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\(\dfrac{x+y-2017z}{z}=\dfrac{y+z-2017x}{x}=\dfrac{z+x-2017y}{y}\)
<=> \(\dfrac{x+y}{z}-2017=\dfrac{z+y}{x}-2017=\dfrac{z+x}{y}-2017\)
<=> \(\dfrac{x+y}{z}=\dfrac{z+y}{x}=\dfrac{z+x}{y}\)
đặt x+y+z = t
=> \(\dfrac{t-z}{z}=\dfrac{t-x}{x}=\dfrac{t-y}{y}< =>\dfrac{t}{z}-1=\dfrac{t}{x}-1=\dfrac{t}{y}-1\) \(< =>\dfrac{t}{z}=\dfrac{t}{y}=\dfrac{t}{x}\)
=> x=y=z
ta lại có
\(P=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{x}{z}\right)\left(1+\dfrac{z}{y}\right)\)
vì x=y=z => P = \(\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Ta có : từ x - y - z =0
\(\Rightarrow x-z=y\) ; \(-z=y-x\) ; \(y+z=x\)
Lại có \(B=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(\Rightarrow B=\dfrac{x-z}{x}.\dfrac{y-x}{y}.\dfrac{y+z}{z}\)
thay các hằng đẳng thức vừa tìm được vào B
\(\Rightarrow B=\dfrac{y}{x}.\dfrac{-z}{y}.\dfrac{x}{z}=-1\)
vậy B = -1
tik mik nha !!!
Vì \(x-y-z=0\Rightarrow\left[{}\begin{matrix}x-z=y\\y-x=-z\\z+y=x\end{matrix}\right.\) (1)
Thay (1) vào B ta đc:
\(B=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(=\left(\dfrac{x-z}{x}\right)\left(\dfrac{y-x}{y}\right)\left(\dfrac{z+y}{z}\right)\)
\(=\dfrac{y}{x}.\dfrac{-z}{y}.\dfrac{x}{z}\)
\(=-1\)
Vậy \(B=-1.\)
x-y-z=0
\(\Rightarrow x=y+z\)
\(\Rightarrow y=x-z\)
\(\Rightarrow-z=y-z\)
\(B=\left(1-\dfrac{z}{x}\right).\left(1-\dfrac{y}{x}\right).\left(1+\dfrac{y}{z}\right)\)
\(B=\left(\dfrac{x-z}{x}\right).\left(\dfrac{y-x}{y}\right).\left(\dfrac{z+y}{z}\right)\)
\(B=(\dfrac{y}{x}).\left(\dfrac{-z}{y}\right).\left(\dfrac{x}{z}\right)\)
\(B=\dfrac{\left(y.x.-z\right)}{\left(y.x.z\right)}\Rightarrow B=-1\)
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\)
\(\Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\)
\(\Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}\\\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x+y+z\right)=y\left(x+y+z\right)\\y\left(x+y+z\right)=z\left(x+y+z\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x+y+z\right)=0\\\left(y-z\right)\left(x+y+z\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\x+y+z=0\end{matrix}\right.\\\left[{}\begin{matrix}y=z\\x+y+z=0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=y=z\\x+y+z=0\end{matrix}\right.\)
\(\circledast\) 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 \(A=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)=\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}=\dfrac{-xyz}{xyz}=-1\)
a, \(\left|3x-4\right|+\left|3y+5\right|=0\)
Ta có :
\(\left|3x-4\right|\ge0\forall x;\left|3y+5\right|\ge0\forall x\\ \)
\(\Rightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\forall x\\ \Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{5}{3}\end{matrix}\right.\\ Vậy.........\)
b, \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
Ta có :
\(\left|x+\dfrac{19}{5}\right|\ge0\forall x;\left|y+\dfrac{1890}{1975}\right|\ge0\forall y;\left|z-2004\right|\ge0\forall z \)
\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1890}{1975}\\z=2004\end{matrix}\right.\\ Vậy............\)
c, \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
Ta có : \(\left|x+\dfrac{9}{2}\right|\ge0\forall x;\left|y+\dfrac{4}{3}\right|\ge0\forall y;\left|z+\dfrac{7}{2}\right|\ge0\forall z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\\ Vậy............\)
d, \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
Ta có :
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x;\left|y-\dfrac{1}{5}\right|\ge0\forall y;\left|x+y+z\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=0-\dfrac{1}{5}+\dfrac{3}{4}=\dfrac{11}{20}\end{matrix}\right.\\ Vậy.......\)
e, Câu cuối bn làm tương tự như câu a, b, c nhé!
Vào đây:
Câu hỏi của Phạm Đức Minh - Toán lớp 7 | Học trực tuyến
Ta có: \(x-y-z=0\)
\(\Rightarrow x-y=z\)
\(x-z=y\)
\(y+z=x\)
\(\Rightarrow B=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(=\dfrac{x-z}{x}.\dfrac{-\left(y-x\right)}{y}.\dfrac{z+y}{z}\)
\(=\dfrac{y}{x}.-\dfrac{z}{y}.\dfrac{z}{x}=-1\)
\(\Rightarrow B=-1\)