chơi ăn gian wa

ta co: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}.\)

\(\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=0\)

=> x + y + z = 0

Lai co: x³ + y³ +z³ - 3xyz = (x+y+z).(x²+y²+z² - xy - yz - zx)

             x³ + y³ + z³ - 3xyz = 0

=> x³ + y³ + z³ = 3xyz

ta co: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}.\)

=> 1/xy + 1/yz + 1/xz = 0

=> x + y + z = 0

Lai co: x³ + y³ +z³ - 3xyz = (x+y+z).(x²+y²+z² - xy - yz - zx)

             x³ + y³ + z³ - 3xyz = 0

=> x³ + y³ + z³ = 3xyz

Cho \(x+y=1\)

Ta có :

\(2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)

\(=2\left(x+y\right)\left(x^2+y^2-xy\right)-3\left[\left(x+y\right)^2-2xy\right]\)

\(=2.1.\left[\left(x+y\right)^2-3xy\right]-3\left[1-2xy\right]\)

\(=2\left[1-3xy\right]-3-\left(1-2xy\right)\)

\(=2-6xy-3+6xy\)

\(=1\)

ĐK: \(x;y;z\ne0\)

\(\frac{y+z}{x}+\frac{x+z}{y}+\frac{x+y}{z}+3=\left(\frac{y+z}{x}+1\right)+\left(\frac{x+z}{y}+1\right)+\left(\frac{x+y}{z}+1\right)-3+3\)

\(=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0\left(đpcm\right)\)

bạn cảm ơn ai vay có bn ấy có giup bn làm đau

mk chua hok den nen ko co bit lam

Biến đổi \(\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{x^4-x-y^4+y}{\left(y^3-1\right)\left(x^3-1\right)}=\frac{\left(x^4-y^4\right)-\left(x-y\right)}{xy\left(y^2+y+1\right)\left(x^2+x+1\right)}\)

(Do x+y=1 => \(\hept{\begin{cases}y-1=-x\\x-1=-y\end{cases}}\))

\(=\frac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)-\left(x-y\right)}{xy\left(x^2y^2+y^2x+y^2+yx^2+xy+y+x^2+x+1\right)}\)

\(=\frac{\left(x-y\right)\left(x^3+y^3-1\right)}{xy\left[x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+2\right]}\)

\(=\frac{\left(x-y\right)\left(x^2-x+y^2-y\right)}{xy\left[x^2y^2+\left(x+y\right)^2+2\right]}=\frac{\left(x-y\right)\left[x\left(x-1\right)+y\left(y-1\right)\right]}{xy\left(x^2y^2+3\right)}\)

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

\(\Rightarrow\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\left(đpcm\right)\)