\(y.0.25=3.6+\frac{7}{20}\)

\(y.0.25=\frac{79}{20}=3.95\)

\(y=3.95:0.25=\frac{79}{5}=15.8\)

y x 0,125 = 3,6 + 2.8 : 8

y x 0,125 = 3,6 + 0,35

y x 0,125 = 3,95

y = 3,95 : 0,125

y = 31,6

k mik nha 

0.25 x 4 x 3.6 : 0.125=1x3,6x1:8=0,45

5.79 x 0.24 - 0.12 x 1.58=0,24x5=1,2

nhớ k mình nhé

a: \(A=\dfrac{3^3\cdot2^3+3^3\cdot2^2+3^3\cdot1}{-13}=\dfrac{27\left(2^3+2^2+1\right)}{-13}=-27\)

b: \(B=\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^3\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)

\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9}{2^{10}\cdot3^7+2^{10}\cdot5\cdot3^8}\)

\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\right)}{2^{10}\cdot3^7\left(1+5\cdot3\right)}=\dfrac{2}{3}\cdot\dfrac{4+27}{1+15}=\dfrac{2}{3}\cdot\dfrac{31}{16}=\dfrac{31}{24}\)

c: \(C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{35}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot2^6-7\right)}=\dfrac{10-9}{5\cdot64-7}=\dfrac{1}{313}\)

a) \(16^{^3}:8^2=\left(8.2\right)^3:8^2=8^3.2^3:8^2=\left(8^3:8^2\right).2^3=8.8=64\)

b)\(8^3.\left(0,125\right)^3=\left(8.0,125\right)^3=1^3=1\)

c)\(7^{^{200}}.\left(\frac{1}{7}\right)^{200}=\left(7.\frac{1}{7}\right)^{200}=1^{200}=1\)

d)\(4.\left(0,25\right)^3.64=4.\left(0,25\right)^3.4^3=4.\left(0,25.4\right)^3=4.1=4\)

e)....

cậu có thể tham khảo bài làm trên đây ạ, chúc cậu hok tốt ^^

1/8x15+0.125x55+125/1000

=1/120+6.875+1/8

=7.0083

ta có: 10/9 < x < 11/2

=> 1 < x < = 5

=> x = {2;3;4;5}

\(\frac{7}{9}+\frac{16}{48}< x< \frac{43}{8}+0.125\)

\(\frac{10}{9}< x< \frac{11}{2}\)

\(1.11< x< 5.5\)

Vậy \(x=2;3;4;5.\)

Ta có:7/9+16/48=7/9+1/3=10/9

         43/8+0.125=43/8+1/8=11/2

         =>10/9<x<11/2

        =>9/9<10/9<x<11/2<12/2

        =>1<x<6

Vì x là số tự nhiên nên x=2,3,4,5

Vậy x=2,3,4,5.

doi lat nhe

x=(6và3/5-0.125*8+2vvà 2/15*0.03)*25/3

X=(33/5-1+32/15*3/100)*25/3

x=(28/5+8/125)*25/3

=708/125*25/3

=236/5

317.14