\(\frac{1}{b}+\frac{1}{c}=\frac{1}{d}\Leftrightarrow\frac{b+c}{bc}=\frac{1}{d}\Leftrightarrow d=\frac{bc}{b+c}\)

Ta có

\(HD\perp AB;AC\perp AB\) => HD//AC \(\Rightarrow\frac{BD}{BC}=\frac{HD}{AC}=\frac{d}{b}\Rightarrow d=\frac{b.BD}{BC}\) (*)

Xét tg ABC có AD là phân giác của \(\widehat{A}\) nên

\(\frac{BD}{AB}=\frac{CD}{AC}\) (Trong tam giác, đường phân giác của một góc chia cạnh đối diện thành hai đoạn tỉ lệ với hai cạnh kề của hai đoạn ấy)

\(\Rightarrow\frac{BD}{c}=\frac{CD}{b}=\frac{BD+CD}{b+c}=\frac{BC}{b+c}\Rightarrow BC=\frac{BD.\left(b+c\right)}{c}\) Thay vào (*)

\(d=\frac{b.BD}{\frac{BD.\left(b+c\right)}{c}}=\frac{b.BD.c}{BD.\left(b+c\right)}=\frac{bc}{b+c}\Leftrightarrow\frac{1}{b}+\frac{1}{c}=\frac{1}{d}\left(dpcm\right)\)

a, \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{-6}{x-1}\)

\(=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{4x^2-3x+17+2x^2-3x+1-6x^2+6x+6}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(=\frac{24}{\left(x-1\right)\left(x^2+x+1\right)}\)

b, \(\frac{-3}{x+1}+\frac{2}{x^2-x+1}+\frac{6+3x^2}{x^3+1}\)

\(=\frac{-3\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{6+3x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

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

Bài 1 : 

\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)

\(=x^2-4x+4-x+9=x^2-5x+13\)

Bài 2 : 

a, \(P=\frac{1-4x^2}{4x^2-4x+1}=\frac{\left(1-2x\right)\left(2x+1\right)}{\left(2x-1\right)^2}\)

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

b, Thay x = -4 ta được : 

\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)