1/1000

\(=\dfrac{1993}{7+1986}=\dfrac{1993}{1993}=1\)

c= \(\dfrac{1994x1993-1992x1993}{1992x1993+1994x7+1986}\)

c=\(\dfrac{1993x\left(1994-1992\right)}{1992x1993+1994x7+1986}\)

c= \(\dfrac{1993x2}{1992x1993+1994x7+1986}\)

c=\(\dfrac{2}{1992x1+1994x7+1986}\)

c=\(\dfrac{1}{13903161}\)

576,(6)

13840 : 24 = 576,(6)

576,(6)

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3mol\Rightarrow m_C=3,6g\)

\(n_{H_2O}=\dfrac{7,2}{18}=0,4mol\Rightarrow n_H=0,4\cdot2=0,8\Rightarrow m_H=0,8g\)

Nhận thấy: \(m_C+m_H=4,4=m_A\)

\(\Rightarrow A\) chỉ chứa hai nguyên tố C và H.

Gọi CTHH là \(C_xH_y\).

\(\Rightarrow x:y=n_C:n_H=0,3:0,8=3:8\)

\(\Rightarrow C_3H_8\)

Gọi CTĐGN là \(\left(C_3H_8\right)_n\)

Mà \(M=44\)g/mol\(\Rightarrow44n=44\Rightarrow n=1\)

Vậy CTPT là \(C_3H_8\)

A không làm mất màu dung dịch brom.

12/7 x 14/3 = 168/21 = 8

\(=\dfrac{12}{7}\times\dfrac{14}{3}=8\)

\(\dfrac{12}{7}.\dfrac{14}{3}=\dfrac{168}{21}\)

\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\)

\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}=\dfrac{2\sqrt{a}}{\sqrt{a}}=2\)

Câu 12: A

Câu 13: C

Câu 14: D

ý c đúng nha bn

C

do \(\dfrac{123}{124}=\dfrac{15375}{15500}\)

\(\dfrac{124}{125}=\dfrac{15376}{15500}\)

mà 15375<15376

vậy \(\dfrac{123}{124}< \dfrac{124}{125}\)

=1400

412 + 988

=1400

tính nhanh nha mn^^