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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:
Ta có: \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{y+t+x}=\dfrac{t}{y+x+z}\)
\(\Rightarrow\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{y+t+x}+1=\dfrac{t}{y+x+z}+1\)
\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{y+t+x}=\dfrac{x+y+z+t}{y+x+z}\)+) Xét \(x+y+z+t=0\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\x+t=-\left(y+z\right)\end{matrix}\right.\)
\(\Rightarrow A=-1\)
+) Xét \(x+y+z+t\ne0\Rightarrow x=y=z=t\)
\(\Rightarrow A=1\)
Vậy A = -1 hoặc A = 1
Ta có:\(\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{y+t+x}+1=\dfrac{t}{y+x+z}+1\)\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{t+x+y}=\dfrac{x+y+z+t}{x+y+z}\)
Nếu x+y+z+t\(\ne\)0 thì y+z+t=z+t+x=t+x+y=x+y+z
=>x=y=z=t nên P=1+1+1+1=4
Nếu X+y+z+t=0 thì P=-4
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\)
\(\Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\)
\(\Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}\\\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x+y+z\right)=y\left(x+y+z\right)\\y\left(x+y+z\right)=z\left(x+y+z\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x+y+z\right)=0\\\left(y-z\right)\left(x+y+z\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\x+y+z=0\end{matrix}\right.\\\left[{}\begin{matrix}y=z\\x+y+z=0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=y=z\\x+y+z=0\end{matrix}\right.\)
\(\circledast\) 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 \(A=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)=\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}=\dfrac{-xyz}{xyz}=-1\)
+) Nếu \(x+y+z\ne0\)
Theo t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}=\dfrac{\left(y+z-x\right)+\left(z+x-y\right)+\left(x+y-z\right)}{x+y+z}=\dfrac{x+y+z}{x+y+z}=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y+z-x}{x}=1\\\dfrac{x+z-y}{y}=1\\\dfrac{x+y-z}{z}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+z-x=x\\x+z-y=y\\x+y-z=z\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+z=2x\\x+z=2y\\x+y=2z\end{matrix}\right.\)
\(\Leftrightarrow B=\left(\dfrac{x+y}{y}\right)\left(\dfrac{y+z}{z}\right)\left(\dfrac{x+z}{x}\right)\)
\(\Leftrightarrow B=\dfrac{2z}{y}.\dfrac{2x}{z}.\dfrac{2y}{x}=2\)
+) Nếu \(x+y+z\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)
\(\Leftrightarrow B=\dfrac{-z}{y}.\dfrac{-x}{z}.\dfrac{-y}{x}=-1\)
Vậy ..
Hằng à,t chưa thấy đứa này ngu như mày
\(\dfrac{2x.2y.2z}{xyz}=2\) thì học hành cái qq j
\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{1}{3}=\dfrac{x+y}{\left(x+y\right)+2\left(z+t\right)}\)
\(\Rightarrow\left(x+y\right)+2\left(z+t\right)=3\left(x+y\right)\)
\(\Rightarrow2\left(z+t\right)=2\left(x+y\right)\Rightarrow\dfrac{x+y}{z+t}=1\)
Chứng minh tương tự ta được:
\(\dfrac{y+z}{x+t}=1;\dfrac{z+t}{x+y}=1;\dfrac{t+x}{y+z}=1\)
\(\Rightarrow P=1+1+1+1=4\)
+Xét x+y+z+t=0
\(\Rightarrow\)\(\left\{{}\begin{matrix}z+t=-\left(x+y\right)\\x+t=-\left(y+z\right)\\x+y=-\left(z+t\right)\\y+z=-\left(t+x\right)\end{matrix}\right.\)
Khi đó M=-4
+Xét x+y+z+t\(\ne\)0
ADTC dãy tỉ số bằng nhau ta có
\(\dfrac{x}{y+z+t}\)=\(\dfrac{y}{x+y+t}\)=\(\dfrac{z}{x+y+t}\)=\(\dfrac{z}{x+y+t}\)=\(\dfrac{x+y+z+t}{3.\left(x+y+z+t\right)}\)=\(\dfrac{1}{3}\)
+Với\(\dfrac{x}{y+z+t}\)=\(\dfrac{1}{3}\)
\(\Rightarrow\)3x=y+z+t
\(\Rightarrow\)4x=x+y+z+t
Chứng minh tương tự ta có
4y=x+y+z+t
4z=x+y+z+t
4t=x+y+z+t
Do đó x=y=z=t
Khi đó M=4
\(\text{Ta có : }\dfrac{x}{y+z}=\dfrac{y}{x+z}=\dfrac{z}{y+x}\\ \Rightarrow\dfrac{y+z}{x}=\dfrac{x+z}{y}=\dfrac{y+x}{z}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\dfrac{y+z}{x}=\dfrac{x+z}{y}=\dfrac{y+x}{z}\\ =\dfrac{\left(y+z\right)+\left(x+z\right)+\left(y+x\right)}{x+y+z}\\ =\dfrac{y+z+x+z+y+x}{x+y+z}\\ =\dfrac{\left(y+y\right)+\left(z+z\right)+\left(x+x\right)}{x+y+z}\\ =\dfrac{2y+2z+2x}{x+y+z}\\ =\dfrac{2\left(x+y+z\right)}{x+y+z}\\ =2\\ \)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z}{x}=2\\\dfrac{x+z}{y}=2\\\dfrac{y+x}{z}=2\end{matrix}\right.\Rightarrow\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{y+x}{z}=2+2+2=6\)
Vậy \(\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{y+x}{z}=6\)