Ta có : \(\frac{a-\left(c-b\right)}{b-c}+\frac{b-\left(a-c\right)}{c-a}+\frac{c-\left(b-a\right)}{a-b}=3\)

\(\Leftrightarrow\frac{a+\left(b-c\right)}{b-c}-1+\frac{b+\left(c-a\right)}{c-a}-1+\frac{c+\left(a-b\right)}{a-b}-1=0\)

\(\Leftrightarrow\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)

\(\Rightarrow\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)=0\)

\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{a+c}{\left(b-c\right)\left(a-b\right)}+\frac{b+c}{\left(c-a\right)\left(a-b\right)}=0\)

\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a^2-b^2+c^2-a^2+b^2-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)

Từ gt ta có : \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)0

Từ đó suy ra điều phải chứng minh

Ta có : \(\frac{ab+1}{b}=\frac{bc+1}{c}=\frac{ac+1}{a}\Leftrightarrow a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}\)

Từ \(a+\frac{1}{b}=b+\frac{1}{c}\Rightarrow a-b=\frac{1}{c}-\frac{1}{b}\Leftrightarrow a-b=\frac{b-c}{bc}\)(1)

Tương tự : \(b+\frac{1}{c}=c+\frac{1}{a}\Leftrightarrow b-c=\frac{c-a}{ac}\) (2) ; \(c+\frac{1}{a}=a+\frac{1}{b}\Leftrightarrow c-a=\frac{a-b}{ab}\)(3)

Nhân (1) , (2), (3) theo vế :

\(\left(a-b\right)\left(b-c\right)\left(c-a\right)=\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{a^2b^2c^2}\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(1-\frac{1}{a^2b^2c^2}\right)=0\)

Vì abc khác 1 nên\(a^2b^2c^2\ne1\) \(\Rightarrow1-\frac{1}{a^2b^2c^2}\ne0\)

Do đó \(\left(a-b\right)\left(b-c\right)\left(c-a\right)=0\Rightarrow\)a = b hoặc b = c hoặc c = a

• Với a = b , từ giả thiết ta có b = c => a = b = c
• Với b = c , từ giả thiết ta có c = a => a = b = c
• Với c = a , từ giả thiết ta có a = b => a = b = c

Vậy a = b = c 

a. Ta thấy góc DAB = góc DBC (gt) và góc ABD = góc BDC (So le trong) nên \(\Delta DAB\sim\Delta CBD\left(g-g\right)\)

b. Ta có: \(\frac{DA}{BC}=\frac{AB}{BD}\Rightarrow\frac{3}{4}=\frac{5}{BD}\Rightarrow BD=\frac{20}{3}\)

\(\frac{AB}{BC}=\frac{BD}{DC}\Rightarrow DC=\frac{4.20}{3}:3=\frac{80}{9}\)

c. Ta thấy \(\frac{S_{ABD}}{S_{BDC}}=\left(\frac{3}{4}\right)^2=\frac{9}{16}\Rightarrow\frac{S_{ABD}}{S_{ABCD}}=\frac{9}{25}\Rightarrow S_{ABCD}=\frac{125}{9}\left(cm^2\right)\)

Chúc em học tốt :)

a) Xét VP : \(4\left(p-b\right)\left(p-c\right)=4\left(\frac{a+b+c}{2}-b\right)\left(\frac{a+b+c}{2}-c\right)\)

\(=\left(a+c-b\right)\left(a+b-c\right)=\left[a+\left(c-b\right)\right].\left[a-\left(c-b\right)\right]\)

\(=a^2-\left(b-c\right)^2=a^2-b^2-c^2+2bc=VT\)

b) Xét VT : \(p^2+\left(p-a\right)^2+\left(p-b\right)^2+\left(p-c\right)^2\)

\(=\left(\frac{a+b+c}{2}\right)^2+\left(\frac{a+b+c}{2}-a\right)^2+\left(\frac{a+b+c}{2}-b\right)^2+\left(\frac{a+b+c}{2}-c\right)^2\)

\(=\frac{\left(a+b+c\right)^2+\left(b+c-a\right)^2+\left(a+c-b\right)^2+\left(a+b-c\right)^2}{4}\)

\(=\frac{4\left(a^2+b^2+c^2\right)+2\left(ab+bc+ac+bc-ac-ab+ac-bc-ab+ab-bc-ac\right)}{4}\)

\(=\frac{4\left(a^2+b^2+c^2\right)}{4}=a^2+b^2+c^2=VP\)