a: 20inch=50,8cm=50,80cm

b: 30inch=76,20cm

Câu 2: 

uses crt;

var a:array[1..100]of integer;

i,n,t:integer;

begin

clrscr;

write('Nhap n='); readln(n);

for i:=1 to n do

begin

write('A[',i,']='); readln(a[i]);

end;

t:=0;

for i:=1 to n do 

if (4<a[i]) and (a[i]<15) then t:=t+a[i];

writeln(t);

readln;

end.

Bài 2:

\(a,\dfrac{-1}{3}.\dfrac{5}{7}=\dfrac{-5}{21}\\ b,\dfrac{1}{2}.\dfrac{-3}{4}=\dfrac{-3}{8}\\ c,\dfrac{19}{7}.\dfrac{7}{15}=\dfrac{19.1}{1.15}=\dfrac{19}{15}\\ d,\dfrac{5}{11}:\left(-9\right)=\dfrac{5}{11}.\dfrac{-1}{9}=\dfrac{-5}{99}\\ e,-1:\dfrac{3}{5}=-1.\dfrac{5}{3}=-\dfrac{5}{3}\\ f,\dfrac{-2}{9}:\dfrac{2}{9}:\dfrac{-9}{14}=-\left(\dfrac{2}{9}:\dfrac{2}{9}\right).\dfrac{-14}{9}=-1.\dfrac{-14}{9}=\dfrac{14}{9}\\ k,\left(-\dfrac{2}{7}\right)^2=\left(-1\right)^2.\left(\dfrac{2}{7}\right)^2=1.\dfrac{2^2}{7^2}=\dfrac{4}{49}\\ l,\dfrac{1}{3}.\left(\dfrac{1}{3}\right)^3=\left(\dfrac{1}{3}\right)^4=\dfrac{1^4}{3^4}=\dfrac{1}{243}\\ m,\left(-\dfrac{1}{2}\right)^2.\left(\dfrac{1}{2}\right)^3=\left(-1\right)^2.\left(\dfrac{1}{2}\right)^{2+3}=1.\left(\dfrac{1}{2}\right)^5=1.\dfrac{1^5}{2^5}=\dfrac{1}{32}\)

Bài 1:

\(a,\dfrac{1}{-8}+\dfrac{-5}{8}=\dfrac{-1}{8}+\dfrac{-5}{8}=\dfrac{-\left(1+5\right)}{8}=-\dfrac{6}{8}=\dfrac{-6:2}{8:2}=-\dfrac{3}{4}\\ b,\dfrac{1}{7}+\dfrac{-3}{7}=\dfrac{1-3}{7}=-\dfrac{2}{7}\\ c,\dfrac{-12}{35}+\dfrac{-7}{35}=\dfrac{-\left(12+7\right)}{35}=\dfrac{-19}{35}\\ d,\dfrac{1}{6}+\dfrac{2}{5}=\dfrac{1.5+2.6}{6.5}=\dfrac{5+12}{30}=\dfrac{17}{30}\\ e,\dfrac{3}{5}+\dfrac{-7}{4}=\dfrac{3.4-7.5}{5.4}=\dfrac{12-35}{20}=\dfrac{-23}{20}\\ g,-2-\left(-\dfrac{1}{5}\right)=-2+\dfrac{1}{5}=\dfrac{-2.5+1}{5}=\dfrac{-9}{5}\\ h,\dfrac{2}{3}-\left(-1\right)=\dfrac{2}{3}+1=\dfrac{5}{3}\\ i,4-\dfrac{2}{3}=\dfrac{4.3-2}{3}=\dfrac{12-2}{3}=\dfrac{10}{3}\\ j,\dfrac{3}{4}-2=\dfrac{3-2.4}{4}=\dfrac{-5}{4}\\ k,-1-\left(-\dfrac{2}{3}\right)=-1+\dfrac{2}{3}=\dfrac{-1.3+2}{3}=\dfrac{-3+2}{3}=-\dfrac{1}{3}\)

Câu 22:

\(n_{CH_3COOH}=\dfrac{45}{60}=0,75\left(mol\right)\)

PT: \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H₂SO₄, t^o)

Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,75\left(mol\right)\)

Mà: H = 80%
\(\Rightarrow n_{CH_3COOC_2H_5\left(TT\right)}=0,75.80\%=0,6\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5\left(TT\right)}=0,6.88=52,8\left(g\right)\)

Câu 23:

m_CaCO3 = 20.80% = 16 (g) \(\Rightarrow n_{CaCO_3}=\dfrac{16}{100}=0,16\left(mol\right)\)

PT: \(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\)

a, Theo PT: \(n_{CH_3COOH}=2n_{CaCO_3}=0,32\left(mol\right)\Rightarrow m_{CH_3COOH}=0,32.60=19,2\left(g\right)\)

b, \(n_{CO_2}=n_{CaCO_3}=0,16\left(mol\right)\Rightarrow V_{CO_2}=0,16.22,4=3,584\left(l\right)\)

16 C

17 A

18 B

19 A

20 C

21 B

22 B

23 C

24 B

25 D

26 D

27 B

28 C

29 A

30 D

31 C

Ta có: \(7x+4=x-2m\left(1\right)\)

Thay \(x=6\) vào \(\left(1\right)\) ta có:

\(7.6+4=6-2m\)

\(\Rightarrow46=6-2m\)

\(\Rightarrow-2m=40\)

\(\Rightarrow m=-20\)