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a ) \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2+2.0=0\)
\(\Leftrightarrow a^2+b^2+c^2=0\)
Do \(a^2\ge0;b^2\ge0;c^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=0\) ( * )
Thay * vào biểu thức M , ta được :
\(M=\left(0-1\right)^{1999}+0^{2000}+\left(0+1\right)^{2001}\)
\(=-1^{1999}+0+1^{2001}\)
\(=-1+0+1\)
\(=0\)
Vậy \(M=0\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc+ac+ab-1}{abc}=0\)
\(\Leftrightarrow bc+ac+ab-1=0\)
\(\Leftrightarrow bc+ac+ab=1\)
Mà \(a^2+b^2+c^2=1\)
\(\Rightarrow bc+ac+ab=a^2+b^2+c^2\)
\(\Rightarrow2bc+2ac+2ab=2a^2+2b^2+2c^2\)
\(\Rightarrow2a^2+2b^2+2c^2-2bc-2ac-2ab=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Do \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Mà \(P=\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}\)
\(\Rightarrow P=\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\)
\(\Rightarrow P=1+1+1=3\)
Vậy \(P=3\)
Ta có: \(\left\{{}\begin{matrix}abc=1\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3b^3c^3=1\\\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{-1}{c}\end{matrix}\right.\)
\(a^3b^3+b^3c^3+c^3a^3=a^3b^3c^3\left(\dfrac{a^3b^3+b^3c^3+c^3a^3}{a^3b^3c^3}\right)=1.\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\)
\(\Rightarrow S=\left(a^3b^3+b^3c^3+c^3a^3\right)\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)^2\)
Lại có:
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3+\dfrac{1}{c^3}-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2-\dfrac{1}{c}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{1}{c^2}\right)-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(=\dfrac{-3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{-3}{ab}\left(\dfrac{-1}{c}\right)=\dfrac{3}{abc}=3\)
\(\Rightarrow S=\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)^2=3^2=9\)
Đặt \(ab=x;bc=y;ca=z\) thì có \(x^3+y^3+z^3=3xyz\) dễ nhé
Lời giải:
$(a^2+b^2)^3=(a^3+b^2)^3$
$\Leftrightarrow a^2+b^2=a^3+b^2$
$\Leftrightarrow a^2=a^3$
$\Leftrightarrow a^2(a-1)=0$
Vì $a\neq 0$ nên $a-1=0\Rightarrow a=1$
Do đó: $s=\frac{b}{a}+\frac{a}{b}=b+\frac{1}{b}$ còn giá trị cụ thể thì không xác định được bạn nhé.