Question

Giúp mình với.

Answer 1

\(\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right).\left(x+y+z\right)=x+y+z\) vì

\(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1\)

\(\Leftrightarrow\dfrac{x^2}{y+z}+\dfrac{x\left(y+z\right)}{y+z}+\dfrac{y^2}{z+x}+\dfrac{y\left(z+x\right)}{z+x}+\dfrac{z^2}{x+y}+\dfrac{z\left(x+y\right)}{x+y}=x+y+z\Leftrightarrow\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}+x+y+z=x+y+z\Leftrightarrow\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=0\left(đpcm\right)\)

Answer 2

ahjhj bó tay

Answer 3

kq=0 nhé Dũng :)

Answer 4

cai nay.... cai nay ..... cai nay.... cai nay...... cai nay .......... cai nay....... thôi mk chịu