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\(=\left(a+b-c\right)\left(a-b\right)^2\) nha !
P/S:Ko có mục đích xấu,đăng lên cho bạn thôi.
Bài 1:
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ac=0\Leftrightarrow bc=-ab-ac\)
\(\dfrac{a^2}{a^2+2bc}=\dfrac{a^2}{a^2+bc-ab-ac}=\dfrac{a^2}{\left(a-c\right)\left(a-b\right)}\)
CMTT: \(\left\{{}\begin{matrix}\dfrac{b^2}{b^2+2ca}=\dfrac{b^2}{\left(b-c\right)\left(b-a\right)}\\\dfrac{c^2}{c^2+2ab}=\dfrac{c^2}{\left(b-c\right)\left(a-c\right)}\end{matrix}\right.\)
\(M=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
Bài 2:
\(a^3+b^3+c^3-3abc=\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3abc-3a^2b-3ab^2\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)(do \(a+b+c=0\))
\(\Rightarrow A=\dfrac{0}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}=0\)
chị giải thích cho em cái đoạn này với ạ
\(\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
a) \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2\right)-\left(a+b+c\right)\left(ab+bc+ac\right)\)
\(=a^3+ab^2+ac^2+a^2b+b^3+c^2b+a^2c+b^2c+c^3-a^2b-abc-a^2c-ab^2-b^2c-abc-abc-bc^2-ac^2\)
\(=a^3+b^3+c^3-3abc\left(đpcm\right)\)
b) Bạn chỉ cần nhân bung cả 2 vế ra là được á .
c) \(2\left(a+b+c\right)\left(\dfrac{b}{2}+\dfrac{c}{2}-\dfrac{a}{2}\right)\)
\(=2\left(a+b+c\right)\left(\dfrac{b+c-a}{2}\right)\)
\(=\left(a+b+c\right)\left(b+c-a\right)\)
\(=ab+ac-a^2+b^2+bc-ab+bc+c^2-ac\)
\(=2bc+b^2+c^2-a^2\left(đpcm\right)\)
a^3+b^3+c^3-3abc=(a+b)^3-3a^2.b-3a.b^2-3abc=[(a+b)^3+c^3]-3ab(a+b+c)=(a+b+c).[(a+b)^2-c.(a+b)+c^2]-3ab(a+b+c)=(a+b+c).(a^2+2ab+b^2-ac-bc+c^2-3ab)=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
1) a3+b3+c3-3abc = (a+b)3-3ab(a+b)+c3-3abc
= (a+b+c)(a2+2ab+b2-ab-ac+c2) -3ab(a+b+c)
= (a+b+c)( a2+b2+c2-ab-bc-ca)
nhanh lên với ak
Ta có :
a^3+b^3+c^3-3abc
=(a+b)^3+c^3-3ab(a+b) - 3abc
=(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)
=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
=> 2(a^3+b^3+c^3-3abc)= (a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ca)
=(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]