14,35 × 63 + 14,35 × 36 + 14,35

= 14,35 x 63 + 14,35 x 36 + 14,35 x 1

= 14,35 x ( 63 + 36 + 1 )

= 14,56 x 100

= 1456

\(14,35\times63+14,35\times36+14,35\)

\(=14,35\times\left(63+36+1\right)\)

\(=14,35\times100\)

\(=1435\)

\(15=3.5\)           \(65=5.13\)

\(ƯCLN\left(15;65\right)=5\)

tổ phó khổ quá 

14,35:(0,5+0,125)

=14,35:0,625

=22,96

= 14,35 : (0,5 + 0,125)
= 14,35 : 0,625

= 22,96

\(\left[\left(2,18-1,38\right).1,25\right]:\left[\left(17,9-15,65\right):1,125\right]:\left(-0,4\right)\)

\(=\left[0,18.1,25\right]:\left[2,25:1,125\right]:\left(-0,4\right)\)

\(=\left[0,225\right]:\left[2\right]:\left(-0,4\right)\)

\(=\left[0,225\right]:\left[2\right]:\left(-0,4\right)=-0,045\)

[ (2,18 - 1,38).1,25]: [(17,9 - 15,65): 1,125]:(-0,4)

= [0,8.1,25]: [(2,25): 1,125]:(-0,4)

= 1:[ (2,25): 1,125]:(-0,4)

= 1:(2):(-0,4)

= 0,5: -(0,4)

= - \(\dfrac{5}{4}\)

a) S

b) S

c) Đ

d) S

a, S

b, S

c, Đ

d, S

\(HCl + AgNO_3 \to AgCl + HNO_3\\ n_{HCl} = n_{AgCl} = \dfrac{14,35}{143,5} = 0,1(mol)\\ \Rightarrow C\%_{HCl} = \dfrac{0,1.36,5}{10}.100\% = 36,5\%\)

bạn xem lại đi ạ. vừa nãy mk có giải cho bạn rồi ạ!!!

\(=14,35\times\left(2+8\right)=143,5\)

bài 1 : a, 88 : 0,1 = 8,8.           b, 181,404 : 0,65 = 279,08 dư 0,002.

bài 2 : a, S.   b, S.  c, S.    d, Đ.

tick nha.