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\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left(a-b\right)-b^2\left(b-c\right)+c^2\left(a-b\right)\)
\(=\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a+b-b-c\right)=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(ab^2-ac^2-b^3+bc^2\)
\(=b^2\left(a-b\right)-c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(b^2-c^2\right)=\left(a-b\right)\left(b-c\right)\left(b+c\right)\)
Vậy \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
\(=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(b+c\right)}=\frac{a-c}{b+c}\)
Có a2(b-c) + b2(c-a) + c2(a-b)
= a2(b-c) - b2(a-c) + c2(a-b)
= a2(b-c) - b2(b-c+a-b) + c2(a-b)
= a2(b-c) - b2(b-c) - b2(a-b) + c2(a-b)
=[a2(b-c) - b2(b-c)] - [b2(a-b) - c2(a-b)]
=(b-c)(a2-b2) - (a-b)(b2-c2)
=(b-c)(a-b)(a+b) - (a-b)(b-c)(b+c)
=(b-c)(a-b)[(a+b)-(b+c)]
=(b-c)(a-b)(a-c)
Có ab2 - ac2 - b3 + bc2
= (ab2-ac2) - (b3-bc2)
=a(b2-c2) - b(b2-c2)
=(b2-c2)(a-b)
=(b-c)(b+c)(a-b)
Có a2(b-c) + b2(c-a) + c2(a-b) / ab2 - ac2 - b3 + bc2
= (b-c)(a-b)(a-c) / (b-c)(b+c)(a-b)
= (a-c) / (b+c)
Mình tính thử a ,b ,c bằng nhau đó
Mình nghĩ là 0,037037037037037037
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)
\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(A=\frac{a+b}{a^3+b^3}=\frac{a+b}{\left(a+b\right)\left(a^2-ab+b^2\right)}=\frac{1}{a^2-ab+b^2}\)
\(C=\frac{2ab-b}{8a^3-1}=\frac{b\left(2a-1\right)}{\left(2a-1\right)\left(4a^2+2a+1\right)}=\frac{b}{4a^2+2a+1}\)
Câu b xem lại đề đi nhé
Sửa đề cho dễ đọc
\(1P=\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)
\(\Leftrightarrow1P=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a^2+b^2+c^2-ab-bc-ca}=a+b+c\)