Ta có

\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}=\frac{a^2+ab-bc-ab}{\left(a+b\right)\left(a+c\right)}=\frac{a\cdot\left(a+b\right)-b\cdot\left(c+a\right)}{\left(a+b\right)\left(c+a\right)}=\frac{a}{a+c}-\frac{b}{a+b}\left(1\right)\)

tương tự

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

\(\frac{c^2-ab}{\left(c+a\right)\left(b+c\right)}=\frac{c}{c+b}-\frac{a}{a+b}\left(3\right)\)

Cộng (1);(2) và (3) ta có

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

thank bạn nha

quy đồng là ra

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

Đặt \(x=\frac{b}{c-a},y=\frac{c}{a-b},z=\frac{a}{b-c}\) , suy ra : \(P=-xy-yz-xz\)

Lại có : \(\left(x-1\right)\left(y-1\right)\left(z-1\right)=\left(x+1\right)\left(y+1\right)\left(z+1\right)\)

\(\Rightarrow xy+yz+xz=-1\Rightarrow P=1\)

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

\(=3x+\frac{3}{x}=3\left(x+\frac{1}{x}\right)\)

\(=\frac{-bc\left(b-c\right)}{\left(a-b\right)\left(c-a\right)\left(b-c\right)}+\frac{-ca\left(c-a\right)}{\left(b-c\right)\left(a-b\right)\left(c-a\right)}+\frac{-ab\left(a-b\right)}{\left(c-a\right)\left(b-c\right)\left(a-b\right)}\)

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

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

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

\(=\frac{\left(a-c\right)\left[b\left(b-a\right)-c\left(b-a\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

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

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

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

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

\(P=\frac{0}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\)

\(P=0\)

Xét: \(f\left(x\right)=\frac{x^2-bc}{\left(x+b\right)\left(x+c\right)}+\frac{b^2-xc}{\left(b+c\right)\left(b+x\right)}+\frac{c^2-xb}{\left(c+x\right)\left(c+b\right)}\)

\(\Rightarrow f\left(a\right)=P\)

Ta có: \(f\left(b\right)=\frac{b^2-bc}{2b\left(b+c\right)}+\frac{b^2-bc}{2b\left(b+c\right)}+\frac{c^2-b^2}{\left(c+b\right)\left(c+b\right)}\)

\(\Rightarrow f\left(b\right)=\frac{2b\left(b-c\right)}{2b\left(b+c\right)}+\frac{\left(c-b\right)\left(c+b\right)}{\left(c+b\right)\left(c+b\right)}=\frac{b-c}{b+c}+\frac{c-b}{c+b}=0\left(1\right)\)

Chứng minh tương tự ta cũng có: \(f\left(c\right)=0\left(2\right)\)

Từ (1) và (2) suy ra \(f\left(x\right)=0\left(\forall x\right)\Rightarrow f\left(a\right)=0\left(\forall x\right)\)

Vậy A =0

mình ghi sai rồi \(P=1\)mới đúng chứ kg phải A =1

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

Tương tự:

\(b^2+ab-c^2-ac=\left(b-c\right)\left(a+b+c\right)\)

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

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

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

cảm ơn b nha ^^