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Câu hỏi của không cần biết - Toán lớp 8 - Học toán với OnlineMath
Theo bài ra ta có:
\(a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(=\dfrac{bc+ac+ab}{abc}=bc+ac+ab\)
Ta lại có:
\(\left(a.b.c-1\right)+\left(a+b+c\right)-\left(bc+ca+ab\right)=0\)
\(=>\left(a-1\right)\left(b-1\right)\left(c-1\right)=0\)
\(=>\left[{}\begin{matrix}a-1=0\\b-1=0\\c-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\)
CHÚC BẠN HỌC TỐT.........
\(a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\\ \Leftrightarrow a+b+c=\dfrac{bc+ac+ab}{abc}\\ \Leftrightarrow a+b+c=bc+ac+ab\\ \Leftrightarrow a+b+c-ab-bc-ac+abc-1=0\\ -a\left(b-1\right)-c\left(b-1\right)+ac\left(b-1\right)+\left(b-1\right)=0\\ \Leftrightarrow\left(b-1\right)\left(-a-c+ac+1\right)=0\\ \Leftrightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)\\ \Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\)
Ta có: \(a+\dfrac{1}{b}=-4\)
\(\Rightarrow\left(a+\dfrac{1}{b}\right)^3=\left(-4\right)^3\)
\(\Rightarrow a^3+3.a^2.\dfrac{1}{b}+3.a.\dfrac{1}{b^2}+\dfrac{1}{b^3}=-64\)
\(\Rightarrow a^3+\dfrac{3a^2}{b}+\dfrac{3a}{b^2}+\dfrac{1}{b^3}=-64\)
\(\Rightarrow a^3+\dfrac{1}{b^3}=-64-\dfrac{3a^2}{b}-\dfrac{3a}{b^2}\)
\(\Rightarrow a^3+\dfrac{1}{b^3}=-64-\dfrac{3a}{b}\left(a+\dfrac{1}{b}\right)\)
\(\Rightarrow a^3+\dfrac{1}{b^3}=-64-3.\left(-4\right).\left(-4\right)\)
\(\Rightarrow a^3+\dfrac{1}{b^3}=-112\)
\(P=a^3+\dfrac{1}{b^3}\\ =\left(a+\dfrac{1}{b}\right)\left(a^2+\dfrac{a}{b}+\dfrac{1}{b^2}\right)\\ =-4\left(a^2+\dfrac{2a}{b}+\dfrac{1}{b^2}-\dfrac{a}{b}\right)\\ =-4\left[\left(a+\dfrac{1}{b}\right)^2-\dfrac{a}{b}\right]\\ =-4\left[\left(-4\right)^2-\left(-4\right)\right]\\ =-80\)
\(a+\dfrac{1}{b}=\dfrac{a}{b}\Leftrightarrow\dfrac{ab+1}{b}=\dfrac{a}{b}\Leftrightarrow ab+1=a\left(1\right)\)
\(\dfrac{a}{b}=-4\Leftrightarrow a=-4b\left(2\right)\)
Thay (2) vào (1), ta được:
\(-4b^2+1=-4b\)
\(\Rightarrow-4b^2+4b+1=0\)
\(\Rightarrow-4\left(b^2+b-\dfrac{1}{4}\right)=0\)
\(\Rightarrow-4\left(b^2+2\cdot b\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+2=0\)
\(\Rightarrow-4\left(b+\dfrac{1}{2}\right)^2=-2\)
\(\Rightarrow\left(b+\dfrac{1}{2}\right)^2=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}b+\dfrac{1}{2}=\sqrt{\dfrac{1}{2}}\\b+\dfrac{1}{2}=-\sqrt{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}b=\dfrac{-1+\sqrt{2}}{2}\\b=\dfrac{-1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}b=\dfrac{-1+\sqrt{2}}{2}\\a=2-2\sqrt{2}\end{matrix}\right.\\\left\{{}\begin{matrix}b=\dfrac{-1-\sqrt{2}}{2}\\a=2+2\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy ..................................
Bạn vào đây tham khảo sau đó áp dụng vào bài của bạn nhé: Câu hỏi của Võ Khánh Lê - Toán lớp 0 | Học trực tuyến
Ta có :\(\left(\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{2}{ab}+\dfrac{2}{bc}-\dfrac{2}{ac}\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left(\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}\right)^2+\dfrac{2}{ab}-\dfrac{2}{bc}+\dfrac{2}{ac}\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=1+2\left(\dfrac{c-a+b}{abc}\right)\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=1+2\left(\dfrac{c-\left(a-b\right)}{abc}\right)\left(1\right)\)
Theo đề ra : a=b+c
\(\Leftrightarrow c=a-b\)
\(\Leftrightarrow c-\left(a-b\right)=0\)
\(\left(1\right)\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=1+2\left(\dfrac{0}{abc}\right)=1\)
\(Hay\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=1\left(đpcm\right)\)
`1/a+1/b+1/c=1/(a+b+c)`
`<=>(a+b)/(ab)+(a+b)/(c(a+b+c))=0`
`<=>(a+b)(ab+ac+bc+c^2)=0`
`<=>(a+b)(a+c)(b+c)=0`
`=>` $\left[ \begin{array}{l}a=-b\\b=-c\\c=-a\end{array} \right.$
`=>` PT luôn tồn tại 2 số đối nhau