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TH1: \(x+y+z+t\ne0\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)\(\dfrac{x}{y+z+t}=\dfrac{1}{3}\Rightarrow3x=y+z+t\Rightarrow4x=x+y+z+t\\ \dfrac{y}{z+t+x}=\dfrac{1}{3}\Rightarrow3y=x+z+t\Rightarrow4y=x+y+z+t\\ \dfrac{z}{t+x+y}=\dfrac{1}{3}\Rightarrow3z=x+y+t\Rightarrow4z=x+y+z+t\\ \dfrac{t}{x+y+z}=\dfrac{1}{3}\Rightarrow3t=x+y+z\Rightarrow4t=x+y+z+t\)
\(\Rightarrow4x=4y=4z=4t\\
\Rightarrow x=y=z=t\)
\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =1+1+1+1\\ =4\)
TH1: \(x+y+z+t=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\t+x=-\left(y+z\right)\end{matrix}\right.\)
\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =\dfrac{-\left(z+t\right)}{z+t}+\dfrac{-\left(t+x\right)}{t+x}+\dfrac{-\left(x+y\right)}{x+y}+\dfrac{-\left(y+z\right)}{y+z}\\ =-1-1-1-1\\ =-4\)
Lời giải:
Nếu $x+y+z+t=0$ thì:
$P=\frac{-(z+t)}{z+t}+\frac{-(t+x)}{t+x}+\frac{-(x+y)}{x+y}+\frac{-(y+z)}{y+z}$
$=-1+(-1)+(-1)+(-1)=-4$
Nếu $x+y+z+t\neq 0$ thì áp dụng TCDTSBN:
$\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}=\frac{x+y+z+t}{3(x+y+z+t)}=\frac{1}{3}$
$\Rightarrow 3x=y+z+t; 3y=z+t+x; 3z=t+x+y; 3t=x+y+z$
$\Rightarrow x=y=z=t$
$\Rightarrow P=1+1+1+1=4$
TH1: \(x+y+z+t\ne0\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)\(\dfrac{x}{y+z+t}=\dfrac{1}{3}\Rightarrow3x=y+z+t\Rightarrow4x=x+y+z+t\\ \dfrac{y}{z+t+x}=\dfrac{1}{3}\Rightarrow3y=x+z+t\Rightarrow4y=x+y+z+t\\ \dfrac{z}{t+x+y}=\dfrac{1}{3}\Rightarrow3z=x+y+t\Rightarrow4z=x+y+z+t\\ \dfrac{t}{x+y+z}=\dfrac{1}{3}\Rightarrow3t=x+y+z\Rightarrow4t=x+y+z+t\)
\(\Rightarrow4x=4y=4z=4t\\ \Rightarrow x=y=z=t\)
\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =1+1+1+1\\ =4\)
TH2: \(x+y+z+t=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\t+x=-\left(y+z\right)\end{matrix}\right.\)
\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =\dfrac{-\left(z+t\right)}{z+t}+\dfrac{-\left(t+x\right)}{t+x}+\dfrac{-\left(x+y\right)}{x+y}+\dfrac{-\left(y+z\right)}{y+z}\\ =-1-1-1-1\\ =-4\)
Ta có :
\(\dfrac{x}{y+z+t}=\dfrac{y}{x+z+t}=\dfrac{z}{x+y+t}=\dfrac{t}{x+y+z}\)\(\Rightarrow\dfrac{x}{y+z+t}+1=\dfrac{y}{x+z+t}+1=\dfrac{z}{x+y+t}+1\)\(=\dfrac{t}{x+y+z}+1\)
\(\Rightarrow\dfrac{x+y+z+t}{x+y+t}=\dfrac{x+y+z+t}{x+z+t}=\dfrac{x+y+z+t}{x+y+z}\)
\(=\dfrac{x+y+z+t}{x+y+z}\)
* Nếu \(x+y+z+t=0\)
\(\Rightarrow x+y=-\left(z+t\right)\)
\(y+z=-\left(t+x\right)\)
Thay vào A ta được: \(P=-1+-1=-2\)
*Nếu \(x+y+z+t\ne0\)
\(\Rightarrow x+y+t=x+y+z\Rightarrow t=z\)
Làm tương tự tự ta suy ra được \(x=y=z=t\)
=> \(x+y=z+t\)
\(y+z=t+x\)
Thay vào A ta được A= 1+1=2
Vậy... tik mik nha !!!
Xét:
\(\dfrac{x}{y+z+t}+1=\dfrac{y}{x+t+z}+1=\dfrac{z}{t+x+y}+1=\dfrac{t}{x+y+z}+1\)
\(\Leftrightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{t+x+y}=\dfrac{x+y+z+t}{x+y+z}\)
+ TH1: Nếu \(x+y+z+t\ne0\Rightarrow x=y=z=t\Rightarrow P=4\)
+ TH2: Nếu \(x+y+z+t=0\Rightarrow P=-4\)
Vậy \(\left[{}\begin{matrix}P=4\\P=-4\end{matrix}\right.\)
Theo dãy tỉ số = nhau ta có :
\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3x+3y+3z+3t}=\dfrac{1}{3}\)
\(\dfrac{x}{y+z+t}=\dfrac{1}{3}\Leftrightarrow3x=y+z+t\) (1)
\(\dfrac{y}{z+t+x}=\dfrac{1}{3}\Leftrightarrow3y=z+t+x\) (2)
\(\dfrac{z}{t+x+y}=\dfrac{1}{3}\Leftrightarrow3z=t+x+y\) (3)
\(\dfrac{t}{x+y+z}=\dfrac{1}{3}\Leftrightarrow3t=x+y+z\) (4)
Từ (1) và (2) => 3x + 3y = x + y + 2(z+t) => 2(x+y) = 2(z+t) => x + y = z + t (5)
Từ (2) và (3) => 3y + 3z = y + z + 2(t + x) => 2(y+z) = 2(t+x) = > y + z = t + x
Vậy P = \(\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}=4\)
Khi nào cần gì thì hỏi nhé ok