Bài này mình làm 2 cách cho bạn dễ hiểu nha
C1:\(P=\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1\Leftrightarrow x\left(z+x\right)\left(x+y\right)+y\left(y+z\right)\left(x+y\right)+z\left(z+x\right)\left(y+z\right)=\left(y+z\right)\left(x+y\right)\left(z+x\right) \)\(\Leftrightarrow x^2\left(y+z\right)+y^2\left(x+z\right)+z^2\left(x+y\right)+x^3+y^3+z^3+3xyz=x^2\left(y+z\right)+y^2\left(x+z\right)+z^2\left(x+y\right)+2xyz\)
\(\Leftrightarrow x^3+y^3+z^3+xyz=0\)
\(\Rightarrow\left(x^3+y^3+z^3+xyz\right)\left(x+y+z\right)=0 \)
Ta cũng thấy Q=\(Q=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=\dfrac{x^2\left(z+x\right)\left(x+y\right)+y^2\left(y+z\right)\left(x+y\right)+z^2\left(y+z\right)\left(z+x\right)}{\left(y+z\right)\left(x+z\right)\left(x+y\right)}=\dfrac{\left(x^3+y^3+z^3+xyz\right)\left(x+y+z\right)}{\left(y+z\right)\left(x+z\right)\left(x+y\right)}=0\)
C2 nè :
\(P=\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1\)
\(P=\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)\left(x+y+z\right)=x+y+z .\)
\(\Leftrightarrow\dfrac{x^2+x\left(y+z\right)}{y+z}+\dfrac{y^2+y\left(x+z\right)}{z+x}+\dfrac{z^2+z\left(x+y\right)}{x+y}=x+y+z.\)
\(\Leftrightarrow\dfrac{x^2}{y+z}+x+\dfrac{y^2}{z+x}+y+\dfrac{z^2}{x+y}+z=x+y+z \left(ĐPCM\right)\)
Q = \(\dfrac{x^2}{y+z}\) + \(\dfrac{y^2}{x+z}\) + \(\dfrac{z^2}{x+y}\)
= \(\dfrac{x\left[\left(x+y+z\right)-\left(y+z\right)\right]}{y+z}\) + \(\dfrac{y\left[\left(x+y+z\right)-\left(x+z\right)\right]}{x+z}\) + \(\dfrac{z\left[\left(x+y+z\right)-\left(x+y\right)\right]}{x+y}\)
= \(\dfrac{x\left(x+y+z\right)-x\left(y+z\right)}{y+z}\) + \(\dfrac{y\left(x+y+z\right)-y\left(x+z\right)}{x+z}\) + \(\dfrac{z\left(x+y+z\right)-z\left(x+y\right)}{x+y}\)
= \(\dfrac{x\left(x+y+z\right)}{y+z}\) - x + \(\dfrac{y\left(x+y+z\right)}{x+z}\) - y + \(\dfrac{z\left(x+y+z\right)}{x+y}\) - z
= (x + y + z)\(\left[\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\right]\) - (x + y + z)
= (x + y +z) . P - (x + y + z)
= ( x + y +z) .1 - (x + y +z)
= 0 (đpcm)