Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. a + b + c = 0 \(\Rightarrow\)a + b = -c \(\Rightarrow\)( a + b )2 = ( -c )2 \(\Rightarrow\)a2 + b2 - c2 = -2ab
Tương tự : b2 + c2 - a2 = -2bc ; c2 + a2 - b2 = -2ac
Ta có : \(\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}\)
\(=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ac}=\frac{-1}{2}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(=\frac{-1}{2}\left(\frac{a+b+c}{abc}\right)=0\)
2. tương tự
3,4 . có ở dưới, câu hỏi của Quyết Tâm chiến thắng
Cho mình sửa đề một chút nha 
\(A=\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}\)(*)
Theo bài ra , ta có :
\(\left(+\right)a+b+c=0\)
\(\Leftrightarrow a+b=-c\)
\(\Leftrightarrow\left(a+b\right)^2=c^2\)
\(\Leftrightarrow a^2+2ab+b^2=c^2\)
\(\Leftrightarrow a^2+b^2-c^2=-2ab\) (1)
\(\left(+\right)a+b+c=0\)
\(\Leftrightarrow b+c=-a\)
\(\Leftrightarrow\left(b+c\right)^2=a^2\)
\(\Leftrightarrow b^2+2bc+c^2=a^2\)
\(\Leftrightarrow b^2+c^2-a^2=-2bc\) (2)
\(\left(+\right)a+b+c=0\)
\(\Leftrightarrow a+c=-b\)
\(\Leftrightarrow\left(a+c\right)^2=b^2\)
\(\Leftrightarrow a^2+2ac+c^2=b^2\)
\(\Leftrightarrow a^2+c^2-b^2=-2ac\) (3)
Thay (1) , (2) , (3) vào (*) ta được
\(A=\dfrac{ab}{a^2+b^2-c^2}+\dfrac{bc}{b^2+c^2-a^2}+\dfrac{ca}{c^2+a^2-b^2}\)
\(=\dfrac{ab}{-2ab}+\dfrac{bc}{-2bc}+\dfrac{ca}{-2ca}=-\dfrac{1}{2}+-\dfrac{1}{2}+-\dfrac{1}{2}=-\dfrac{3}{2}\)
Vậy \(A=-\dfrac{3}{2}\)
Chúc bạn học tốt =))
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-bc-ca+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\Leftrightarrow a+b+c=0\) vì \(a\ne b\ne c\)
\(\Rightarrow P=\frac{ab^2}{\left(a+b\right)^2-c^2-2ab}+...\)
\(=\frac{ab^2}{\left(a+b+c\right)\left(a+b-c\right)-2ab}+..\)
\(=\frac{ab^2}{-2ab}+\frac{bc^2}{-2bc}+\frac{ca^2}{-2ca}=-\left(\frac{a+b+c}{2}\right)=0\)
Ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\)
\(\Rightarrow\frac{bcx+acy+abz}{abc}=0\)
\(\Rightarrow bcx+acy+abz=0\)
Lại có:\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\)
\(\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2.\frac{bcx+acy+abz}{xyz}=4\)(bình phương hai vế)
\(\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}=4\)(Vì \(bcx+acy+abz=0\))
Từ (1) \(\Rightarrow bcx+acy+abz=0\)
Gọi \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\left(2\right)\)
Từ (2) \(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{ab}{xy}+\frac{ac}{xz}+\frac{bc}{yz}\right)=0\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=4-\left(\frac{abz+acy+bcx}{xyz}\right)\)
\(=4\)
\(b,\frac{ab}{a^2+b^2+c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)
Từ \(a+b+c=0\Rightarrow a+b=-c\Rightarrow a^2+b^2-c^2=-2ab\)
Tương tự \(b^2+c^2-a^2=-2bc\)và \(c^2+a^2-b^2=-2ac\)
\(\Rightarrow\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ca}=\frac{1}{-2}+\frac{1}{-2}+\frac{1}{-2}\)
\(=-\frac{3}{2}\)
a, \(M=\frac{a}{ab+a+2}+\frac{b}{bc+b+1}+\frac{2c}{ca+2c+2}\)
\(=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{bc}{1+bc+b}\)
\(=\frac{1+b+bc}{1+b+bc}=1\)
\(\Rightarrow M=1\)
a) \(A=\frac{a^2}{cb}+\frac{b^2}{ca}+\frac{c^2}{ab}\)
\(A=\frac{a^2.a+b^2.b+c^2.c}{abc}\)
\(A=\frac{a^3+b^3+c^3}{abc}\left(1\right)\)
Ta lại có: \(a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Leftrightarrow\left(a+b\right)^3=-c^3\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3a^2b-3ab^2\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(-c\right)\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\left(2\right)\)
Lấy (2) thay vào (1), ta được:
\(\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3\)
a) cho a+b+c=0a+b+c=0 và abc khác 0 Tính a2(a2−b2−c2)+b2(b2−c2−a2)+c2(c2−b2−a2)
b) B mình k biết