1. B

2. A

3. D

Giải thích vì sao chọn căn cứ vapf nào chọn

\(a=\lim\limits_{x\rightarrow-3}\dfrac{x+3}{\left(x+3\right)\left(x-3\right)}=\lim\limits_{x\rightarrow-3}\dfrac{1}{x-3}=-\dfrac{1}{6}\)

\(b=\lim\limits_{x\rightarrow2}\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\lim\limits_{x\rightarrow2}\dfrac{x+3}{x+2}=\dfrac{5}{4}\)

\(c=\lim\limits_{x\rightarrow4}\dfrac{\left(x-4\right)\left(x+4\right)}{\left(x+5\right)\left(x-4\right)}=\lim\limits_{x\rightarrow4}\dfrac{x+4}{x+5}=\dfrac{8}{9}\)

\(d=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\dfrac{x+2}{x-1}=4\)

\(e=\lim\limits_{x\rightarrow2}\dfrac{x+7-9}{\left(x-2\right)\left(\sqrt{x+7}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\left(x-2\right)\left(\sqrt{x+7}+3\right)}=\lim\limits_{x\rightarrow2}\dfrac{1}{\sqrt{x+7}+3}=\dfrac{1}{6}\)

\(f=\lim\limits_{x\rightarrow1}\dfrac{x+3-4}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{\sqrt{x+3}+2}=\dfrac{1}{4}\)

\(h=\lim\limits_{x\rightarrow-3}\dfrac{x+7-4}{\left(x+3\right)\left(\sqrt{x+7}+2\right)}=\lim\limits_{x\rightarrow-3}\dfrac{x+3}{\left(x+3\right)\left(\sqrt{x+7}+2\right)}=\lim\limits_{x\rightarrow-3}\dfrac{1}{\sqrt{x+7}+2}=\dfrac{1}{4}\)

Bài 1:

a, 

= lim_x->-3 \(\dfrac{x+3}{\left(x+3\right)\left(x-3\right)}\)

= lim_x->3  x-3

= -3 -3

= -6

b, 

= lim_x->2 \(\dfrac{\left(x-2\right)\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}\)

= lim_x->2  \(\dfrac{x+3}{x+2}\)

= \(\dfrac{5}{4}\)

c,

= lim_x->4   \(\dfrac{\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+5\right)}\)

= lim_x->4   \(\dfrac{\left(x+4\right)}{\left(x+5\right)}\)

= \(\dfrac{8}{9}\)

d,

= lim_x->2   \(\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x-1\right)}\)

= lim_x->2   \(\dfrac{\left(x+2\right)}{\left(x-1\right)}\)

= 4

2:

uses crt;

var a:array[1..30]of integer;

i,dem:integer;

begin

clrscr;

for i:=1 to 30 do

begin

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

end;

dem:=0;

for i:=1 to n do 

  if a[i] mod 5<>0 then inc(dem);

writeln('So phan tu khong chia het cho 5 la: ',dem);

readln;

end.

bài nào???

đâu tui hok thấy ??????

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bài 3

a: Xét ΔABC có 

M là trung điểm của bC

I là trung điểm của AC

Do đó: MI là đường trung bình của ΔABC

Suy ra: MI=AB/2=3(cm)

\(2CH_4 \xrightarrow{làm\ lạnh\ nhanh,t^o}C_2H_2 + 3H_2\\ C_2H_2 + H_2 \xrightarrow{t^o,PbCO_3} C_2H_4\\ C_2H_4 + H_2O \xrightarrow{H^+} C_2H_5OH\\ C_2H_5OH \xrightarrow{t^o,xt} C_2H_4 + H_2O\\ C_2H_2 + H_2O \xrightarrow{xt} CH_3CHO\\ C_2H_5OH + CuO \xrightarrow{t^o} CH_3CHO + Cu + H_2O\)