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

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

Auto giải thích thêm câu b) (để tránh bị các thành phần spammer bắt bẻ)

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

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

Cái đầu ko rút gọn được

Cái sau:

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

Tham khảo:

Cho a≠b≠c, a+b≠c và c2+2ab-2ac-2bc=0 Hãy rút gọn \(B=\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\) - Hoc24

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

\(=\dfrac{a^2+a^2-2ac+c^2}{b^2+b^2-2bc+c^2}\)

\(=\dfrac{2a^2-2ac+c^2}{2b^2-2bc+c^2}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\Rightarrow\left\{{}\begin{matrix}bc=-ab-ac\\ab=-bc-ac\\ac=-ab-bc\end{matrix}\right.\)

\(M=\dfrac{1}{a^2+bc-ab-ac}+\dfrac{1}{b^2+ac-ab-bc}+\dfrac{1}{c^2+ab-bc-ac}\)

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

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

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

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

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

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

\(=a^2\)

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

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

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

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

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

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

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

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

a) sai đề

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

c) xem lại đề có j ib lại tui

phần a đúng đề mà