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ĐKXĐ: a, b, c \(\ne\) 0
Xét 2 trường hợp:
+) a + b + c = 0: Khi đó \(\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)
\(\Rightarrow M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)
+) a + b + c \(\ne\) 0: Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b+b+c+c+a}{c+a+b}=2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)
\(\Rightarrow M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}=\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}=8\)
a,ta có: \(a^3+b^3+c^3=3abc\)
<=>\(a^3+b^3+c^3-3abc=0\)
<=>\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
<=>\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
<=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
<=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
<=>\(\left(a+b+c\right)2\left(a^2-ab+b^2-ac-bc+c^2\right)=0\)
<=>\(\left(a+b+c\right)\left(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\right)=0\)
=>a=b,a=c,b=c
=>a=b=c
thay a=b=c vào P ta đc
\(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
2b)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
<=> \(\dfrac{ab+bc+ca}{abc}=\dfrac{1}{a+b+c}\)
<=> (ab+bc+ca)(a+b+c)=abc
<=> (ab+bc+ca)(a+b+c)-abc=0
<=> (a+b)(b+c)(c+a) = 0
<=> a+b=0 hoặc b+c=0 hoặc c+a=0
<=> a=-b hoặc b=-c hoặc c = -a
sau đó thay vào cái cần c/m
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 \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Rightarrow a-b=0;b-c=0;c-a=0\) \(\Rightarrow a=b=c\)
\(\Rightarrow P=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1+1+1=3\)
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)
Vì a, b, c là các số dương \(\Rightarrow a=b=c=0\) ( loại )
\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Rightarrow a=b=c\) ( tự chứng minh )
\(\Rightarrow M=\left(\dfrac{a}{b}-1\right)+\left(\dfrac{b}{c}-1\right)+\left(\dfrac{c}{a}-1\right)=0\)
Vậy M = 0
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 \(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)