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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:
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3+\dfrac{1}{c^3}-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2-\dfrac{1}{c}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{1}{c^2}\right)-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(=\dfrac{-3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{-3}{ab}\left(\dfrac{-1}{c}\right)=\dfrac{3}{abc}=3\)
\(\Rightarrow S=\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)^2=3^2=9\)
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)
\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{ab+ac+bc}{abc}\right)=1\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+ac+bc\right)-abc=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+ac+bc\right)+c\left(ab+ac+bc\right)-abc=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+ac+bc\right)+c^2\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-b\\a=-c\\b=-c\end{matrix}\right.\)
Đến đây thì nghi ngờ bạn chép sai đề biểu thức R, lẽ ra phải là dấu nhân mới tính được, nếu ko thì kết quả vẫn còn 2 ẩn
\(R=\left(a^{2017}+b^{2017}\right)\left(b^{2019}+c^{2019}\right)\left(c^{2021}+a^{2021}\right)\)
Thế này mới chính xác, kết quả \(R=0\)
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 : \(a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)
\(\Leftrightarrow\left(ab\right)^3+\left(bc\right)^3+\left(ac\right)^3=3ab.bc.ac\)
Đặt \(ab=x;bc=y;ac=z\) . Khi đó , ta có :
\(x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x^3+y^3+3x^2y+3y^2x\right)+z^3-3x^2y-3y^2x-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2-3xy\right]=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y+z=0\\x^2+y^2+z^2-xy-yz-xz=0\end{matrix}\right.\)
Với \(x+y+z=0\Rightarrow ab+ac+bc=0\)
Với \(x^2+y^2+z^2-xy-yz-xz=0\)
\(\Rightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)
\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)=0\)
\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
Lí luận tổng này \(\ge0\) ( làm tắt )
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\x-z=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=y\\y=z\end{matrix}\right.\) \(\Rightarrow x=y=z\)
\(\Rightarrow ab=ac=bc\)
....
Đến bước này chịu , bạn xem đề có sai không ?
Đặt \(ab=x;bc=y;ca=z\) thì có \(x^3+y^3+z^3=3xyz\) dễ nhé