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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:
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}\\ =8\)
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:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b+b+c+c+a}{c+a+b}=2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)
\(\Rightarrow M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}=\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}=8\)
Bài này mình làm một lần ở trường rồi nhưng không có điện thoại chụp được:((
Ta có: \(\dfrac{a^3}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^3}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^3}{\left(c-a\right)\left(c-b\right)}=\dfrac{a^3\left(c-b\right)+b^3\left(a-c\right)-c^3\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}\)\(=\dfrac{a^3\left(c-b\right)+b^3a-b^3c-c^3a+c^3b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=\dfrac{a^3\left(c-b\right)-a\left(c^3-b^3\right)+bc\left(c^2-b^2\right)}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=\dfrac{a^3\left(c-b\right)-a\left(c-b\right)\left(a^2+bc+b^2\right)+bc\left(c-b\right)\left(c+b\right)}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}\)\(=\dfrac{\left(c-b\right)\left(a^3-ac^2-abc-ab^2+bc^2+b^2c\right)}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}=\dfrac{\left(c-b\right)\left[a\left(a^2-b^2\right)-c^2\left(a-b\right)-bc\left(a-b\right)\right]}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}\)\(=\dfrac{\left(c-b\right)\left[a\left(a-b\right)\left(a+b\right)-c\left(a-b\right)-bc\left(a-b\right)\right]}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=\dfrac{\left(c-b\right)\left(a-b\right)\left(a^2+ab-c-bc\right)}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}\)
\(\dfrac{\left(c-b\right)\left(a-b\right)\left[a^2-c^2+ab-bc\right]}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=\dfrac{\left(c-b\right)\left(a-b\right)\left[\left(a-c\right)\left(a+c\right)+b\left(a-c\right)\right]}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=\dfrac{\left(c-b\right)\left(a-b\right)\left(a-c\right)\left(a+b+c\right)}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}\)\(=a+b+c\)
Vì a, b, c là các số nguyên
=> a+b+c là các số nguyên
=> Đpcm.
Đấy mình làm chi tiết tiền tiệt lắm luôn, không hiểu thì mình chịu rồi, trời lạnh mà đánh máy nhiều thế này buốt tay lắm luôn:vv
Ta có:
\(\dfrac{b-c}{1\left(a-b\right)\left(a-c\right)}+\dfrac{c-a}{\left(b-c\right)\left(b-a\right)}+\dfrac{a-b}{\left(c-a\right)\left(c-b\right)}\)
\(=\dfrac{c-b}{1\left(a-b\right)\left(c-a\right)}+\dfrac{a-c}{\left(b-c\right)\left(a-b\right)}+\dfrac{b-a}{\left(c-a\right)\left(b-c\right)}\)
Quy đồng rút gọn ta được
\(=\dfrac{2\left(ab+bc+ca-a^2-b^2-c^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{2\left[\left(a-b\right)\left(b-c\right)+\left(b-c\right)\left(c-a\right)+\left(c-a\right)\left(a-b\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=2\left(\dfrac{1}{a-b}+\dfrac{1}{b-c}+\dfrac{1}{c-a}\right)\)
PS: Hôm qua đi chơi nên nay mới giải nhé.
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
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
\(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
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 \(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)