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\(a\ne\pm b\) => \(a\pm b\ne0\)
Như vậy: \(a\left(a+b\right)\left(b+c\right)=b\left(b+c\right)\left(b+a\right)\)
<=> \(a\left(a+b\right)=b\left(b+c\right)\)
<=> \(a^2+ab-b^2-bc=0\)
<=> \(\left(a-b\right)\left(a+b+c\right)=0\)
<=> \(a+b+c=0\) đpcm
a(a+b)(a+c)=b(b+c)(b+a)\(\Leftrightarrow\)a(a+c)=b(b+c) \(\Leftrightarrow\) a(a+c)-b (b=c) =0 \(\Leftrightarrow\) a2-b2+ac-bc=0 \(\Leftrightarrow\) ( a - b) ( a + b)+c ( a-b )=0 \(\Leftrightarrow\) ( a-b)( a+b+c)=0 \(\Leftrightarrow\) a+b+c=0(do a\(\ne\) \(\mp\)b)
Lời giải:
$a+b+c=0\Rightarrow a=-(b+c)\Rightarrow a^2=(b+c)^2$
$\Rightarrow b^2+c^2-a^2=b^2+c^2-(b+c)^2=-2bc$
$\Rightarrow \frac{1}{b^2+c^2-a^2}=\frac{1}{-2bc}=\frac{-1}{2bc}$
Hoàn toàn tương tự với các phân thức khác và cộng theo vế:
\(\text{VT}=\frac{-1}{2bc}+\frac{-1}{2ac}+\frac{-1}{2ab}=\frac{-(a+b+c)}{2abc}=\frac{-0}{2abc}=0\) (đpcm)
Em(mình) thử nhé, ko chắc đâu
3/ Ta có \(\left(a+b\right)\left(b+c\right)\left(c+a\right)=ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)
\(=\left[ab\left(a+b\right)+abc\right]+\left[bc\left(b+c\right)+abc\right]+\left[ca\left(c+a\right)+ca\right]-abc\)
\(=\left(a+b+c\right)ab+\left(a+b+c\right)bc+\left(a+b+c\right)ca-abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)= -abc
Suy ra \(P=\frac{-abc}{abc}=-1\)
Vậy..
`(a+b+c)^2=3(ab+bc+ca)`
`<=>a^2+b^2+c^2+2ab+2bc+2ca=3(ab+bc+ca)`
`<=>a^2+b^2+c^2=ab+bc+ca`
`<=>2a^2+2b^2+2c^2=2ab+2bc+2ca`
`<=>(a-b)^2+(b-c)^2+(c-a)^2=0`
`VT>=0`
Dấu "=" xảy ra khi `a=b=c`
`a^3+b^3+c^3=3abc`
`<=>a^3+b^3+c^3-3abc=0`
`<=>(a+b)^3+c^3-3abc-3ab(a+b)=0`
`<=>(a+b)^3+c^3-3ab(a+b+c)=0`
`<=>(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0`
`**a+b+c=0`
`**a^2+b^2+c^2=ab+bc+ca`
`<=>a=b=c`
Vì \(c^2+2\left(ab-ac-bc\right)=0\) nên :
\(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a^2+\left(a-c\right)^2+\left(c^2+2ab-2ac-2bc\right)}{b^2+\left(b-c\right)^2+\left(c^2+2ab-2ac-2bc\right)}\)
\(=\frac{2a^2+2c^2-4ac+2ab-2bc}{2b^2+2c^2-4bc+2ab-2ac}=\frac{\left(a-c\right)^2+b\left(a-c\right)}{\left(b-c\right)^2+a\left(b-c\right)}\)
\(=\frac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(b-c+a\right)}=\frac{a-c}{b-c}\) \(\left(b\ne c,a+b\ne0\right)\)
\(A=\frac{b^2c^2}{a}+\frac{c^2a^2}{b}+\frac{a^2b^2}{c}=\frac{a^3b^3+b^3c^3+c^3a^3}{abc}=\frac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{abc}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)
\(\Rightarrow\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3=3.ab.bc.ca=3a^2b^2c^2\)
Vậy \(A=\frac{3a^2b^2c^2}{abc}=3abc\left(a,b,c\ne0\right)\)
Ta có:
\(a\left(a+b\right)\left(a+c\right)=b\left(b+c\right)\left(b+a\right)\)
\(\Leftrightarrow a\left(a+b\right)\left(a+c\right)-b\left(b+c\right)\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-b^2+ac-bc\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)\left(a+b+c\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\a-b=0\\a+b+c=0\end{matrix}\right.\)
Vì \(a\ne\pm b\Rightarrow a+b+c=0\) (đpcm)