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+, Nếu x+y+z+t = 0 => M = -1 + (-1) + (-1) + (-1) = -4
+, Nếu x+y+z+t khác 0 thì :
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
x/y+z+t = y/x+z+t = z/x+t+y = t/x+y+z = x+y+z+t/3x+3y+3z+3t = 1/3
=> x=1/3.(y+z+t) ; y=1/3.(z+x+t) ; z=1/3.(x+y+t) ; t=1/3.(x+y+z)
=> x=y=z=t
=> M = 1+1+1+1 = 4
Tk mk nha
\(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{x+t+y}=\frac{t}{x+y+z}\)
\(\Rightarrow\frac{x}{y+z+t}+1=\frac{y}{z+t+x}+1=\frac{z}{x+t+y}+1=\frac{t}{x+y+z}+1\)
\(\Rightarrow\frac{x+y+z+t}{y+z+t}=\frac{x+y+z+t}{z+t+x}=\frac{x+y+z+t}{x+t+y}=\frac{x+y+z+t}{x+y+z}\)
+) Xét x + y + z + t= 0 => x + y = -(z+t) ; y + z = -(x+t); z+t = -(x+y); t+x = -(y+z)
\(\Rightarrow M=\frac{-\left(z+t\right)}{z+t}+\frac{-\left(x+t\right)}{t+x}+\frac{-\left(x+y\right)}{x+y}+\frac{-\left(y+z\right)}{y+z}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
+) Xét x+y+z+t khác 0 => x=y=z=t
\(\Rightarrow M=1+1+1+1=4\)
Ta có: \(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{x+t+y}=\frac{t}{x+y+z}\)
Thêm 1 vào mỗi phân số ta được:
\(\frac{x}{y+z+t}+1=\frac{y}{z+t+x}+1=\frac{z}{x+t+y}+1=\frac{t}{x+y+z}+1\)
\(\Rightarrow\frac{x+y+z+t}{y+z+t}=\frac{x+y+z+t}{z+t+x}=\frac{x+y+z+t}{x+t+y}=\frac{x+y+z+t}{x+y+z}\)
- Nếu x + y + z + t \(\ne\) 0 thì x = y = z = t
\(\Rightarrow P=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}=\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}=1+1+1+1=4\)
- Nếu x + y + z + t = 0 thì x + y = -(z + t)
y + z = -(t + x)
z + t = -(x + y)
t + x = -(y + z)
\(\Rightarrow P=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}=\frac{-\left(z+t\right)}{z+t}+\frac{-\left(t+x\right)}{t+x}+\frac{-\left(x+y\right)}{x+y}+\frac{-\left(y+z\right)}{y+z}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
xét x+y+z+t=0
=>x+y=-(z+t)
y+z=-(t+x)
\(\Rightarrow M=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}=\frac{x+y}{-\left(x+y\right)}+\frac{y+z}{-\left(y+z\right)}+\frac{z+t}{-\left(z+t\right)}+\frac{t+x}{-\left(t+x\right)}\)
\(=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
xét x+y+z+t\(\ne0\)
\(\frac{x}{y+z+t}+\frac{y}{x+z+t}+\frac{z}{x+y+t}+\frac{t}{x+y+z}=\frac{x+y+z+t}{3\left(x+y+z+t\right)}=\frac{1}{3}\)
=>3x=x+y+z
=>4x=x+y+z+t
3y=x+z+t
=>4y=x+y+z+t
3z=x+y+t
=>4z=x+y+z+t
3t=x+y+z+t
=>4t=x+y+z+t
=>4x=4y=4z=4t
=>x=y=z=t
\(\Rightarrow P=\frac{x+y}{z+t}+\frac{y+z}{x+t}+\frac{z+t}{x+y}+\frac{x+t}{y+z}=\frac{x+y}{x+y}+\frac{x+y}{x+y}+\frac{x+y}{x+y}+\frac{x+y}{x+y}\)
=1+1+1+1=4
Vậy P=-4 khi \(x+y+z+t=0\)
P=4 khi \(x+y+z+t\ne0\)
TA CÓ : ( x / y + z + t ) + 1 = ( y / z +t + x ) + 1 = ( t / x + y + z ) + 1
Suy ra : x+y+z+t / y+z+t = x+y+z+t / z+t+x = x+y+z+t / t+x+y = x+y+z+t / x+y+z
do x+y+z+t khác 0 suy ra x=y=z=t suy ra M= 1+1+1+1 =4
tích đúng nha