\(A=\frac{a^2+bc}{b+ac}+\frac{b^2+ca}{c+ab}+\frac{c^2+ab}{a+bc}\)

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

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

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

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

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

Tương tự:

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

Cộng lại:

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

P/S:Đây nha ! Bài này lớp 7 chắc xem xong key cũng chắc là đang khó hiểu nhỉ ? Đưa giấy bút ra rồi nháp vài cái là hiểu ngay thôi !

1. Cho \(4a^2+b^2=5ab\) và 2a>b>0

Tính \(A=\frac{ab}{4a^2-b^2}\)

2.Cho \(2x^2+2y^2=5xy\)và x>y>0

Tính \(A=\frac{x+y}{x-y}\)

3.Cho \(a^3+b^3+c^3=3ab\)

Tính \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)

4. Cho \(a+b+c=0\left(a,b,c\ne0\right)\)

Rút gọn: \(A=\frac{ab}{a^2+b^2-c}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)

5.Cho \(a\ne b,b\ne c,c\ne a\)và ab+bc+ac =1

Tính \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(a+c\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)

Lm đc càng nhiều càng tốt nha. Giúp mk vs nha!!