\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\) (ĐK: \(x\ne0\))

\(\Leftrightarrow\frac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\frac{3}{x\left(x^4+x^2+1\right)}\)

\(\Leftrightarrow\frac{x^3+1-\left(x^3-1\right)}{x^4+x^2+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)

\(\Rightarrow2x=3\Leftrightarrow x=\frac{3}{2}\left(tm\right)\).

Đặt \(x-7=t\).

Phương trình ban đầu tương đương với: 

\(\left(t+1\right)^4+\left(t-1\right)^4=16\)

\(\Leftrightarrow t^4+4t^3+6t^2+4t+1+t^4-4t^3+6t^2-4t+1=16\)

\(\Leftrightarrow t^4+6t^2-7=0\)

\(\Leftrightarrow\left(t^2+7\right)\left(t^2-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=1\Rightarrow x=8\\t=-1\Rightarrow x=6\end{cases}}\)

\(\frac{a^2+b^2}{a-b}\ge2\sqrt{2}\)

\(\Leftrightarrow\frac{a^2+\left(\frac{1}{a}\right)^2}{a-\frac{1}{a}}\ge2\sqrt{2}\)

\(\Leftrightarrow\frac{a^4+1}{a^3-a}\ge2\sqrt{2}\)

\(\Leftrightarrow a^4-2\sqrt{2}a^3+2\sqrt{2}a+1\ge0\)

\(\Leftrightarrow\left(a^2-\sqrt{2}a-1\right)^2\ge0\)( đúng )

Dấu = xảy ra khi:

\(a^2-\sqrt{2}a-1^2=0\)

\(\Rightarrow a=\sqrt{2+\sqrt{3}}\)

\(\Rightarrow b=\sqrt{2-\sqrt{3}}\)

\(x^2+x-2=9y\)

Với \(y\ne0\)suy ra \(x^2+x-2⋮9\).

- \(x=3k\left(k\inℤ\right)\Rightarrow x^2+x-2=9k^2+3k-2⋮̸3\).

- \(x=3k+1\left(k\inℤ\right)\Rightarrow x^2+x-2=9k^2+9k=9y\).

\(\Leftrightarrow y=k^2+k\).

Suy ra nghiệm \(\left(x,y\right)=\left(3k+1,k^2+k\right),\left(k\inℤ\right)\).

- \(x=3k+2\left(k\inℤ\right)\Rightarrow x^2+x-2=9k^2+15k+4⋮̸3\).

Với \(y=0\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\).

Vậy phương trình có nghiệm \(\left(x,y\right)=\left(3k+1,k^2+k\right),\left(k\inℤ\right)\).

\(x^2+5y^2-4xy-5y+4=0\)

\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(y^2-4y+4\right)-y=0\)

\(\Leftrightarrow\left(x-2y\right)^2+\left(y-2\right)^2-y=0\)

.....Làm nốt

\(a^3+2b^3+c^3\ge b^2\left(a+c\right)+b\left(a^2+c^2\right)\)

\(\Leftrightarrow a^3+2b^3+c^3-b^2\left(a+c\right)-b\left(a^2+c^2\right)\ge0\)

\(\Leftrightarrow\left(a^3+b^3-b^2a-ab^2\right)+\left(c^3+b^3-b^2c-bc^2\right)\ge0\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2+\left(b+c\right)\left(b-c\right)^2\ge0\)( đúng )
Vậy ta có ĐPCM

\(x.\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)

\(\Leftrightarrow\left(x^2+x-1+1\right)\left(x^2+x-1-1\right)=24\)

\(\Leftrightarrow\left(x^2+x-1\right)^2-1-24=0\)

\(\Leftrightarrow\left(x^2+x-1\right)^2=25\)

\(\Leftrightarrow x^2+x-1=\pm5\)

\(TH1:x^2+x-1=5\) 

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)

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

\(TH2:x^2+x-1=-5\)

\(\Leftrightarrow x^2+x+4=0\)

Vì  \(x^2+x+4=x^2+2\cdot\frac{1}{2}\cdot x+\frac{1}{4}+\frac{15}{4}=\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0\)

Mà \(x^2+x+4=0\)

=> pt vô nghiệm

Vậy \(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

x( x - 1 )( x + 1 )( x + 2 ) = 24

<=> [ x( x + 1 ) ][ ( x - 1 )( x + 2 ) ] = 24

<=> ( x² + x )( x² + x - 2 ) - 24 = 0

Đặt t = x² + x

pt <=> t( t - 2 ) - 24 = 0

<=> t² - 2t - 24 = 0

<=> t² - 6t + 4t - 24 = 0

<=> t( t - 6 ) + 4( t - 6 ) = 0

<=> ( t - 6 )( t + 4 ) = 0

<=> ( x² + x - 6 )( x² + x + 4 ) = 0

<=> ( x² - 2x + 3x - 6 )( x² + x + 4 ) = 0

<=> [ x( x - 2 ) + 3( x - 2 ) ]( x² + x + 4 ) = 0

<=> ( x - 2 )( x + 3 )( x² + x + 4 ) = 0

Vì x² + x + 4 = ( x + 1/2 )² + 15/4 ≥ 15/4 > 0 ∀ x

=> x - 2 = 0 hoặc x + 3 = 0

<=> x = 2 hoặc x = -3

Vậy ... 

Áp dụng định lý Bezout ta được:

$f\left(x\right)$chia cho x+1 dư 2 $\Rightarrow f\left(-1\right)=2$

Vì bậc của đa thức chia là 3 nên $f\left(x\right)=\left(x+1\right)\left(x^2+1\right)q\left(x\right)+a{x}^2+bx+c$

$=\left(x^2+1\right)\left(x+1\right)q\left(x\right)+\left(a{x}^2+a\right)-a+bx+c$

$=\left(x^2+1\right)\left(x+1\right)q\left(x\right)+a\left(x^2+1\right)+bx+c-a$

$=\left(x^2+1\right)\left[\left(x+1\right)q\left(x\right)+a\right]+bx+c-a$

Vì $f\left(-1\right)=4$nên $a-b+c=4\left(1\right)$

Vì f(x) chia cho $x^2+1$dư 2x+3 nên

$\text{hept}\{\begin{array}{c}b=2\\ c-a=3\end{array}\left(2\right)$

Từ (1) và (2) $\Rightarrow \text{hept}\{\begin{array}{c}a+c=6\\ b=2\\ c-a=3\end{array}\iff \text{hept}\{\begin{array}{c}a=\frac{3}{2}\\ b=2\\ c=\frac{9}{2}\end{array}$

Vậy dư f(x) chia cho $\left(x+1\right)\left(x^2+1\right)$là $\frac{3}{2}{x}^2+2x+\frac{1}{2}$