Question

Cho \(\left(x_1p-y_1q\right)^{2n}+\left(x_2p-y_2q\right)^{2n}+...+\left(x_mp+y_mq\right)\)

CMR:\(\frac{x_1+x_2+...+x_n}{y_1+y_2+...+y_2}=\frac{q}{p}\)

Answer 1

Ta có: \(\left(x_1p-y_1q\right)^{2n}\ge0;\left(x_2p-y_2q\right)^{2n}\ge0;....;\left(x_mp+y_mq\right)^{2n}\ge0\)

=>(x₁p-y₁q)²ⁿ+(x₂p-y₂q)²ⁿ+...+(x_mp-y_mq)²ⁿ > hoặc = 0

Mà theo đề (x₁p-y₁q)²ⁿ+(x₂p-y₂q)²ⁿ+...+(x_mp-y_mq)²ⁿ < hoặc = 0

nên: (x₁p-y₁q)²ⁿ+(x₂p-y₂q)²ⁿ+...+(x_mp-y_mq)²ⁿ=0

=>x₁p-y₁q=0 =>x₁=y₁q/p

    x₂p-y₂q=0 =>x₂=y₂q/p

........

    x_mp-y_mq=0 =>x_m=y_mq/p

Suy ra: \(\frac{x_1+x_2+...+x_n}{y_1+y_2+....+y_n}=\frac{\frac{y_1q}{p}+\frac{y_2q}{p}+...+\frac{y_mq}{p}}{y_1+y_2+...+y_m}=\frac{\frac{q}{p}\left(y_1+y_2+....+y_m\right)}{y_1+y_2+...+y_m}=\frac{q}{p}\)

=>điều phải chứng minh

Answer 2

ah quen!thieu dieu kien Cho......\(\le0\)voi moi m,n thuocN*