Question

[Nguyễn Khánh Ly]

1. Cho \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\)chứng minh rằng \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\)

2. Giả sử \(a_1,b_1,c_1,a_2,b_2,c_2\)là các số khác 0 thỏa mãn đk : \(\dfrac{a_1}{a_2}+\dfrac{b_1}{b_2}+\dfrac{c_1}{c_2}=0\)và\(\dfrac{a_2}{a_1}+\dfrac{b_2}{b_1}+\dfrac{c_2}{c_1}=1\)

CMR : \(\dfrac{a_2^2}{a^2_1}+\dfrac{b_2^2}{b_1^2}+\dfrac{c_2^2}{c_1^2}=1\)

Mình còn không hiểu đề bài cho lắm vậy nên mong mọi người giúp mình

Thanks

Answer 1

[Nguyễn Dương Huy]

1. Ta có \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\)

\(\Rightarrow\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c\)

\(\Rightarrow\dfrac{a^2}{b+c}+\left(b+c\right)\left(\dfrac{a}{b+c}\right)+\dfrac{b^2}{c+a}+\left(c+a\right)\left(\dfrac{b}{c+a}\right)+\dfrac{c^2}{a+b}+\left(a+b\right)\left(\dfrac{c}{a+b}\right)=a+b+c\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+a+b+c=a+b+c\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\) (đpcm).

Answer 2

[Nguyễn Dương Huy]

2. Ta có: \(\dfrac{a_1}{a_2}+\dfrac{b_1}{b_2}+\dfrac{c_1}{c_2}=0\)

\(\Rightarrow\dfrac{a_1b_2c_2+b_1a_2c_2+c_1a_2b_2}{a_2b_2c_2}=0\)

\(\Rightarrow a_1b_2c_2+b_1a_2c_2+c_1a_2b_2=0\)

Lại có: \(\dfrac{a_2}{a_1}+\dfrac{b_2}{b_1}+\dfrac{c_2}{c_1}=1\)

\(\Rightarrow\left(\dfrac{a_2}{a_1}+\dfrac{b_2}{b_1}+\dfrac{c_2}{c_1}\right)^2=1\)

\(\Rightarrow\dfrac{a_2^2}{a_1^2}+\dfrac{b_2^2}{b_1^2}+\dfrac{c_2^2}{c_1^2}+2\left(\dfrac{a_2b_2}{a_1b_1}+\dfrac{b_2c_2}{b_1c_1}+\dfrac{a_2c_2}{a_1c_1}\right)=1\)

Mặt khác: \(\dfrac{a_2b_2}{a_1b_1}+\dfrac{b_2c_2}{b_1c_1}+\dfrac{a_2c_2}{a_1c_1}=\dfrac{a_1b_2c_2+b_1a_2c_2+c_1a_2b_2}{a_1b_1c_1}=0\)

Vậy \(\dfrac{a_2^2}{a_1^2}+\dfrac{b_2^2}{b_1^2}+\dfrac{c_2^2}{c_1^2}=1\) (đpcm)