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

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

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

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

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

\(3x^2=3\)

\(\Rightarrow x^2=1\Rightarrow x=\left\{1;-1\right\}\)

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

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

\(\Leftrightarrow\)2x^2-2x=4x-x^2+3

\(\Leftrightarrow\)3x^2-6x=3

\(\Leftrightarrow\)3x(x-2)=3

\(\orbr{\begin{cases}3x=3\\x-2=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}\)

Vậy pt có 2 nghiệm x=1 và x=5

\(\frac{x^2+5}{25-x^2}=\frac{3}{x+5}+\frac{x}{x-5}\)

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

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

\(\Rightarrow x^2+5=3\left(5-x\right)-x\left(5+x\right)\)

\(\Leftrightarrow x^2+5=15-3x-5x-x^2\)

\(\Leftrightarrow15-3x-5x-x^2-x^2-5=0\)

\(\Leftrightarrow10-8x-2x^2=0\)

\(\Leftrightarrow2x^2+8x-10=0\)

\(\Leftrightarrow2\left(x^2+4x-5\right)=0\)

\(\Leftrightarrow2\left(x^2+5x-x-5\right)=0\)

\(\Leftrightarrow x^2-x+5x-5=0\)

\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}}\)

PTĐTTNT???

\(x^4+x^2y^2+y^4\)

\(=\text{ }\left[\left(x^2\right)^2+2.x^2.y^2+\left(y^2\right)^2\right]-x^2y^2\)

\(=\left(x^2+y^2\right)^2-\left(xy\right)^2\)

\(=\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)