Cho a,b,c khác 0 và 1/a+ 1/b+1/c = 0 CMR $\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}$ 11/08/2021 Bởi Valerie Cho a,b,c khác 0 và 1/a+ 1/b+1/c = 0 CMR $\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}$
Giải thích các bước giải: Ta có: $\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}$ $\leftrightarrow \sqrt{a+b}-\sqrt{a+c}=\sqrt{b+c}$ $\leftrightarrow (\sqrt{a+b}-\sqrt{a+c})^2=(\sqrt{b+c})^2$ $\leftrightarrow a+b+a+c-2\cdot\sqrt{a+b}\cdot\sqrt{a+c}=b+c$ $\leftrightarrow 2a=2\cdot\sqrt{(a+b)(a+c)}$ $\leftrightarrow \sqrt{(a+b)(a+c)}=a$ $\leftrightarrow (a+b)(a+c)=a^2$ $\leftrightarrow a^2+ab+ac+bc=a^2$ $\leftrightarrow ab+ac+bc=0$ $\leftrightarrow \dfrac{ab+ac+bc}{abc}=0$ $\leftrightarrow \dfrac1a+\dfrac1b+\dfrac1c=0$ (đúng) $\to$ đpcm Bình luận
Xét $\sqrt[]{a+b} = \sqrt[]{a+c} + \sqrt[]{b+c}$ $\to a+b= a+c+b+c+2\sqrt[]{(a+c).(b+c)}$ $\to a+b=a+b+2c+2\sqrt[]{(a+c).(b+c)}$ $\to -c = \sqrt[]{(a+c).(b+c)}$ $\to (-c)^2 = (a+c).(b+c)$ $\to c^2 = ab+ac+bc+c^2$ $\to ab+bc+ca= 0 $ $\to \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} = 0$ ( Đúng ) $\to đpcm$ Bình luận
Giải thích các bước giải:
Ta có:
$\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}$
$\leftrightarrow \sqrt{a+b}-\sqrt{a+c}=\sqrt{b+c}$
$\leftrightarrow (\sqrt{a+b}-\sqrt{a+c})^2=(\sqrt{b+c})^2$
$\leftrightarrow a+b+a+c-2\cdot\sqrt{a+b}\cdot\sqrt{a+c}=b+c$
$\leftrightarrow 2a=2\cdot\sqrt{(a+b)(a+c)}$
$\leftrightarrow \sqrt{(a+b)(a+c)}=a$
$\leftrightarrow (a+b)(a+c)=a^2$
$\leftrightarrow a^2+ab+ac+bc=a^2$
$\leftrightarrow ab+ac+bc=0$
$\leftrightarrow \dfrac{ab+ac+bc}{abc}=0$
$\leftrightarrow \dfrac1a+\dfrac1b+\dfrac1c=0$ (đúng)
$\to$ đpcm
Xét $\sqrt[]{a+b} = \sqrt[]{a+c} + \sqrt[]{b+c}$
$\to a+b= a+c+b+c+2\sqrt[]{(a+c).(b+c)}$
$\to a+b=a+b+2c+2\sqrt[]{(a+c).(b+c)}$
$\to -c = \sqrt[]{(a+c).(b+c)}$
$\to (-c)^2 = (a+c).(b+c)$
$\to c^2 = ab+ac+bc+c^2$
$\to ab+bc+ca= 0 $
$\to \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} = 0$ ( Đúng )
$\to đpcm$