Question

a)Cho x;y;z\(\ne\)0 thỏa mãn \(\frac{x-y-z}{x}=\frac{-x+y-z}{y}=\frac{-x-y+z}{z}\)

Tính \(A=\left(1+\frac{y}{x}\right)\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)\)

b)Tính \(C=4x^2y+2xy^3-y^2\)biết \(\left|x+3y-1\right|=-3\left|y+3\right|\)

Answer 1

a) theo tính chất của dãy tỉ số bằng nhau có 

\(\frac{x-y-z}{x}=\frac{-x+y-z}{y}=\frac{-x-y+z}{z}=\frac{x-y-z-x+y-z-x-y+z}{x+y+z}=\frac{-\left(x+y+z\right)}{x+y+z}=-1\)

=> x - y - z = - x  => 2.x = y + z

    y - x - z = - y  => 2.y = x+z

    z - x - y = - z => 2.z = x+y

Ta có: \(A=\left(1+\frac{y}{x}\right)\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)=\frac{x+y}{x}.\frac{y+z}{y}.\frac{z+x}{z}=\frac{2z}{x}.\frac{2x}{y}.\frac{2y}{z}=\frac{2xyz}{xyz}=2\)

b) Vì \(\left|x+3y-1\right|\ge0\); \(-3\left|y+3\right|\le0\)

=> \(\left|x+3y-1\right|=-3\left|y+3\right|\) khi \(\left|x+3y-1\right|=-3\left|y+3\right|=0\)

=> x+ 3y - 1 = 0 và y + 3 = 0

=> x = 1 - 3y và y = -3 => x = 1- 3(-3) = 10; y = -3

=> C = 4.10².(-3) + 2.10.(-3)² - (-3)² = -1029