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P= \(\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{a^2+c^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)
=
\(\dfrac{a+b+c}{\left(b^2+c^2-a^2\right)\left(a+b+c\right)}+\dfrac{a+b+c}{\left(a^2+c^2-b^2\right)\left(a+b+c\right)}+\dfrac{a+b+c}{\left(a^2+b^2-c^2\right)\left(a+b+c\right)}\)
= 0+0+0 = 0
Vậy P= 0
Ngu vãi ko bt đúng không nx
a+b+c=0
=>a+b=-c; a+c=-b; b+c=-a
\(a^2-b^2-c^2\)
\(=a^2-\left(b^2+c^2\right)\)
\(=a^2-\left\lbrack\left(b+c\right)^2-2bc\right\rbrack=a^2-\left(a^2-2bc\right)=2bc\)
\(b^2-a^2-c^2\)
\(=b^2-\left(a^2+c^2\right)\)
\(=b^2-\left\lbrack\left(a+c\right)^2-2ac\right\rbrack=b^2-\left(b^2-2ac\right)=2ac\)
\(c^2-a^2-b^2\)
\(=c^2-\left\lbrack a^2+b^2\right\rbrack\)
\(=c^2-\left\lbrack\left(a+b\right)^2-2ab\right\rbrack=c^2-\left(c^2-2ab\right)=2ab\)
Ta có: \(\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}\)
\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}\)
\(=\frac{\left(a+b\right)^3-3ab\left(a+b\right)+c^3}{2bac}=\frac{\left(-c\right)^3-3ab\cdot\left(-c\right)+c^3}{2abc}=\frac{3abc}{2abc}=\frac32\)
Lời giải:
Ta có:
$a(a-b)+b(b-c)+c(c-a)=a^2+b^2+c^2-ab-bc-ac$
$=\frac{3}{2}(a^2+b^2+c^2)-[\frac{1}{2}(a^2+b^2+c^2)+ab+bc+ac]$
$=\frac{3}{2}(a^2+b^2+c^2)-\frac{1}{2}(a^2+b^2+c^2+2ab+2bc+2ac)$
$=\frac{3}{2}(a^2+b^2+c^2)-\frac{1}{2}(a+b+c)^2$
$=\frac{3}{2}(a^2+b^2+c^2)$
$\Rightarrow P=\frac{a^2+b^2+c^2}{\frac{3}{2}(a^2+b^2+c^2)}=\frac{2}{3}$
Câu hỏi của Jungkookie - Toán lớp 7 - Học toán với OnlineMath
a: \(x^2-8x+5\)
\(=x^2-8x+16-11\)
\(=\left(x-4\right)^2-11\ge-11\forall x\)
Dấu '=' xảy ra khi x-4=0
=>x=4
b: \(a^3+b^3+c^3=3bac\)
=>\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
=>\(\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)=0\)
=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
=>\(a^2+b^2+c^2-ab-ac-bc=0\)
=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
=>a=b=c
\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\)
\(=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3}{3^2}=\frac13\)
\(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
\(A=\dfrac{a^2+b^2+c^2}{a^2+b^2-2ab+b^2+c^2-2bc+a^2+c^2-2ca}=\)
\(=\dfrac{a^2+b^2+c^2}{2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)}=\)
\(=\dfrac{-2\left(ab+bc+ca\right)}{-4\left(ab+bc++ca\right)-2\left(ab+bc+ca\right)}=\dfrac{1}{3}\)