Chứng minh rằng: Nếu $m$ $\neq$ $0$; $n$ $\neq$; $m$ $\neq$ $-n$ thì
$\sqrt{\frac{1}{m^{2}} + \frac{1}{n^{2}}+ \frac{1}{(m+n)^{2} } }$ $=$$|$$\frac{1}{m}$ $+$$\frac{1}{n}$ $-$ $\frac{1}{m+n}$ $|$
Chứng minh rằng: Nếu $m$ $\neq$ $0$; $n$ $\neq$; $m$ $\neq$ $-n$ thì
$\sqrt{\frac{1}{m^{2}} + \frac{1}{n^{2}}+ \frac{1}{(m+n)^{2} } }$ $=$$|$$\frac{1}{m}$ $+$$\frac{1}{n}$ $-$ $\frac{1}{m+n}$ $|$
Đáp án:
Giải thích các bước giải:
Xét `(|1/m+1/n-1/(m+n)|)^2(m ne0;n ne0;m ne -n)`
`=1/m^2+1/n^2+1/(m+n)^2+2/(mn)-2/(m(m+n))-2/(n(m+n))`
`=1/m^2+1/n^2+1/(m+n)^2+2(1/(mn)-1/(m(m+n))-1/(n(m+n)))`
`=1/m^2+1/n^2+1/(m+n)^2+2((m+n-n-m)/(mn(m+n)))`
`=1/m^2+1/n^2+1/(m+n)^2+2(0/(mn(m+n)))`
`=1/m^2+1/n^2+1/(m+n)^2(dpcm)`
Đáp án:
Giải thích các bước giải:
$\quad \sqrt{\dfrac{1}{m^2} + \dfrac{1}{n^2} + \dfrac{1}{(m+n)^2}}\qquad (m\ne 0;\ n\ne 0;\ m\ne -n)$
$= \sqrt{\dfrac{1}{m^2} + \dfrac{1}{n^2} + \dfrac{1}{(m+n)^2} + 2\cdot \dfrac{m + n – m – n}{mn(m+n)}}$
$= \sqrt{\dfrac{1}{m^2} + \dfrac{1}{n^2} + \dfrac{1}{(m+n)^2} + 2\cdot \left[\dfrac{m +n}{mn(m+n)} – \dfrac{m}{mn(m+n)} – \dfrac{n}{mn(m+n)}\right]}$
$= \sqrt{\dfrac{1}{m^2} + \dfrac{1}{n^2} + \dfrac{1}{(m+n)^2} + 2\cdot \left[\dfrac{1}{mn} – \dfrac{1}{n(m+n)} – \dfrac{1}{m(m+n)}\right]}$
$= \sqrt{\dfrac{1}{m^2} + \dfrac{1}{n^2} + \dfrac{1}{(m+n)^2} +\dfrac{2}{mn} – \dfrac{2}{n(m+n)} – \dfrac{2}{m(m+n)}}$
$= \sqrt{\left(\dfrac1m + \dfrac1n – \dfrac{1}{m+n}\right)^2}$
$=\left|\dfrac1m + \dfrac1n – \dfrac{1}{m+n}\right|$