Question

giả sử a1, b1, c1, a2, b2, c2 là các số khác 0 thỏa mãn các 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: \(\dfrac{a^2_2}{a_1^2}+\dfrac{b_2^2}{b_1^2}+\dfrac{c_2^2}{c_1^2}=1\)

Answer 1

Ta có:

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

\(\Rightarrow\dfrac{a^2_2}{a^2_1}+\dfrac{b_2^2}{b_1^2}+\dfrac{c_2^2}{c_1^2}+2\left(\dfrac{a_2b_2}{a_1b_1}+\dfrac{b_2c_2}{b_1c_1}+\dfrac{c_2a_2}{a_1c_1}\right)=1\)

\(\Rightarrow\dfrac{a_2^2}{a^2_1}+\dfrac{b^2_2}{b^2_1}+\dfrac{c^2_2}{c^2_1}+2\left(\dfrac{a_2b_2c_1+b_2c_2a_1+c_2a_2b_1}{a_1b_1c_1}\right)=1\)(1)

Theo giả thiết:

\(\dfrac{a_1}{a_2}+\dfrac{b_1}{b_2}+\dfrac{c_1}{c_2}=0\Leftrightarrow\dfrac{a_1b_2c_2+b_1a_2c_2+c_1a_2b_2}{a_2b_2c_2}=0\)(2)

Từ (1) và (2) suy ra đpcm

Answer 2

Đặt \(\dfrac{a_1}{a_2}=p;\dfrac{b_1}{b_2}=q;\dfrac{c_1}{c_2}=r\), có:

\(p+q+r=0\) (1)

\(\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}=1\) (2)

Từ (2) => \(\dfrac{1}{p^2}+\dfrac{1}{q^2}+\dfrac{1}{r^2}+2\dfrac{p+q+r}{pqr}=1\)

Kết hợp với (1), ta được: \(\dfrac{1}{p^2}+\dfrac{1}{q^2}+\dfrac{1}{r^2}=1\Rightarrow\dfrac{a^2_2}{a^2_1}+\dfrac{b^2_2}{b_1^2}+\dfrac{c_2^2}{c^2_1}=1\left(đpcm\right)\)