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Xét \(x+y+z=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+z=-x\\z+x=-y\\x+y=-z\end{matrix}\right.\)
\(\Rightarrow A=\left(2-1\right)\left(2-1\right)\left(2-1\right)=1\)
Xét \(x+y+z\ne0\) thì ta có:
\(\dfrac{x}{y+z+3x}=\dfrac{y}{z+x+3y}=\dfrac{z}{x+y+3z}=\dfrac{x+y+z}{5x+5y+5z}=\dfrac{x+y+z}{5\left(x+y+z\right)}=\dfrac{1}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}5x=y+z+3x\\5y=z+x+3y\\5z=x+y+3z\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=y+z\\2y=z+x\\2z=x+y\end{matrix}\right.\)
\(\Rightarrow A=\left(2+2\right)\left(2+2\right)\left(2+2\right)=64\)
Vậy \(\left[{}\begin{matrix}A=1\\A=64\end{matrix}\right.\)
Nếu bị lỗi thì bạn có thể xem đây nhé:
b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Đặt \(x=15k;y=20k;z=24k\)
Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
Áp dụng tích chất dãy tỉ số bằng nhau ta có :
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}=\dfrac{x+y+z}{x+y+z}=1\\ \Rightarrow\left\{{}\begin{matrix}y+z-x=x\\z+x-y=y\\x+y-z=z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z=2x\\z+x=2y\\x+y=2z\end{matrix}\right.\)
\(\Rightarrow\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)=\dfrac{x+y}{y}.\dfrac{y+z}{z}.\dfrac{x+z}{x}=\dfrac{2z}{y}.\dfrac{2x}{z}.\dfrac{2y}{x}=8\)
Vào đây nhé: Câu hỏi của Vũ Ngọc Minh Anh - Toán lớp 7 | Học trực tuyến
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\)
\(\Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\)
\(\Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}\\\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x+y+z\right)=y\left(x+y+z\right)\\y\left(x+y+z\right)=z\left(x+y+z\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x+y+z\right)=0\\\left(y-z\right)\left(x+y+z\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\x+y+z=0\end{matrix}\right.\\\left[{}\begin{matrix}y=z\\x+y+z=0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=y=z\\x+y+z=0\end{matrix}\right.\)
\(\circledast\) Với \(x=y=z\) thì \(A=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
\(\circledast\) Với \(x+y+z=0\) thì\(\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)
Khi đó \(A=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)=\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}=\dfrac{-xyz}{xyz}=-1\)
Lời giải:
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=t\Rightarrow \left\{\begin{matrix} x=2t\\ y=3t\\ z=4t\end{matrix}\right.\)
Ta có: \(|x-y|=\frac{z^2}{12}\Leftrightarrow |2t-3t|=\frac{16t^2}{12}\)
\(\Leftrightarrow 3|-t|=4t^2\)
Nếu \(t\geq 0\Rightarrow 4t^2=3|-t|=3t\)
\(\Leftrightarrow t(4t-3)=0\Leftrightarrow \left[\begin{matrix} t=0\\ t=\frac{3}{4}\end{matrix}\right.\)
+) \(t=0\rightarrow x=y=z=0\rightarrow yz-x=0\)
+) \(t=\frac{3}{4}\Rightarrow x=\frac{3}{2}; y=\frac{9}{4}; z=3\) \(\rightarrow yz-x=\frac{21}{4}\)
Nếu \(t<0\Rightarrow 4t^2=3|-t|=-3t\)
\(\Leftrightarrow t(4t+3)=0\Leftrightarrow t=-\frac{3}{4}\)
\(\Rightarrow x=\frac{-3}{2}; y=\frac{-9}{4}; z=-3\rightarrow yz-x=\frac{33}{4}\)
Từ các TH trên suy ra \((yz-x)_{\max}=\frac{33}{4}\)\
\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\\\left|x-y\right|=\dfrac{z^2}{12}\end{matrix}\right.\) sử dụng t/c dãy tỷ bằng nhau
\(z=0\Rightarrow x=y=0=>yz-x=0\)
\(z\ne0\Rightarrow\dfrac{yz-x}{3z-2}=\dfrac{z}{4}\Rightarrow yz-x=\dfrac{z}{4}\left(3z-2\right)=\dfrac{3z^2-2z}{4}\) (1)
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x-y}{-1}=\dfrac{z}{4}\Rightarrow\left|x-y\right|=\dfrac{\left|z\right|}{4}=\dfrac{z^2}{12}\)\(\Rightarrow\left[{}\begin{matrix}z=0\\z=\pm3\end{matrix}\right.\)(2)
(1) và (2) =>\(Max\left(yz-x\right)=\dfrac{3.\left(-3\right)^2-2\left(-3\right)}{4}=\dfrac{33}{4}\)