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a, Ta thấy : \(\left\{{}\begin{matrix}\left(2a+1\right)^2\ge0\\\left(b+3\right)^2\ge0\\\left(5c-6\right)^2\ge0\end{matrix}\right.\)\(\forall a,b,c\in R\)
\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\ge0\forall a,b,c\in R\)
Mà \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\le0\)
Nên trường hợp chỉ xảy ra là : \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2=0\)
- Dấu " = " xảy ra \(\left\{{}\begin{matrix}2a+1=0\\b+3=0\\5c-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=-3\\c=\dfrac{6}{5}\end{matrix}\right.\)
Vậy ...
b,c,d tương tự câu a nha chỉ cần thay số vào là ra ;-;
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`VT = (b-c)/((a-b)(a-c)) + (c-a)/((b-c)(b-a)) +(a-b)/((c-a)(c-b)) = 2/(a-b) + 2/(b-c) + 2/(c-a)`
`=-((a-b-a+c)/((a-b)(a-c))+(b-c-b+a)/((b-c)(b-a))+(c-a-c+b)/((c-a)(c-b)))`
`=-((a-b)/((a-b)(a-c))-(a-c)/((a-b)(a-c))+(b-c)/((b-c)(b-a))-(b-a)/((b-c)(b-a))+(c-a)/((c-a)(c-b))-(c-b)/((c-a)(c-b)))`
`= 1/(c-a)+1/(a-b)+1/(a-b)+1/(b-c)+1/(b-c)+1/(c-a)`
`=2/(a-b)+2/(b-c)+2/(c-a)=VP(đpcm)`
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\(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{2a+2b+2c}{a+b+c}=2\\ \Rightarrow\left\{{}\begin{matrix}b+c-5=2a\\a+c+2=2b\\a+b+3=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c=a+5\\a+b+c=b-2\\a+b+c=c-3\end{matrix}\right.\)
Lại có \(\dfrac{1}{a+b+c}=2\Rightarrow a+b+c=\dfrac{1}{2}\Rightarrow\left\{{}\begin{matrix}a+5=\dfrac{1}{2}\\b-2=\dfrac{1}{2}\\c-3=\dfrac{1}{2}\end{matrix}\right.\)
Từ đó tự giải ra
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}=k\Rightarrow a=3k;b=4k;c=5k\)
\(\Rightarrow4\left(a-b\right)\left(b-c\right)=4\left(3k-4k\right)\left(4k-5k\right)\)
\(=4.\left[\left(3-4\right).k\right].\left[\left(4-5\right).k\right]\)
\(=4.\left[-k\right].\left[-k\right]=4k^2\left(1\right)\)
\(\Rightarrow\left(a-c\right)^2=\left(3k-5k\right)^2=\left[\left(3-5\right).k\right]^2=\left[-2k\right]^2=4k^2\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)
Vậy \(4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\left(dpcm\right)\)