\(\left(m+n+q\right)^2=m^2+n^2+q^2\)

<=>\(m^2+n^2+q^2+2\left(mn+nq+qm\right)=m^2+n^2+q^2\)

<=>\(mn+nq+qm=0\)

<=>\(\frac{mn+nq+qm}{mnq}=0\)

<=>\(\frac{mn}{mnq}+\frac{nq}{mnq}+\frac{qm}{mnq}=0\)

<=>\(\frac{1}{q}+\frac{1}{m}+\frac{1}{n}=0\)

<=>\(\frac{1}{m}+\frac{1}{n}=-\frac{1}{q}\)

<=>\(\left(\frac{1}{m}+\frac{1}{n}\right)^3=\left(-\frac{1}{q}\right)^3\)

<=>\(\frac{1}{m^3}+\frac{3}{mn}\left(\frac{1}{m}+\frac{1}{n}\right)+\frac{1}{n^3}=-\frac{1}{q^3}\)

<=>\(\frac{1}{m^3}+\frac{1}{n^3}+\frac{1}{q^3}=-\frac{3}{mn}\cdot\left(-\frac{1}{q}\right)=\frac{3}{mnq}\) (đpcm)

\(\Leftrightarrow\frac{x^2+x+1+2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Leftrightarrow x^2+x+1+2x^2-2x=3x^2\)

\(\Leftrightarrow x^2+x+1+2x^2-2x-3x^2=-1\)

\(\Leftrightarrow-x=-1\)

\(\Leftrightarrow x=-1\)

Vậy .................

ĐKXĐ : \(x\ne1\)

Ta có : \(\frac{1}{x-1}+\frac{2x}{x^2+x+1}=\frac{3x^2}{x^3-1}\)

\(\Leftrightarrow\frac{x^2+x+1+2x\left(x-1\right)}{x^3-1}=\frac{3x^2}{x^3-1}\)

=>\(x^2+x+1+2x^2-2x=3x^2\)

\(\Rightarrow3x^2+1-x=3x^2\)

\(\Rightarrow1-x=0\)

<=> x=1 ( không thỏa mãn Đkxđ)

Vậy x=\(\varnothing\)

Ai giúp mk được mk cho 2 tk luôn!

.......... điểm danh nè!!!!!!!!!!!

me chưa ngủ

a) Ta có: \(A=1999.2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1< 2000^2\)

Vậy A < 2000²

c) \(E=26^2-24^2=\left(26-24\right)\left(26+24\right)=2.50\)

    \(F=27^2-25^2=\left(27-25\right)\left(27+25\right)=2.52\)

Vì 50 < 52 => 2.50 < 2.52

=> E < F

\(263^2+74.263+37^2=263^2+2.37.263+37^2=\left(263+37\right)^2=300^2=90000\)

Áp dụng hằng đẳng thức thức 1: (a+b)²=a²+2ab+b²

Cảm ơn bạn rất nhiều!