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Từ \(a^2-b=b^2-c\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=b-c\)
\(\Leftrightarrow a+b=\frac{b-c}{a-b}\)
\(\Rightarrow a+b+1=\frac{b-c}{a-b}+1=\frac{a-c}{a-b}\)
Tương tự ta có:
\(\hept{\begin{cases}b+c+1=\frac{b-a}{b-c}\\c+a+1=\frac{c-b}{c-a}\end{cases}}\)
\(\Rightarrow\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)=\frac{a-c}{a-b}.\frac{b-a}{b-c}.\frac{c-b}{c-a}=-1\)
Vì a + b + c = 0
<=> (a + b + c)2 = 0
<=> a2 + b2 + c2 = -2(ab + bc + ca)
Khi đó \(\frac{9\left(a^2+b^2+c^2\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}=\frac{-18\left(ab+bc+ca\right)}{2\left(a^2+b^2+c^2-ab-bc-ca\right)}\)
\(=\frac{-18\left(ab+bc+ca\right)}{-6\left(ab+bc+ca\right)}=3\)
Ta có:
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ca=0\)
Ta lại có:
\(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ca}+\frac{c^2}{c^2+2ab}\)
\(=\frac{a^2}{a^2-ab+bc-ca}+\frac{b^2}{b^2-ab-bc+ca}+\frac{c^2}{c^2+ab-bc-ca}\)
\(=\frac{a^2}{\left(b-a\right)\left(c-a\right)}+\frac{b^2}{\left(a-b\right)\left(c-b\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)
\(=-\left(\frac{a^2}{\left(a-b\right)\left(c-a\right)}+\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(b-c\right)}\right)\)
\(=-\left(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\right)\)
\(=-\frac{\left(a-b\right)\left(c-b\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)
đặt x=a-b;y=b-c;z=c-a
ta có x+y+z=0
nên ta có ĐPCM
\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)
<=> \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
<=> \(2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=0\)
<=> \(\frac{z}{xyz}+\frac{y}{xyz}+\frac{x}{xyz}=0\)
<=> \(\frac{x+y+z}{xyz}=0\) (luôn đúng )
https://olm.vn/hoi-dap/detail/48946023107.html vào trang đó coi rồi
ta có a+b+c=0 => a+b=-c => a^2 +b^2 =c^2-2ab
tương tự a^2 + c^2 =b^2-2ac
b^2 + c^2 =a^2-2bc
thế cào A= -1/2ab + -1/2ac + -1/2bc = -(c+a+b)/2abc=0 (vì a+b+c=0 )
ta có:a^3+b^3+c^3=3abc
<=>(a+b)^3+c^3-3ab(a+b)-3abc=0
<=>(a+b+c)[(a+b)^2+(a+b)c+c^2]-3ab(a+b...
<=>(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0
<=>1/2(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]...
do a,b,c doi mot khac nhau nen PT<=>a+b+c=0(DPCM)
lộn nha không phải cái trang đó đâu cái này này
TA có: \(\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b} = 0 \quad (1)\)
=>\(\frac{a}{b-c}+\frac{b}{c-a}=\frac{-c}{a-b}\)
=>\(\frac{a(c-a) + b(b-c)}{(b-c)(c-a)} = -\frac{c}{a-b}\)
=>\(\frac{ac - a^2 + b^2 - bc}{(b-c)(c-a)}=-\frac{c}{a-b}\)
=>\(\frac{(b^2 - a^2) - c(b - a)}{(b-c)(c-a)}=-\frac{c}{a-b}\)
=>\(\frac{(b - a)(a + b - c)}{(b-c)(c-a)}=-\frac{c}{a-b}\)
=>\(\frac{-(a - b)(a + b - c)}{(b-c)(c-a) \cdot (a - b)} = -\frac{c}{(a - b)^2}\)
=>\(\frac{-(a + b - c)}{(b-c)(c-a)}=-\frac{c}{(a - b)^2}\)
=>\(\frac{c}{(a-b)^2}=\frac{a + b - c}{(b-c)(c-a)}\quad\)(2)
Chứng minh tương tự, ta cũng có:
\(\frac{a}{(b-c)^2}=\frac{b + c - a}{(c-a)(a-b)}\quad\left(3\right)\)
\(\frac{b}{(c-a)^2}=\frac{c + a - b}{(a-b)(b-c)}\quad\left(4\right)\)
Từ (2),(3),(4) suy ra \(S=\frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=\frac{b + c - a}{(c-a)(a-b)}+\frac{c + a - b}{(a-b)(b-c)}+\frac{a + b - c}{(b-c)(c-a)}\)
\(=\frac{(b+c-a)(b-c) + (c+a-b)(c-a) + (a+b-c)(a-b)}{(a-b)(b-c)(c-a)}\)
\(=\frac{b^2-c^2-ab+ac+c^2-a^2-bc+ab+a^2-b^2-ac+bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
=0