program sosanh;

uses crt;

var a,b,c : real;

begin

      readln(a,b);

      c:= a-b;

      if c>0 then writeln(' a lon hon b');

      if c<0 then writeln('a nho hon b');

      if c=0 then writeln('a bang b');

      readln;

end.

Dốt tin nên chỉ làm dc vậy, đúng thì 5 sao nhé

B

Áp dụng định lý bơ du ta được : 

\(2^3+3.2^2+2a+5=8+12+2a+5=25+2a\)

Vậy \(f\left(x\right)=25+2a\)

Sửa lại đề: \(\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}+\frac{1}{1-x}\)

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

\(=\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2-1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

\(=\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{x}{x^2+x+1}\)

BT <=> 

\(A=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)

\(=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)

\(=\frac{x^2-9-x-3}{MTC}=\frac{x^2-x-12}{MTC}\)

A = \(\frac{x+2}{x+3}\)\(-\frac{5}{X^2+X-6}\)\(+\frac{1}{2-X}\)

A= \(\frac{x+2}{x+3}\)\(-\frac{5}{\left(X-2\right)\left(X+3\right)}\)\(-\frac{1}{X-2}\)

A = \(\frac{\left(X+2\right)\left(X-2\right)}{\left(X-2\right)\left(X+3\right)}\)\(-\frac{5}{\left(X-2\right)\left(X+3\right)}\)\(-\frac{X+3}{\left(X-2\right)\left(X+3\right)}\)

A= \(\frac{\left(X+2\right)\left(X-2\right)-5-\left(X+3\right)}{\left(X-2\right)\left(X+3\right)}\)

A= \(\frac{X-4-5-X-3}{\left(X-2\right)\left(X+3\right)}\)

A= \(-\frac{12}{\left(X-2\right)\left(X+3\right)}\)

Sửa đề : \(x^2-y^2+x-y\)

\(=\left(x-y\right)\left(x+y\right)+\left(x-y\right)=\left(x-y\right)\left(x+y+1\right)\)

hay sửa như này =)) 

\(x^2-y^2-x-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-1\right)\)

\(x^2-y^2+x-y\)

\(=\left(x-y\right)\left(x+y\right)+\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y+1\right)\)