Question

\(\frac{x}{z+t+y}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)

Chứng minh p nguyên p= \(\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)

Answer 1

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

Answer 2

bạn tự làm tiếp đi nhé

Answer 3

\(\frac{x}{y+z+t}=\frac{y}{x+z+t}=\frac{z}{x+y+t}=\frac{t}{x+y+z}\)

\(\Leftrightarrow\frac{x}{y+z+t}+1=\frac{y}{x+z+t}+1=\frac{z}{x+y+t}+1=\frac{t}{x+y+z}+1\)

\(\Leftrightarrow\frac{x}{x+y+z+t}=\frac{y}{x+y+z+t}=\frac{z}{x+y+z+t}=\frac{t}{x+y+z+t}\)

\(\Rightarrow x=y=z=t\) Thay vào p ta được

\(p=\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}=1+1+1+1=4\)

=> p là số nguyên (đpcm)