Rút gọn biểu thức: 1/a^2 x 1/b^2 x 1/c^2= (1/a + 1/b +1/c)^2 . Cho a+b+c=0 27/07/2021 Bởi Kennedy Rút gọn biểu thức: 1/a^2 x 1/b^2 x 1/c^2= (1/a + 1/b +1/c)^2 . Cho a+b+c=0
Ta có: $a + b + c = 0$ $\Rightarrow (a + b + c)^2 = 0$ $\Rightarrow a^2 + b^2 + c^2 + 2(ab + bc + ca) = 0$ $\Rightarrow a^2 +b^2 +c^2 = -2(ab + bc + ca)$ Ta được: $\dfrac{1}{a^2}\cdot \dfrac{1}{b^2}\cdot \dfrac{1}{c^2} = \left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right)^2$ $\Leftrightarrow \dfrac{1}{a^2b^2c^2} = \dfrac{(ab + bc + ca)^2}{a^2b^2c^2}$ $\Leftrightarrow (ab + bc + ca)^2 = 1$ $\Leftrightarrow \left[\begin{array}{l}ab + bc + ca = 1\\ab+ bc+ ca = -1\end{array}\right.$ $\Leftrightarrow \left[\begin{array}{l}a^2 + b^2 + c^2 = -2\quad \text{(vô lí)}\\a^2 +b^2 +c^2 = 2\end{array}\right.$ Vậy $\dfrac{1}{a^2}\cdot \dfrac{1}{b^2}\cdot \dfrac{1}{c^2} = \left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right)^2 = a^2 +b^2 +c^2$ với $a +b + c = 0$ Bình luận
Ta có: $a + b + c = 0$
$\Rightarrow (a + b + c)^2 = 0$
$\Rightarrow a^2 + b^2 + c^2 + 2(ab + bc + ca) = 0$
$\Rightarrow a^2 +b^2 +c^2 = -2(ab + bc + ca)$
Ta được:
$\dfrac{1}{a^2}\cdot \dfrac{1}{b^2}\cdot \dfrac{1}{c^2} = \left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right)^2$
$\Leftrightarrow \dfrac{1}{a^2b^2c^2} = \dfrac{(ab + bc + ca)^2}{a^2b^2c^2}$
$\Leftrightarrow (ab + bc + ca)^2 = 1$
$\Leftrightarrow \left[\begin{array}{l}ab + bc + ca = 1\\ab+ bc+ ca = -1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}a^2 + b^2 + c^2 = -2\quad \text{(vô lí)}\\a^2 +b^2 +c^2 = 2\end{array}\right.$
Vậy $\dfrac{1}{a^2}\cdot \dfrac{1}{b^2}\cdot \dfrac{1}{c^2} = \left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right)^2 = a^2 +b^2 +c^2$ với $a +b + c = 0$