Giúp mình câu này với

Số mol \(0,6.10^{23}\text{ phân tử }O_2=0,6.16.2.10^{23}=19,2.10^{23}\)

Suy ra \(M_A=\frac{m_A}{n_A}=\frac{8}{19,2.10^{23}}\left(\text{g/mol}\right)\)

P/s: Ko chắc

a, điều kiện xác định là \(x\ne1;x\ne-1\)

\(\frac{3x+3}{x^2-1}\)

\(=\frac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{3}{x-1}\)

b, để \(\frac{3x+3}{x^2-1}=-2\Rightarrow\frac{3}{x-1}=-2\)

\(\Rightarrow-2x+2=3\)

\(\Rightarrow-2x=1\)

\(\Rightarrow x=-\frac{1}{2}\)

a. ĐKXĐ: x² - 1\(\ne\)0 (=) x \(\ne\)\(\pm\)1

b. \(\frac{3x+3}{x^2-1}\)

\(=\frac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{3}{x+1}\)với x \(\pm\)1

c. \(\frac{3}{x+1}=-2\)

\(\Rightarrow\)\(\left(x+1\right).\left(-2\right)=3\)

\(-2x-2=3\)

\(-2x=5\)

\(x=-\frac{5}{2}\)(t/m đk)

= bao nhiu

\(2^3\)hay \(2x^3\)

\(f\left(x\right)=x^7\)

           \(=x^7+x^6+x^5+x^4-x^6-x^5-x^4-x^3+x^3+x^2+x+1-x^2-x-1\)

            \(=x^4\left(x^3+x^2+x+1\right)-x^3\left(x^3+x^2+x+1\right)+\left(x^3+x^2+x+1\right)-\left(x^2+x+1\right)\)

             \(=\left(x^3+x^2+x+1\right)\left(x^4-x^3+1\right)-\left(x^2+x+1\right)\)

Vì \(\left(x^3+x^2+x+1\right)\left(x^4-x^3+1\right) ⋮ x^3+x^2+x+1 \forall x\) mà \(-\left(x^2+x+1\right)\) có bậc 2 , \(x^3+x^2+x+1\) có bậc 3

=> đa thức dư cần tìm là \(-\left(x^2+x+1\right)\)

program bai1;

uses crt;

var a,b: integer;

begin

clrscr;

write('nhap so a ='); readln(a);

write('nhap so b ='); readln(b);

writeln(' hai so vua nhap la so: ' ,a,' va ',b);

if a> b then writeln('so lon nhat la : ' ,a) else

writeln('so lon nhat la : ' ,b);

readln;

end.

program ho_tro;

uses crt;

var a,b,max: real;

begin

clrscr;

writeln(' nhap a va b:'); readln(a,b);

max:=a;

max:=b;

if  max<a then max:=a;

if max<b then max:=b;

writeln(' so lon nhat la :',max);

readln;

end.

lần sau nếu có hỏi j về tin thì qua bên h mik sẽ giúp