Câu hỏi của sjfdksfdkjlsjlfkdjdkfsl

Question

Chứng minh rằng m=n=p biết m²+n²+p²-mp-np-mn=0

☆MĭηɦღAηɦ❄ · Accepted Answer

Ta có : $\left(a-b\right)^2=\left(a-b\right)\left(a-b\right)=a^2-ab-ab+b^2=a^2-2ab+b^2$

$\Rightarrow m^2+n^2+p^2-mp-np-mn=0$

$\Leftrightarrow2m^2+2n^2+2p^2-2mp-2np-2mn=0$

$\Leftrightarrow\left(m^2-2mp+p^2\right)+\left(n^2-2mn+m^2\right)+\left(p^2-2np+n^2\right)=0$

$\Leftrightarrow\left(m-p\right)^2+\left(n-m\right)^2+\left(p-n\right)^2=0$

$\Leftrightarrow\hept{\begin{cases}m-p=0\n-m=0\p-n=0\end{cases}\Leftrightarrow}m=n=p\left(ĐPCM\right)$

Câu trả lời:

Ta có : $\left(a-b\right)^2=\left(a-b\right)\left(a-b\right)=a^2-ab-ab+b^2=a^2-2ab+b^2$

$\Rightarrow m^2+n^2+p^2-mp-np-mn=0$

$\Leftrightarrow2m^2+2n^2+2p^2-2mp-2np-2mn=0$

$\Leftrightarrow\left(m^2-2mp+p^2\right)+\left(n^2-2mn+m^2\right)+\left(p^2-2np+n^2\right)=0$

$\Leftrightarrow\left(m-p\right)^2+\left(n-m\right)^2+\left(p-n\right)^2=0$

$\Leftrightarrow\hept{\begin{cases}m-p=0\n-m=0\p-n=0\end{cases}\Leftrightarrow}m=n=p\left(ĐPCM\right)$

Nguyen Minh Hieu · Answer

Ta có: m^2 + n^2 + p^2 - mp - np - mn = 0 => m^2 + n^2 + p^2 = mp + np + mn

=> mp = m^2 => m = p;

=> mn = n^2 => m = n;

=> np = p^2 => n = p.

Vậy m = n = p.