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Ta có: \(A=a\left(a^2-bc\right)+b\left(b^2-ac\right)+c\left(c^2-ab\right)=0\)
\(\Rightarrow A=a^3+b^3+c^3-3abc=0\) \(\Rightarrow A=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Rightarrow A=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow A=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Vì \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-ac-bc=0\)
Xét \(M=a^2+b^2+c^2-ab-ac-bc=0\)
\(\Rightarrow2M=2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Rightarrow2M=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Rightarrow a-b=0;b-c=0;c-a=0\) \(\Rightarrow a=b=c\)
\(\Rightarrow P=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1+1+1=3\)
oh no bài thứ nhất là dạng chứng minh cs đúng ko ,
ko thể nào là dạng tìm a,b,c đc-.-
TH1: \(a+b+c=0\Rightarrow P=\dfrac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=-1\)
TH2: \(a+b+c\ne0\)
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{a+c-b}{b}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{2a.2b.2c}{abc}=8\)
TH1 : a + b + c ≠ 0
Áp dụng t/c dãy tỉ số bằng nhau ta có
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b+b+c+a+c}{a+b+c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\a+c=2b\end{matrix}\right.\)
Khi đó \(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}=\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}=8\)
TH2 : a + b + c = 0
\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Khi đó \(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)
Xét 2 TH sau:
TH1: a+b+c=0
Khi đó:
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\\ =-1\)
TH2: a+b+c khác 0
Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
Suy ra: a+b=2c; b+c=2a; c+a=2b
Do đó:
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}\\ =8\)
Xét 2 TH sau:
TH1: a+b+c=0
Khi đó:
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\\ =-1\)
TH2: a+b+c khác 0
Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
Suy ra: a+b=2c; b+c=2a; c+a=2b
Do đó:
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}\\ =8\)
\(P=\dfrac{1}{bc\left(b+c\right)+2023}+\dfrac{1}{ca\left(c+a\right)+2023}+\dfrac{1}{ab\left(a+b\right)+2023}\left(abc=2023\right)\)
\(\Leftrightarrow P=\dfrac{1}{bc\left(b+c\right)+abc}+\dfrac{1}{ca\left(c+a\right)+abc}+\dfrac{1}{ab\left(a+b\right)+abc}\)
\(\Leftrightarrow P=\dfrac{1}{bc\left(a+b+c\right)}+\dfrac{1}{ca\left(a+b+c\right)}+\dfrac{1}{ab\left(a+b+c\right)}\)
\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\)
\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left[\dfrac{a^2bc+b^2ca+c^2ab}{\left(abc\right)^2}\right]\)
\(\Leftrightarrow P=\dfrac{1}{\left(a+b+c\right)}\left[\dfrac{abc\left(a+b+c\right)}{\left(abc\right)^2}\right]\)
\(\Leftrightarrow P=\dfrac{1}{abc}=\dfrac{1}{2023}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)
Do đó:
\(\left\{{}\begin{matrix}a+b-c=c\\b+c-a=a\\c+a-b=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)
Thay a+b=2c;b+c=2a và c+a=2b vào biểu thức \(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(a+b\right)}{abc}\), ta được:
\(P=\dfrac{2a\cdot2b\cdot2c}{abc}=\dfrac{8abc}{abc}=8\)
Vậy: P=8
Ta có: \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\) = \(\dfrac{a+b-c+b+c-a+c+a-b}{a+b+c}\) (t/c dãy tỉ số bằng nhau)
hay \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=1\) (1)
Ta cũng có: \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{a+b-c+b+c-a}{a+c}\) (t/c dãy tỉ số bằng nhau)
hay \(\dfrac{a+b-c}{c}=\dfrac{2b}{a+c}\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{2b}{a+c}=1\) \(\Leftrightarrow\) a + c = 2b (*)
Tương tự ta cũng có: a + b = 2c (**); b + c = 2a (***)
Thay (*); (**); (***) vào P ta được:
P = \(\dfrac{2a.2b.2c}{abc}\) = 2.2.2 = 8
Vậy P = 8
Chúc bn học tốt!