d: 2/3=20/30

6/5=36/30

7/10=21/30

1/15=2/30

45 + 55 = 100

ko khó nha

k nhé 

45+55=100

nhé bạn

45+55=100

Dễ òm !!!

x là 1 nha bn

x+99=-55+55

x+99=0

      x=0-99

      x=-99

x là 10

33x55x(22x3-66)

=33x55x(66-66)

=33x55x0

=0

k mk nha

33x55x(22x3-66)

=33x55x(66-66)

=55x55x0

=0

thấy đúng thì tích nha

33 x 35 x ( 22 x 3 - 66 )

= 33 x 35 x ( 66 - 66 )

= 33 x 35 x 0

= 0

x+100=155-55

x+100=100

        x=100-100

        x=0

vay x=0

 x+100=155-55

=>x+100=100

=>x=0 

Vậy x=0

nhớ tích minh nha

x + 100 = 155 - 55

x + 100 = 100

x = 100 - 100

x = 0

nha bn

a. 11 x11= 121 ; 22 x 22=484 , b.44 x 44 = 1936 ; 55 x 55 = 3025

a, 121

b, 484

c, 1089

d, 1936

e, 3025

HT và $$$

a) 121

    484

b) 1089

    1936

    3025

@Nghệ Mạt

#cua

\(n_{O_2} = \dfrac{0,72}{22,4} = \dfrac{9}{280}(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = \dfrac{9}{140}(mol)\\ m_{KMnO_4} = \dfrac{9}{140}.158 = 10,16(gam)\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = \dfrac{3}{140}(mol)\\ m_{KClO_3} = \dfrac{3}{140}.122,5 = 2,625(gam)\)

\(n_{O_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)

\(0.6..............................................0.3\)

\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)

\(0.2..........................0.3\)

\(m_{KMnO_4}=0.6\cdot158=94.8\left(g\right)\)

\(m_{KClO_3}=0.2\cdot122.5=24.5\left(g\right)\)

\(n_{O_2}=\dfrac{V_{O_2\left(ĐKTC\right)}}{22,4}=\dfrac{0,72}{22,4}=0,03\left(mol\right)\)

- Điều chế O2 bằng KMnO4, ta có PTHH: 

\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

----2----------------------------------------------------1---

----0,06------------------------------------------------0,03---

\(m_{KMnO_4}=n_{KMnO_4}\cdot M_{KMnO_4}=0,06\cdot158=9,48\left(g\right)\)

- Điều chế O2 bằng KClO3, ta có PTHH:

\(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\)

---2-------------------------------------3----

---0,02---------------------------------0,03----

\(m_{KClO_3}=n_{KClO_3}\cdot M_{KClO_3}=0,02\cdot122,5=2,45\left(g\right)\)

x.55+66=100

55x=44

x=4/5

Vậy x=4/5

x.55+66=100

=>x.55=100-66

=>x.55=34

=>x=34:55

=>x=34/55

x . 55 + 66 = 100

x . 55 = 100 - 66 

x . 55 = 34

x = 34 : 55

x = 0,618

Answer : 0,618

x=10 ví 10*11=55*2=110