Bài 1:
\(a+b+c=0\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow c=-\left(a+b\right)^3\) yt
\(N=\dfrac{a^2}{3bc}+\dfrac{b^2}{3ac}+\dfrac{c^2}{3ab}\)
\(=\dfrac{a^3+b^3+c^3}{3abc}\)
Thay \(\Rightarrow c=-\left(a+b\right)^3\) vào biểu thức N ,có :
\(N=\dfrac{a^3+b^3-\left(a+b\right)^3}{3abc}\)
\(=\dfrac{a^3+b^3-a^3-b^3-3ab\left(a+b\right)}{3abc}\)
\(=\dfrac{-3ab.\left(-c\right)}{3abc}\)
\(=\dfrac{3abc}{3abc}=1\)
Vậy N = 1
Bài 1:
a+b+c=0
=>\(\left\{{}\begin{matrix}a^2=b^2+c^2+2bc\\b^2=a^2+c^2+2ac\\c^2=a^2+b^2+2ab\end{matrix}\right.\)
=> \(N=\dfrac{b^2+c^2+2bc}{3bc}+\dfrac{a^2+c^2+2ac}{3ac}+\dfrac{a^2+b^2+2ab}{3ab}=\dfrac{b}{3c}+\dfrac{c}{3b}+\dfrac{2}{3}+\dfrac{a}{3c}+\dfrac{c}{3a}+\dfrac{2}{3}+\dfrac{a}{3b}+\dfrac{b}{3a}+\dfrac{2}{3}=\dfrac{a+c}{3b}+\dfrac{a+b}{3c}+\dfrac{b+c}{3a}+2=\dfrac{-b}{3b}+\dfrac{-c}{3c}+\dfrac{-a}{3a}+2=-\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{3}+2=1\)
Bài 2:
x2+5x+1=0
<=> \(\left[{}\begin{matrix}x1=\dfrac{-5+\sqrt{21}}{2}\\x2=\dfrac{-5-\sqrt{21}}{2}\end{matrix}\right.\)
Ta có :
\(M=\dfrac{1}{2}\left(x^2+1+\dfrac{1}{x^2}\right)=\dfrac{1}{2}\left(\left(x+\dfrac{1}{x}\right)^2-1\right)\)TH1: \(x=\dfrac{-5+\sqrt{21}}{2}\Rightarrow x+\dfrac{1}{x}=\dfrac{-5+\sqrt{21}}{2}+\dfrac{2}{-5+\sqrt{21}}=\dfrac{25+21-10\sqrt{21}+4}{2\left(-5+\sqrt{21}\right)}=\dfrac{-5\left(-10+2\sqrt{21}\right)}{-10+2\sqrt{21}}=-5\)\(\Rightarrow M=\dfrac{1}{2}\left(\left(-5\right)^2-1\right)=12\)
TH2: \(x=\dfrac{-5-\sqrt{21}}{2}\Rightarrow x+\dfrac{1}{x}=\dfrac{-5-\sqrt{21}}{2}+\dfrac{2}{-5-\sqrt{21}}=\dfrac{25+21+10\sqrt{21}+4}{2\left(-5-\sqrt{21}\right)}=-5\)\(\Rightarrow M=12\)
Vậy M=12 (mn có cách nào ngắn hơn thì chia sẻ nhé)
Bài 2 : mk thấy cái giả thiết đề cho cứ sai sai , bn kiểm tra lại đề giúp mk
