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

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

\(=\frac{\left(\cdot a-b-c\right)\left(a^2+b^2+c^2+ac+ab-bc\right)}{4+a^2+b^2+c^2}\)

\(=\frac{2a^2+2b^2+2c^2+2ab-2bc+2ca}{4+a^2+b^2+c^2}\)

\(=\frac{\left(a-b-c\right)^2+a^2+b^2+c^2}{4+a^2+b^2+c^2}=1\)

k mk nha

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: \(\frac{a-c}{b+c}\)

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

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

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

Đầu tiên bạn hãy tự phân tích tử số nha, kết quả là:

    \(a^3+b^3+c^3-3abc=\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)

Ta có: \(a+b+c=3\)

Vậy thay vào biểu thức, ta sẽ được:

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

\(\Leftrightarrow S=\frac{\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)

\(\Leftrightarrow S=\frac{1}{2}\left(a+b+c\right)\)

\(\Leftrightarrow S=\frac{1}{2}.3\)

\(\Leftrightarrow S=\frac{3}{2}\)

Chúc bạn học giỏi và tíck cho mìk vs nha Đỗ Nguyễn Hiền Thảo!

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)

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

mình hướng dẫn thôi được không chứ mình đá bóng bị ngã nên giờ bấm giải chi tiết không nổi

thôi mình sẽ giải chi tiết luôn nhé chứ hướng dẫn khó hiểu lắm

đặt cái vế trái là A. Ta có:

\(A=a\left(\frac{1}{\sqrt{ab+c^2}}+\frac{1}{\sqrt{ac+b^2}}\right)+b\left(\frac{1}{\sqrt{ab+c^2}}+\frac{1}{\sqrt{bc+a^2}}\right)+c\left(\frac{1}{\sqrt{ac+b^2}}+\frac{1}{\sqrt{bc+a^2}}\right)\)

\(\Rightarrow A\ge4\left(\frac{a}{\sqrt{ab+c^2}+\sqrt{ac+b^2}}+\frac{b}{\sqrt{ab+c^2}+\sqrt{bc+a^2}}+\frac{c}{\sqrt{ac+b^2}+\sqrt{bc+a^2}}\right)\)