Question

Cho x+y+z+t=0 và

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

Tính B=\(\frac{2x}{y+z+t}-\frac{3y}{x+z+t}+\frac{4z}{x+y+z}-\frac{5t}{x+y+t}\)

Answer 1

\(B=\frac{2x}{y+z+t}-\frac{3y}{x+z+t}+\frac{4z}{x+y+t}-\frac{5t}{x+y+z}\)

\(B=\frac{2x}{-x}-\frac{3y}{-y}+\frac{4z}{-z}-\frac{5t}{-t}\)

\(B=-2+3-4+5=2\)

Answer 2

giải chi tiết hơn giùm mình dc ko

Answer 3

\(B=\frac{2x}{x+y+z+t-x}-\frac{3y}{x+y+z+t-y}+\frac{4z}{y+z+t+x-z}-\frac{5t}{x+y+z+t-t}\)

Thay x+y+z+t =0.Ta có

\(B=\frac{2x}{-x}-\frac{3y}{-y}+\frac{4z}{-z}-\frac{5t}{-t}=-2+3-4+5\)

B=2