Ko phải ko ai mún giúp bn nhưng mà BÀI này... QUÁ KHÓ

Chúc bn sớm giải dc nha, chứ mik thì chắc là bó tay r đó!!!

bài này mình học là xài hẳng đẳng thức nâng cao đây bạn, có vẻ khó:)

Đặt \(b-c=x,c-a=y,a-b=z\)

\(\Rightarrow x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)

\(\Rightarrow\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3=3\left(b-c\right)\left(c-a\right)\left(a-b\right)\)(1)

Ta có: 

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

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

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

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

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

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

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

Từ (1) và (2) giá trị biểu thức cần tìm là -3.

Chúc bạn học tốt

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

\(=\frac{5\left(...\right)}{2\left(...\right)}=\frac{5}{2}\)

ý B hình như mk giải rồi thì phải

\(A=\frac{a^2.a^7.b^2.c^8}{a^5.b^3.\left(-c\right)^4}=\frac{a^9.b^2.c^8}{a^5.b^3.c^4}=\frac{a^4.c^4}{b}\)

\(B=\left(\frac{3}{7}\right)^5.\frac{\left(25\right)^4.7^5}{15^4.20^2}=\frac{3^5.\left(5^2\right)^4.7^5}{7^5.\left(3.5\right)^4.\left(4.5\right)^2}=\frac{3^5.5^8.7^5}{7^5.3^4.5^4.4^2.5^2}\)

    \(=\frac{3^5.5^8.7^5}{7^5.3^4.5^6.4^2}=\frac{3.5^2}{4^2}=\frac{75}{16}\)

P ơi là Rút gọn mk quên k ghi đề bài

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

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

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

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

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

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

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

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

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

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

Sửa lại: \(\frac{a-c}{b+c}\)

Sửa đề: \(P=\frac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)

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

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

\(P=\frac{\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)}{2.\left(a^2+b^2+c^2-ab-bc-ca\right)}\)

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

\(P=\frac{5\left(a^2+b^2+c^2-ab-ac-bc\right)}{2.\left(a^2+b^2+c^2-ab-bc-ca\right)}\)( a+b+c=0)

\(P=\frac{5}{2}\left[\left(a^2+b^2+c^2-ab-bc-ca\right)\ne0\right]\)