thực hiện phép tính

a,\(x^3+\left[\frac{x\left(2y^3-x^3\right)}{x^3+y^3}\right]^3-\left[\frac{y\left(2x^3-y^3\right)}{x^3+y^3}\right]^3\)

b,\(\frac{\frac{x\left(x+y\right)}{x-y}+\frac{x\left(x+z\right)}{x-z}}{1+\frac{\left(y-z\right)^2}{\left(x-y\right)\left(x-z\right)}}+\frac{\frac{y\left(y+z\right)}{y-z}+\frac{y\left(y+x\right)}{y-x}}{1+\frac{\left(z-x\right)^2}{\left(y-z\right)\left(y-x\right)}}+\frac{\frac{z\left(z+x\right)}{z-x}+\frac{z\left(z+y\right)}{z-y}}{1+\frac{\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}}\)

c,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)

ta có:(vế phải)²\(\le3\left(\frac{x^3}{y+z}+\frac{y^3}{z+x}+\frac{z^3}{x+y}\right)\)

cần chứng minh:

(vế trái)²/3\(\ge\frac{x^3}{y+z}+\frac{y^3}{z+x}+\frac{z^3}{x+y}\)

\(\Leftrightarrow\frac{x}{y+z}\left(\frac{x^3+\frac{1}{3}}{y+z}-x^2\right)+...\ge0\)

\(\Leftrightarrow\frac{x^2}{y+z}\left(x-y\right)\left(x-z\right)+\frac{y^2}{z+x}\left(y-x\right)\left(y-z\right)+\frac{z^2}{x+y}\left(z-x\right)\left(z-y\right)\ge0\)

bđt luôn đúng vì là bđt schur mở rộng

Đề:

Cho các số thực x, y, z thoả mãn x + y + z = 1 và \(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1\)

\(\left(x\ne-y;y\ne-z;z\ne-x\right)\)

Giá trị của biểu thức \(P=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\) là . . .

Giải:

x + y + z = 1

=> x = 1 - (y + z)

y = 1 - (x + z)

z = 1 - (x + y)

Thay x = 1 - (y + z); y = 1 - (x + z) và z = 1 - (x + y) vào P, ta có:

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

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

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

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

\(=1-1\)

\(=0\)

ĐS: 0

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