Lời giải:

Gọi biểu thức đã cho là $A$.

CM vế 1:

Ta có:

$\frac{a+b}{a+b+c}> \frac{a+b}{a+b+c+d}$

$\frac{b+c}{b+c+d}> \frac{b+c}{a+b+c+d}$

$\frac{c+d}{c+d+a}> \frac{c+d}{a+b+c+d}$

$\frac{d+a}{d+a+b}> \frac{d+a}{a+b+c+d}$

Cộng lại: $A> \frac{2(a+b+c+d)}{a+b+c+d}=2>1$

CM vế 2:

Ta thấy $\frac{a+b}{a+b+c}-\frac{a+b+d}{a+b+c+d}=\frac{-cd}{(a+b+c)(a+b+c+d)}< 0$ với $a,b,c,d>0$

$\Rightarrow \frac{a+b}{a+b+c}< \frac{a+b+d}{a+b+c+d}$

Hoàn toàn tương tự với các phân thức còn lại:

$\Rightarrow A< \frac{3(a+b+c+d)}{a+b+c+d}=3$

Ta có đpcm.

nếu \(\frac{a+b}{b+c}\)=\(\frac{c+d}{d+a}\)

  => a =c

program hotrotinhoc;
function toigian(x,y: integer): integer;
var z: integer;
begin
while y<>0 do
begin
z:=x mod y;
x:=y;
y:=z;
end;
toigian:=x;
end;
var i,n,a,b,c,d: integer;
tu,mau: real;
begin
readln(a,b,c,d);
if (c=0) and (d=0) then write(a,' ',b);
if (c=0) and (d<>0) then write(c,'/',d,'=0');
if (c<>0) and (d<>0) then
if b/toigian(a,b)=1 then
writeln(a,'/',b,'=',(a/toigian(a,b))/b/toigian(a,b):1:0) else
writeln(a,'/',b,'=',(a/toigian(a,b)):1:0,'/',b/toigian(a,b):1:0);
if (c<>0) and (d<>0) then
begin
tu:=(a*d+c*b)/toigian(a*d+c*b,b*d);
mau:=(b*d)/toigian(a*d+c*b,d*b);
if (mau=1) or (tu=0) then writeln(a,'/',b,'+',c,'/',d,'=',tu/mau:1:0) else
writeln(a,'/',b,'+',c,'/',d,'=',tu:1:0,'/',mau:1:0);
tu:=(a*d-c*b)/toigian(a*d-c*b,b*d);
mau:=(b*d)/toigian(a*d-c*b,d*b);
if (mau=1) or (tu=0) then writeln(a,'/',b,'-',c,'/',d,'=',tu/mau:1:0) else
writeln(a,'/',b,'-',c,'/',d,'=',tu:1:0,'/',mau:1:0);
tu:=(a*c)/toigian(a*c,b*d);
mau:=(b*d)/toigian(a*c,b*d);
if mau=1 then writeln(a,'/',b,'*',c,'/',d,'=',tu/mau:1:0) else
writeln(a,'/',b,'*',c,'/',d,'=',tu:1:0,'/',mau:1:0);
tu:=(a*d)/toigian(a*d,b*c);
mau:=(b*c)/toigian(a*d,b*c);
if mau=1 then writeln(a,'/',b,'/',c,'/',d,'=',tu/mau:1:0) else
writeln(a,'/',b,'/',c,'/',d,'=',tu:1:0,'/',mau:1:0)
end;
readln
end.

program hotrotinhoc;

function toigian(x,y: integer): integer;

var z: integer;

begin

while y<>0 do

begin

z:=x mod y;

x:=y;

y:=z;

end;

toigian:=x;

end;

var i,n,a,b,c,d: integer;

tu,mau: real;

begin

readln(a,b,c,d);

if (c=0) and (d=0) then write(a,' ',b);

if (c=0) and (d<>0) then write(c,'/',d,'=0');

if (c<>0) and (d<>0) then

if b/toigian(a,b)=1 then

writeln(a,'/',b,'=',(a/toigian(a,b))/b/toigian(a,b):1:0) else

writeln(a,'/',b,'=',(a/toigian(a,b)):1:0,'/',b/toigian(a,b):1:0);

if (c<>0) and (d<>0) then

begin

tu:=(a*d+c*b)/toigian(a*d+c*b,b*d);

mau:=(b*d)/toigian(a*d+c*b,d*b);

if (mau=1) or (tu=0) then writeln(a,'/',b,'+',c,'/',d,'=',tu/mau:1:0) else

writeln(a,'/',b,'+',c,'/',d,'=',tu:1:0,'/',mau:1:0);

tu:=(a*d-c*b)/toigian(a*d-c*b,b*d);

mau:=(b*d)/toigian(a*d-c*b,d*b);

if (mau=1) or (tu=0) then writeln(a,'/',b,'-',c,'/',d,'=',tu/mau:1:0) else

writeln(a,'/',b,'-',c,'/',d,'=',tu:1:0,'/',mau:1:0);

tu:=(a*c)/toigian(a*c,b*d);

mau:=(b*d)/toigian(a*c,b*d);

if mau=1 then writeln(a,'/',b,'*',c,'/',d,'=',tu/mau:1:0) else

writeln(a,'/',b,'*',c,'/',d,'=',tu:1:0,'/',mau:1:0);

tu:=(a*d)/toigian(a*d,b*c);

mau:=(b*c)/toigian(a*d,b*c);

if mau=1 then writeln(a,'/',b,'/',c,'/',d,'=',tu/mau:1:0) else

writeln(a,'/',b,'/',c,'/',d,'=',tu:1:0,'/',mau:1:0);

end;

readln

end.

\(VT=\frac{a+b-\left(b+d\right)}{d+b}+\frac{\left(d+c\right)-\left(b+c\right)}{b+c}+\frac{\left(b+a\right)-\left(a+c\right)}{c+a}+\frac{\left(c+d\right)-\left(a+d\right)}{a+d}\)

\(VT=\frac{a+b}{d+b}-1+\frac{\left(d+c\right)}{b+c}-1+\frac{\left(b+a\right)}{c+a}-1+\frac{\left(c+d\right)}{a+d}-1\)

\(VT=\left(a+b\right).\left(\frac{1}{d+b}+\frac{1}{a+c}\right)+\left(d+c\right).\left(\frac{1}{b+c}+\frac{1}{a+d}\right)-4\)

Chứng minh đc bđt sau: Với x; y > 0 ta có  \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

Áp dụng ta có: \(VT\ge\left(a+b\right).\frac{4}{d+b+a+c}+\left(d+c\right).\frac{4}{b+c+a+d}-4\ge\frac{4.\left(a+b+c+d\right)}{a+b+c+d}-4=0\)

=> ĐPCM

Cộng 4 vào vế trái nhá

\(VT+4=\left(\dfrac{a-d}{d+b}+1\right)+\left(\dfrac{d-b}{b+c}+1\right)+\left(\dfrac{b-c}{c+a}+1\right)+\left(\dfrac{c-a}{a+d}+1\right)\)

\(=\dfrac{a+b}{d+b}+\dfrac{d+c}{b+c}+\dfrac{a+b}{c+a}+\dfrac{c+d}{a+d}\)

\(=\left(a+b\right)\left(\dfrac{1}{d+b}+\dfrac{1}{c+a}\right)+\left(c+d\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+d}\right)\)

\(\ge\left(a+b\right).\dfrac{4}{a+b+c+d}+\left(c+d\right).\dfrac{4}{a+b+c+d}\)

\(=\left(a+b+c+d\right).\dfrac{4}{a+b+c+d}\)\(=4\)

\(\Rightarrow VT\ge0=VP\)(Đpcm)

Ta có: \(\frac{a+b}{b+c}=\frac{c+d}{d+a}.\)

\(\Rightarrow\frac{a+b}{c+d}=\frac{b+c}{d+a}\)

\(\Rightarrow\frac{a+b}{c+d}+1=\frac{b+c}{d+a}+1.\)

\(\Rightarrow\frac{a+b}{c+d}+\frac{c+d}{c+d}=\frac{b+c}{d+a}+\frac{d+a}{d+a}.\)

\(\Rightarrow\frac{a+b+c+d}{c+d}=\frac{b+c+d+a}{d+a}\)

Nếu \(a+b+c+d\ne0.\)

\(\Rightarrow c+d=d+a\)

\(\Rightarrow c=a\left(đpcm1\right).\)

Nếu \(a+b+c+d=0\) thì hợp với đề.

\(\Rightarrow a+b+c+d=0\left(đpcm2\right).\)

Chúc bạn học tốt!

chỉ cần bài 1,2,3 nữa thui ak