Nhân cả hai vế của đẳng thức cho a+b+c ta được

\(\dfrac{a+b+c}{a+b}\)+\(\dfrac{a+b+c}{a+b}\)=\(\dfrac{a+b+c}{c+a}\)=\(\dfrac{a+b+c}{90}\)

=> a+ \(\dfrac{c}{a+b}\)+1+\(\dfrac{a}{b+c}\)+1+\(\dfrac{b}{c+a}\)=\(\dfrac{2007}{90}\)

=>\(\dfrac{a}{b+c}\)+\(\dfrac{b}{c+a}\)+\(\dfrac{c}{a+b}\)=\(\dfrac{2007}{90}\)-3= 22,3-3=19,3

\(\Leftrightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{a+b+c}{90}\Leftrightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}+\dfrac{b}{c+a}=\dfrac{a+b+c}{a+b}\)\(\Leftrightarrow1+\dfrac{c}{a+b}+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1=\dfrac{2007}{90}\)

\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=\dfrac{193}{10}\)

\(\Rightarrow S=\dfrac{193}{10}\)

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

\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)

\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3\)

\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

\(=2007.\frac{1}{90}-3\)

\(=22,3-3\)

\(=19,3\)

Vậy S = 19,3

Ta có \(\dfrac{a-b}{ab+1}+\dfrac{b-c}{bc+1}+\dfrac{c-a}{ca+1}=\dfrac{\left(a-b\right)\left(bc+1\right)\left(ca+1\right)+\left(b-c\right)\left(ca+1\right)\left(ab+1\right)+\left(a-b\right)\left(bc+1\right)\left(ca+1\right)}{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}=\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}\).

\(4\sqrt{2}x\) ạ

Với cả 3 phần thì dấu "=" xảy ra tại a=b=c=1.

a) \(\dfrac{a}{1+b^2}=\dfrac{a\left(1+b^2\right)}{1+b^2}-\dfrac{ab^2}{1+b^2}=a-\dfrac{ab^2}{1+b^2}\)

(Cosi) \(\ge a-\dfrac{ab^2}{2b}=a-\dfrac{ab}{2}\)

Tương tự : \(\dfrac{b}{1+c^2}\ge b-\dfrac{bc}{2};\dfrac{c}{1+a^2}\ge c-\dfrac{ca}{2}\)

\(\Rightarrow P\ge\left(a+b+c\right)-\dfrac{ab+bc+ca}{2}\ge\left(CS\right)\left(a+b+c\right)-\dfrac{\left(a+b+c\right)^2}{6}=3-\dfrac{3^2}{6}=\dfrac{3}{2}\)

b) \(\dfrac{1}{a^2+1}=1-\dfrac{a^2}{a^2+1}\ge\left(CS\right)1-\dfrac{a^2}{2a}=1-\dfrac{a}{2}\)

Tương tự : \(\dfrac{1}{b^2+1}\ge1-\dfrac{b}{2};\dfrac{1}{c^2+1}\ge1-\dfrac{c}{2}\)

\(\Rightarrow P\ge3-\dfrac{a+b+c}{2}=3-\dfrac{3}{2}=\dfrac{3}{2}\)

c)\(P=\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}=\left(\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\right)+\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\ge\dfrac{3}{2}+\dfrac{3}{2}=3\)

b)Ta có:

\(\left|x+\dfrac{1}{1.2}\right|\ge0,\left|x+\dfrac{1}{2.3}\right|\ge0,...,\left|x+\dfrac{1}{99.100}\right|\ge0\)\(\Rightarrow\)\(\left|x+\dfrac{1}{1.2}\right|+\left|x+\dfrac{1}{2.3}\right|+...+\left|x+\dfrac{1}{99.100}\right|\ge0\)\(\Rightarrow100x\ge0\Rightarrow x\ge0\)

\(\Rightarrow x+\dfrac{1}{1.2}+x+\dfrac{1}{2.3}+...+x+\dfrac{1}{99.100}=100x\)\(\Rightarrow x+x+...+x+\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}=100x\)\(\Rightarrow99x+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..+\dfrac{1}{99}-\dfrac{1}{100}=100x\)\(\Rightarrow1-\dfrac{1}{100}=x\)

\(\Rightarrow x=\dfrac{99}{100}\)

\(\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{a}{b}=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{2}{2}-\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{a}{b}=1-\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{3}{3}-\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{2}{3}\)

Vậy phân số tối giản \(\dfrac{a}{b}=\dfrac{2}{3}\)