Rút gọn P = \(\frac{a^3+b^3+c^3}{\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2}\)
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14 tháng 10 2017
Sửa đề cho nó đẹp
\(\frac{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}\)
\(=\frac{3\left(a-b\right)\left(a-c\right)\left(c-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=-3\)
15 tháng 11 2016
Phân tích mẫu thức thành nhân tử :
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2c-ab^2+ac^2-bc^2\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)\)
\(=\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]=\left(b-c\right)\left(a-c\right)\left(a-b\right).\)
Do đó : \(A=\frac{\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3}{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Nhận xét : Nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz.\)
Đặt \(b-c=x,c-a=y,a-b=z\) thì \(x+y+z=0\)
Theo nhận xét trên : \(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3.\)
PA
15 tháng 11 2016
Tử:
(b - c)3 + (c - a)3 + (a - b)3
= (b - c + c - a + a - b)3 - 3(b - c + c - a)(b - c + a - b)(c - a + a - b)
= 0 - 3(b - a)(a - c)(c - b)
= 3(a - b)(a - c)(c - b)
Mẫu:
a2(b - c) + b2(c - a) + c2(a - b)
= a2(b - c) + b2c - ab2 + ac2 - bc2
= a2(b - c) - a(b2 - c2) + bc(b - c)
= a2(b - c) - a(b - c)(b + c) + bc(b - c)
= (b - c)(a2 - ab - ac + bc)
= (b - c)[a(a - b) - c(a - b)]
= (b - c)(a - b)(a - c)
\(A=\frac{3\left(a-b\right)\left(a-c\right)\left(c-b\right)}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}\)
\(=\frac{3\left(c-b\right)}{b-c}\)
S
3 tháng 11 2018
Phân tích mẫu \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2c-ab^2+c^2a-c^2b\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b+c\right)\left(b-c\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)=\left(b-c\right)\left[a\left(a-c\right)-b\left(a-c\right)\right]\)
\(=\left(b-c\right)\left(a-b\right)\left(a-c\right)=-\left(b-c\right)\left(a-b\right)\left(c-a\right)\)
Đặt b - c = x, c - a = y, a - b = z
=> x + y + z = b - c + c - a + a - b = 0
Từ x+y+z=0 => x3+y3+z3=3xyz (tự c/m)
=>\(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3\)
14 tháng 10 2018
+) Xét tử thức: \(a^3\left(b^2-c^2\right)+b^3\left(c^2-a^2\right)+c^2\left(a^2-b^2\right)\)
\(=a^3\left(b^2-c^2\right)+\left(b^3c^2-b^2c^3\right)-\left(a^2b^3-a^2c^3\right)\)
\(=a^3\left(b-c\right)\left(b+c\right)+b^2c^2\left(b-c\right)-a^2\left(b-c\right)\left(b^2+bc+c^2\right)\)
\(=\left(b-c\right)\left(a^3b+a^3c+b^2c^2-a^2b^2-a^2bc-a^2c^2\right)\)
\(=\left(b-c\right)\left[\left(a^3b-a^2bc\right)+\left(a^3c-a^2c^2\right)+\left(b^2c^2-a^2b^2\right)\right]\)
\(=\left(b-c\right)\left[a^2b\left(a-c\right)+a^2c\left(a-c\right)-b^2\left(a-c\right)\left(a+c\right)\right]\)
\(=\left(b-c\right)\left(a-c\right)\left(a^2b+a^2c-ab^2-b^2c\right)\)
\(=\left(b-c\right)\left(a-c\right)\left[ab\left(a-b\right)+c\left(a-b\right)\left(a+b\right)\right]\)
\(=\left(b-c\right)\left(a-c\right)\left(a-b\right)\left(ab+bc+ca\right)\)
+) Xét mẫu thức: \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2c-bc^2-ab^2+ac^2\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b-c\right)\left(b+c\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)=\left(b-c\right)\left[\left(a^2-ac\right)-\left(ab-bc\right)\right]\)
\(=\left(b-c\right)\left[a\left(a-c\right)-b\left(a-c\right)\right]=\left(b-c\right)\left(a-c\right)\left(a-b\right)\)
Từ đó; ta có:
\(\frac{a^3\left(b^2-c^2\right)+b^3\left(c^2-a^2\right)+c^3\left(a^2-b^2\right)}{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}=\frac{\left(b-c\right)\left(a-c\right)\left(a-b\right)\left(ab+bc+ca\right)}{\left(b-c\right)\left(a-c\right)\left(a-b\right)}\)
\(=ab+bc+ca\). KL:...
22 tháng 9 2018
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
11 tháng 11 2017
\(\frac{a^3+b^3-c^3+3abc}{\left(a-b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}=\frac{\left(a+b\right)^3-c^3-3ab\left(a+b\right)+3abc}{2a^2+2b^2+2c^2-2ab+2bc+2ac}\)
\(=\frac{\left(a+b-c\right)\left[\left(a+b\right)^2+c\left(a+b\right)+c^2\right]-3ab\left(a+b-c\right)}{2a^2+2b^2+2c^2-2ab+2bc+2ac}\)
\(=\frac{\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2-3ab\right)}{2\left(a^2+b^2+c^2-ab+bc+ac\right)}\)
\(=\frac{a+b-c}{2}\)
\(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3=\left(a+b\right)^3+c^3+3\left(a+b\right)c\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+bc+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(=a^3+b^3+c^3+3\left(abc+c^2a+b^2c+bc^2+a^2b+ca^2+ab^2+abc\right)\)
\(=a^3+b^3+c^3+3\left[ab\left(a+b+c\right)+bc\left(a+b+c\right)+ca\left(a+b+c\right)\right]\)
\(=a^3+b^3+c^3+3\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\Rightarrow\)\(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b+c\right)\left(ab+bc+ca\right)\)
Lại có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=\left(a-b+b-c+c-a\right)^2\)
\(-2\left[\left(a-b\right)\left(b-c\right)+\left(b-c\right)\left(c-a\right)+\left(c-a\right)\left(a-b\right)\right]\)
\(=-2\left(ab-ca-b^2+bc+bc-ab-c^2+ca+ca-bc-a^2+ab\right)\)
\(=2\left(a^2+b^2+c^2-ab-bc-ca\right)=2\left(a+b+c\right)^2-6\left(ab+bc+ca\right)\)
\(\Rightarrow\)\(P=\frac{\left(a+b+c\right)^3-3\left(a+b+c\right)\left(ab+bc+ca\right)}{2\left(a+b+c\right)^2-6\left(ab+bc+ca\right)}\)
\(=\frac{\left(a+b+c\right)\left[\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\right]}{2\left[\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\right]}=\frac{a+b+c}{2}\)