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Ta có: x-y-z = 0
\(\Rightarrow\) x = y+z
\(\Rightarrow\)y = x-z
\(\Rightarrow\)z = x-y
Thay vào B ta suy ra: \(\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)
= \(\left(1-\frac{x-y}{x}\right)\left(1-\frac{y+z}{y}\right)\left(1+\frac{x-z}{z}\right)\)
= \(\left(\frac{-y}{x}\right).\left(\frac{z}{y}\right).\left(\frac{x}{z}\right)\)
= -y/y
= -1
Vậy B = -1
* Nếu x = y = z = t; vẫn thỏa gt: x/(y+z+t) = y/(x+z+t) = z/(y+x+t) = t(y+z+x) = 1/3
=> P = 2x/2x + 2x/2x + 2x/2x + 2x/2x = 4
* Nếu có ít nhất 2 số khác nhau, giả sử x # y. tính chất tỉ lệ thức:
x/(y+z+t) = y/(x+z+t) = (x-y) /(y+z+t -x-z-t) = (x-y)/(y-x) = -1
=> x = -(y+z+t) => x+y+z+t = 0
=>
{ x+y = -(z+t) ---- { (x+y)/(z+t) = -1
{ y+z = -(t+x) => { (y+z)/(t+x) = -1
{ z+t = -(x+y) ---- { (z+t)/(x+y) = -1
{ t+x = -(z+y) ---- { (t+x)/(z+y) = -1
=> P = -1 -1 -1 -1 = -4
~~~~~~~~~~~~~~~~~
\(A=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)=\frac{\left(x-z\right)\left(y-x\right)\left(y+z\right)}{xyz}=\frac{y.\left(-z\right).x}{xyz}=-1\)
x - y - z = 0
x = y + z
y = x - z
z = x - y => -z = y - x
B = (1 - z/x)(1 - x/y) (1 + y/z)
B = (x/x - z/x)( y/y - x/y) ( z/z + y/z)
B = \(\frac{x-z}{x}\cdot\frac{y-x}{y}\cdot\frac{z+x}{z}=\frac{y}{x}\cdot\frac{-z}{y}\cdot\frac{x}{z}=-1\)
\(a.\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+xy+y^2+zy+zx+zy+z^2=x^2+y^2+z^2+2xy+2zy+2zx\)
\(b.\left(x-y+z\right)\left(x-y-z\right)=x^2-xy-zx-xy+y^2+zy+zx-zy-z^2=x^2+y^2-z^2-2xy\)
\(c.\left(x-1+y\right)\left(x-1-y\right)=x^2-x-xy-x+1+y+xy-y-y^2=x^2-y^2-2x+1\)
a) = \(^{\left(x+y+z\right)^2}\)=\(x^2\)+\(y^2\)+\(z^2\)+ 2xy +2xz+2yz
b) = \(\left(x-y\right)^2\)-\(z^2\)=\(x^2\)- 2xy+\(y^2\)-\(z^2\)
c)= \(\left(x-1\right)^2\)-\(y^2\)= \(x^2\)-2x+1 - \(y^2\)