Question

bài 1: Cho x+y+z=123, \(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}=\dfrac{1}{45}\)

Tính P=\(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{y+x}\)

bài 2: Tìm x;y sao cho các đơn thức sau mang dấu âm:

\(\dfrac{-1}{4}.x^3.y^4;\dfrac{-4}{5}.x^4.y^3;\dfrac{1}{2}xy\)

hộ mình nhé, chiều nay phải có bài rồi

Answer 1

Ta có :\(\dfrac{x}{y+z}=\dfrac{123-\left(y+z\right)}{y+z}\)

\(\dfrac{y}{x+z}=\dfrac{123-\left(x+z\right)}{x+z}\)

\(\dfrac{z}{y+x}=\dfrac{123-\left(y+x\right)}{y+x}\)

\(\Rightarrow P=\dfrac{123-\left(y+z\right)}{y+z}+\dfrac{123-\left(z+x\right)}{z+x}+\dfrac{123-\left(y+x\right)}{y+x}\)\(\Rightarrow P=123\left(\dfrac{1}{y+z}+\dfrac{1}{x+y}+\dfrac{1}{z+x}\right)-3\)

\(\Rightarrow P=123.\dfrac{1}{45}-3\)

\(\Rightarrow P=-\dfrac{4}{15}\)