Với mọi a,b ta có : ( a - b )² \(\ge\)0 

=> a² + b² \(\ge\)2ab => 2 . ( a² + b² ) \(\ge\)( a + b )² = 1

=> a² + b² \(\ge\)\(\frac{1}{2}\)

Dấu " = " xảy ra <=> a = b = \(\frac{1}{2}\)

\(a,\Leftrightarrow x^2+y^2+z^2-2xy+2xz-2yz\ge0\)

\(\Leftrightarrow\left(x-y+z\right)^2\ge0\left(đúng\right)\)

\(b,\Leftrightarrow4a^2+b^2-4ab\ge0\)

\(\Leftrightarrow\left(2a\right)^2+b^2-4ab\ge0\)

\(\Leftrightarrow\left(2a-b\right)^2\ge0\left(đúng\right)\)

P/s: chưa chắc lắm :(

PTĐTTNT?

1.Đặt \(a^2+a=t\)

\(\Rightarrow\left(a^2+a\right)\left(a^2+a+1\right)-2\)

\(=t\left(t+1\right)-2\)

\(=t^2+t-2\)

\(=t^2+2t-\left(t+2\right)\)

\(=t\left(t+2\right)-\left(t+2\right)\)

\(=\left(t+2\right)\left(t-1\right)\)

Sửa đề: 

\(x^4+2011x^2+2010x+2011\)

\(=\left(x^4-x\right)+2011x^2+2011x+2011\)

\(=x\left(x^3-1\right)+2011\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2011\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2011\right)\)

3. \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)

Đặt \(x^2+5x+4=t\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120\)

\(=t\left(t+2\right)-120\)

\(=t^2+2t+1-121\)

\(=\left(t+1\right)^2-11^2\)

\(=\left(t+1-11\right)\left(t+1+11\right)\)

\(=\left(t-10\right)\left(t+12\right)\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+16\right)\)

\(=\left[\left(x^2-x\right)+\left(6x-6\right)\right]\left(x^2+5x+16\right)\)

\(=\left[x.\left(x-1\right)+6\left(x-1\right)\right]\left(x^2+5x+16\right)\)

\(=\left(x-1\right)\left(x+6\right)\left(x^2+5x+16\right)\)

4. \(\left(x^2+x+4\right)^2+8x\left(x^2+x+1\right)+15x^2\)

\(=\left(x^2+x+4\right)^2+2.\left(x^2+x+1\right).4x+\left(4x\right)^2-x^2\)

\(=\left(x^2+x+4+4x\right)^2-x^2\)

\(=\left(x^2+4+5x-x\right)\left(x^2+5x+x+4\right)\)

\(=\left(x^2+4x+4\right)\left(x^2+6x+4\right)\)

\(=\left(x+2\right)^2\left[\left(x^2+2.x.3+3^2\right)-\left(\sqrt{5}\right)^2\right]\)

\(=\left(x+2\right)^2\left[\left(x+3\right)^2-\left(\sqrt{5}\right)^2\right]\)

\(=\left(x+2\right)^2\left(x+3-\sqrt{5}\right)\left(x+3+\sqrt{5}\right)\)

Ta có: \(\left(x-\frac{1}{x}\right):\left(x+\frac{1}{x}\right)=n\Rightarrow\frac{x^2-1}{x}:\frac{x^2+1}{x}=n\Rightarrow\frac{x^2-1}{x^2+1}=n\)

\(\Rightarrow x^2-1=n\left(x^2+1\right)=nx^2+n\Rightarrow x^2-nx^2=n+1\Rightarrow x^2\left(1-n\right)=n+1\Rightarrow x^2=\frac{n+1}{1-n}\left(n\ne1\right)\)

THay vào V ta được: \(V=\left(\frac{n+1}{1-n}-\frac{1}{\frac{n+1}{1-n}}\right):\left(\frac{n+1}{1-n}+\frac{1}{\frac{n+1}{1-n}}\right)=\left(\frac{n+1}{1-n}-\frac{1-n}{n+1}\right):\left(\frac{n+1}{1-n}+\frac{1-n}{n+1}\right)\)

\(=\frac{\left(n+1\right)^2-\left(1-n\right)^2}{\left(n+1\right)\left(1-n\right)}:\frac{\left(n+1\right)^2+\left(1-n\right)^2}{\left(n+1\right)\left(1-n\right)}=\frac{n^2+2n+1-1+2n-n^2}{n^2+2n+1+1-2n+n^2}\)

\(=\frac{4n}{2n^2+2}=\frac{4n}{2\left(n^2+1\right)}=\frac{2n}{n^2+1}\)

Đặt A=\(n^6-1=\left(n^3-1\right)\left(n^3+1\right)=\left(n-1\right)\left(n+1\right)\left(n^2-n+1\right)\left(n^2+n+1\right)\)

Vì \(n⋮3\Rightarrow̸n=3k\pm1\)

Với n=3k+1 thì A=(3k+1-1)(3k+1+1)[(3k+1)^2-3k-1+1].[(3k+1)^2+3k+1+1]

\(=3k\left(3k+2\right)\left(9k^2+6k+1-3k-1+1\right)\left(9k^2+6k+1+3k+1+1\right)\)

\(=3k\left(3k+2\right)\left(9k^2+3k+1\right)\left(9k^2+9k+3\right)\)

\(=9k\left(3k+2\right)\left(9k^2+3k+1\right)\left(3k^2+3k+1\right)⋮9\)

Với n=3k-1 thì A=(3k-1-1)(3k-1+1)[(3k-1)^2-3k+1+1].[(3k-1)^2+3k-1+1]

\(=3k\left(3k-2\right)\left(9k^2-6k+1-3k+1+1\right)\left(9k^2-6k+1+3k-1+1\right)\)

\(=3k\left(3k-2\right)\left(9k^2-9k+3\right)\left(9k^2-3k+1\right)\)

\(=9k\left(3k-2\right)\left(3k^2-3k+1\right)\left(9k^2-3k+1\right)⋮9\)

Từ 2 trường hợp trên => đpcm

tau méch cô hoài nhá

a) Xét tam giác ABD có :

 M là trung điểm của AB

 F là trung điểm của BD

=) MF là đường trung bình của tam giác ABD

=) MF//AD và MF=\(\frac{1}{2}\)AD    (1)

Xét tam giác tam giác ACD có :

 N là trung điểm CD

 E là trung điểm AC

=) NE là đường trung bình của tam giác ACD

=) NE//AD và NE=\(\frac{1}{2}\)AD     (2)

Từ (1) và (2) =) Tứ giác MENF là hình bình hành

a, ĐKXĐ: \(x\ne0;x\ne\pm1\)

\(P=\left(\frac{2x}{x^2-1}+\frac{x-1}{2x+2}\right):\frac{x+1}{2x}=\left(\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{2\left(x+1\right)}\right):\frac{x+1}{2x}\)

\(=\left(\frac{2x.2}{2\left(x-1\right)\left(x+1\right)}+\frac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\right):\frac{x+1}{2x}\)

\(=\frac{4x+x^2-2x+1}{2\left(x-1\right)\left(x+1\right)}:\frac{x+1}{2x}=\frac{x^2+2x+1}{2\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}\cdot\frac{2x}{x+1}=\frac{x}{x-1}\)

b,Để \(P=2\Leftrightarrow\frac{x}{x-1}=2\Leftrightarrow2\left(x-1\right)=x\Leftrightarrow2x-2-x=0\Leftrightarrow x-2=0\Leftrightarrow x=2\left(tmđk\right)\)

Vậy để P=2 <=> x=2

Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}=3\) (abc=1) (tự c/m)

Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)

=>ab+bc=-ca => (ab+bc)³=-c³a³

=>a³b³+b³c³+3a²b².bc+3ab.b²c²=-c³a³

=>a³b³+b³c³+3ab²c(ab+bc)=-c³a³

=>a³b³+b³c³+3ab²c.(-ca)=-c³a³

=>a³b³+b³c³-3a²b²c²=-c³a³

=>a³b³+b³c³+c³a³=3a²b²c² = 3 (do abc=1)

Vậy F=3.3=9