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\(P=\left(\dfrac{x+2y}{y}\right)\left(\dfrac{y+2z}{z}\right)\left(\dfrac{z+2x}{x}\right)\)
Ta có
\(\dfrac{x+2y-z}{z}=\dfrac{y+2z-x}{x}=\dfrac{z+2x-y}{y}=\)
\(=\dfrac{x+2y-z+y+2z-x+z+2x-y}{x+y+z}=\)
\(=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\dfrac{x+2y}{z}-1=\dfrac{y+2x}{x}-1=\dfrac{z+2x}{y}-1=2\)
\(\Rightarrow\dfrac{x+2y}{z}=\dfrac{y+2x}{x}=\dfrac{z+2x}{y}=3\)
\(\Rightarrow P=3.3.3=27\)
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
Ta có :
\(\dfrac{x+y-z}{z}=\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}\\ \Leftrightarrow\dfrac{x+y+z}{z}=\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}\left(cùngcộngthêm2\right)\)
TH1: \(x+y+z\ne0\)
\(\Rightarrow x=y=z\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)\\ =2\cdot2\cdot2=8\)
TH2: \(x+y+z=0\Rightarrow\left\{{}\begin{matrix}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(y+x\right)\end{matrix}\right.\)(*)
\(\Rightarrow P=\left(1+\dfrac{-\left(y+z\right)}{y}\right)\left(1+\dfrac{-\left(z+x\right)}{z}\right)\left(1+\dfrac{-\left(x+y\right)}{z}\right)\\ =\left(1-1-\dfrac{z}{y}\right)\left(1-1-\dfrac{x}{z}\right)\left(1-1-\dfrac{y}{z}\right)\\ =\left(-\dfrac{z}{y}\right)\left(-\dfrac{x}{z}\right)\left(-\dfrac{y}{z}\right)\\ =-1\)
Vậy P=8 hoặc P=-1
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\)
\(\Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\)
\(\Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}\\\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x+y+z\right)=y\left(x+y+z\right)\\y\left(x+y+z\right)=z\left(x+y+z\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x+y+z\right)=0\\\left(y-z\right)\left(x+y+z\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\x+y+z=0\end{matrix}\right.\\\left[{}\begin{matrix}y=z\\x+y+z=0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=y=z\\x+y+z=0\end{matrix}\right.\)
\(\circledast\) 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 \(A=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)=\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}=\dfrac{-xyz}{xyz}=-1\)
Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)
Tương tự:
\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)
\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)
\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}=\dfrac{x+3y-z+y+3z-x+z+3x-y}{x+y+z}=\dfrac{3(x+y+z)-(x+y+z)}{x+y+z}=\dfrac{2(x+y+z)}{x+y+z}=2\)
\(\Rightarrow x=y=z=0\)
\(\Rightarrow \) P không xác định. (?)
Từ \(\dfrac{x+y-z}{x}=\dfrac{y+z-x}{y}=\dfrac{z+x-y}{z}\)
=> \(1+\dfrac{y-z}{x}=1+\dfrac{z-x}{y}=1+\dfrac{x-y}{z}\)
=> \(\dfrac{y-z}{x}=\dfrac{z-x}{y}=\dfrac{x-y}{z}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{y-z}{x}=\dfrac{z-x}{y}=\dfrac{x-y}{z}=\dfrac{y-z+z-x+x-y}{x+y+z}=\dfrac{0}{x+y+z}=0\)
Ta có : \(\dfrac{y-z}{x}=0\)
=> y - z = 0 ; Vì x # 0 => y = z
\(\dfrac{z-x}{y}=0\)
=> z - x = 0 . Vì y # 0 => z = x
=> y = z = x
Ta có: A = \(\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)
A = (1 + 1) (1 + 1) ( 1 + 1)
A = 2 . 2 . 2 = 8
\(\dfrac{x+y-z}{z}=\dfrac{y+z-x}{x}=\dfrac{x-y+z}{y}\)
\(\Rightarrow\dfrac{x+y-z}{z}+2=\dfrac{y+z-x}{x}+2=\dfrac{x-y+z}{y}+2\)
\(\Rightarrow\dfrac{x+y-z}{z}+\dfrac{2z}{z}=\dfrac{y+z-x}{x}+\dfrac{2x}{x}=\dfrac{x-y+z}{y}+\dfrac{2y}{y}\)
\(\Rightarrow\dfrac{x+y-z+2z}{z}=\dfrac{y+z-x+2x}{x}=\dfrac{x-y+z+2y}{y}\)
\(\Rightarrow\dfrac{x+y+z}{z}=\dfrac{y+z+x}{x}=\dfrac{x+z+y}{y}\)
Điều này xảy ra khi và chỉ khi: \(\left[{}\begin{matrix}x+y+z=0\\x=y=z\end{matrix}\right.\)
\(\circledast\)Với \(x+y+z=0\Leftrightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)
Thay vào \(A\) ta có: \(A=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{x}{z}\right)=\left(\dfrac{x+y}{y}\right)\left(\dfrac{y+z}{z}\right)\left(\dfrac{z+x}{z}\right)=\dfrac{-z.-x.-y}{xyz}=\dfrac{-xyz}{xyz}=-1\)
\(\circledast\) Với \(x=y=z\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{y}=1\\\dfrac{y}{z}=1\\\dfrac{x}{z}=1\end{matrix}\right.\)
Thay vào \(A\) ta có:
\(A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
@ Mashiro Shiina
@Akai Haruma
@Nguyễn Thanh Hằng
@Đẹp Trai Không Bao Giờ Sai
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{t}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{t}=\dfrac{x+y+z}{y+z+t}\)
Suy ra: \(\dfrac{x}{y}.\dfrac{y}{z}.\dfrac{z}{t}=\left(\dfrac{x+y+z}{y+z+t}\right)^3\)
\(\Rightarrow\dfrac{x}{t}=\left(\dfrac{z+y+z}{y+z+t}\right)^3\)
Vậy...
P/s: Sai đề rồi bạn ơi~