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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
Ta có: a+b+c=0
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow2\left(ab+bc+ac\right)=0-1=-1\)
hay \(ab+bc+ac=-\dfrac{1}{2}\)
\(\Leftrightarrow\left(ab+bc+ac\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2ab^2c+2abc^2+2a^2bc=\dfrac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(b+c+a\right)=\dfrac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=\dfrac{1}{4}\)
Ta có: \(M=a^4+b^4+c^4\)
\(\Leftrightarrow M=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2-2a^2b^2-2a^2c^2-2b^2c^2\)
\(\Leftrightarrow M=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+a^2c^2+b^2c^2\right)\)
\(\Leftrightarrow M=1^2-2\cdot\dfrac{1}{4}=1-\dfrac{1}{2}=\dfrac{1}{2}\)
Vậy: \(M=\dfrac{1}{2}\)
Ta có : \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)=1\) ( * )
\(\Rightarrow ab+bc+ac=-\dfrac{1}{2}\)
Lại có : \(\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\) ( suy ra từ * )
\(\Rightarrow a^4+b^4+c^4=2\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{2}\)
Vậy ...
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
ĐK : a;b;c khác 0
Thấy : \(a^2+b^2+c^2=\left(a+b+c\right)^2\Leftrightarrow ab+bc+ac=0\) (1)
Ta có : \(P=\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\)
Từ (1) suy ra : \(\left(b+c\right)a=-bc\Leftrightarrow\dfrac{b+c}{a}=\dfrac{-bc}{a^2}\)
CMTT ; ta có : \(\dfrac{c+a}{b}=\dfrac{-ac}{b^2};\dfrac{a+b}{c}=\dfrac{-ab}{c^2}\)
Suy ra : \(P=-\left(\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ac}{b^2}\right)=-\dfrac{a^3b^3+b^3c^3+a^3c^3}{a^2b^2c^2}\) (2)
Đặt : ab = x ; bc = y ; ac = z ; ta có : x + y + z = 0 \(\Rightarrow x^3+y^3+z^3=3xyz\) (3)
Từ (2) và (3) suy ra : \(P=-\dfrac{3xyz}{xyz}=-3\)
Vậy ...
\(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
\(A=\dfrac{a^2+b^2+c^2}{a^2+b^2-2ab+b^2+c^2-2bc+a^2+c^2-2ca}=\)
\(=\dfrac{a^2+b^2+c^2}{2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)}=\)
\(=\dfrac{-2\left(ab+bc+ca\right)}{-4\left(ab+bc++ca\right)-2\left(ab+bc+ca\right)}=\dfrac{1}{3}\)