cho a+b+c=0 và a,b,c khác 0 cmr √(1/a ²+1/b ²+1/c ²) = ║1/a+1/b+1/c . 01/09/2021 Bởi Lydia cho a+b+c=0 và a,b,c khác 0 cmr √(1/a ²+1/b ²+1/c ²) = ║1/a+1/b+1/c .
Giải thích các bước giải: Ta có: $a+b+c=0$ $\to \dfrac{a+b+c}{abc}=0$ $\to\dfrac1b\cdot\dfrac1c+\dfrac1a\cdot \dfrac1c+\dfrac1a\cdot\dfrac1b=0$ $\to 2\left(\dfrac1b\cdot\dfrac1c+\dfrac1a\cdot \dfrac1c+\dfrac1a\cdot\dfrac1b\right)=0$ $\to\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}+ 2\left(\dfrac1b\cdot\dfrac1c+\dfrac1a\cdot \dfrac1c+\dfrac1a\cdot\dfrac1b\right)=\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}$ $\to\left(\dfrac1a+\dfrac1b+\dfrac1c\right)^2=\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}$ $\to \sqrt{\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}}=\sqrt{\left(\dfrac1a+\dfrac1b+\dfrac1c\right)^2}$ $\to \sqrt{\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}}=\Bigg|\dfrac1a+\dfrac1b+\dfrac1c\Bigg|$ Bình luận
Giải thích các bước giải:
Ta có:
$a+b+c=0$
$\to \dfrac{a+b+c}{abc}=0$
$\to\dfrac1b\cdot\dfrac1c+\dfrac1a\cdot \dfrac1c+\dfrac1a\cdot\dfrac1b=0$
$\to 2\left(\dfrac1b\cdot\dfrac1c+\dfrac1a\cdot \dfrac1c+\dfrac1a\cdot\dfrac1b\right)=0$
$\to\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}+ 2\left(\dfrac1b\cdot\dfrac1c+\dfrac1a\cdot \dfrac1c+\dfrac1a\cdot\dfrac1b\right)=\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}$
$\to\left(\dfrac1a+\dfrac1b+\dfrac1c\right)^2=\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}$
$\to \sqrt{\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}}=\sqrt{\left(\dfrac1a+\dfrac1b+\dfrac1c\right)^2}$
$\to \sqrt{\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}}=\Bigg|\dfrac1a+\dfrac1b+\dfrac1c\Bigg|$