5

=12*6-12*5

=12*(6-5)

=12*1

=12

B = ( 1024 . 10 . 15625 + 9765625 . 256 ) : ( 256 . 625+78125 . 32 )

=( 160000000+2500000000 ):( 160000+2500000)

= 2660000000 : 2660000

= 1000

B = ( 1.024 . 10 . 15.625 + 9.765.625 . 256 ) : ( 256 . 625+78125 . 32 )

B =( 160.000.000+2.500.000.000 ):( 160.000+2.500.000)

B = 2.660.000.000 : 2.660.000

B = 1.000

2:

a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)

1:

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)

Bài 1:

\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)

\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)

\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)

\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)

\(A=\dfrac{-16-1}{4}\)

\(A=-\dfrac{17}{4}\)

Bài 2:

\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)

\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)

\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)

\(=\dfrac{1}{3}\cdot-2\)

\(=-\dfrac{2}{3}\)

\(25-\left(44-756+12\right)+\left(44-756+12\right)\)

\(=25-44+756-12+44-756+12\)

\(=25+\left(44-44\right)+\left(756-756\right)+\left(12-12\right)\)

=25

a)54 * 113 +48 *113 -113*2

 =113*(54+49-2)

=113 * 101

 =11413

Mình chưa hiểu đề lắm nhưng chắc ý bạn là như này:
   \(87\frac{23}{59}+74\frac{17}{71}+12\frac{36}{59}+25\frac{5}{71}\)
= \(\left(87\frac{23}{59}+12\frac{36}{59}\right)+\left(25\frac{5}{71}+74\frac{17}{71}\right)\)
= \(\left(87+12\right)+\left(\frac{23}{59}+\frac{36}{59}\right)+\left(25+74\right)+\left(\frac{5}{71}+\frac{17}{71}\right)\)
= \(99+\frac{59}{59}+99+\frac{22}{71}\)
= \(198+1+\frac{22}{71}\)
= \(199\frac{22}{71}\)
Chúc bạn học tốt ^^!