Ta có: \(\frac{x}{x^2+x+1}=\frac{1}{4}\)

<=> 4x=x²+x+1  <=> x²-3x+1=0

\(P=\frac{x^5-4x^3-17x+9}{x^4+3x^2+2x+11}\)

\(P=\frac{\left(x^5-3x^4+x^3\right)+\left(3x^4-9x^3+3x^2\right)+\left(4x^3-12x^2+4x\right)+\left(9x^2-27x+9\right)+14x}{\left(x^4-3x^3+x^2\right)+\left(3x^3-9x^2+3x\right)+\left(11x^2-33x+11\right)+32x}\)

\(P=\frac{x^3\left(x^2-3x+1\right)+3x^2\left(x^2-3x+1\right)+4x\left(x^2-3x+1\right)+9\left(x^2-3x+1\right)+14x}{x^2\left(x^2-3x+1\right)+3x\left(x^2-3x+1\right)+11\left(x^2-3x+1\right)+32x}\)

\(P=\frac{\left(x^2-3x+1\right)\left(x^3+3x^2+4x+9\right)+14x}{\left(x^2-3x+1\right)\left(x^2+3x+11\right)+32x}\)

Mà x²-3x+1=0  => \(P=\frac{0+14x}{0+32x}=\frac{14x}{32x}=\frac{14}{32}=\frac{7}{16}\)

Đáp số: \(P=\frac{7}{16}\)

ĐK:\(x\in R\)

\(\frac{5}{x^2-2x+2}-\frac{8}{x^2-2x+5}=3\)

\(\Leftrightarrow-\frac{3x^4-12x^3+36x^2-48x+21}{\left(x^2-2x+2\right)\left(x^2-2x+5\right)}=0\)

\(\Leftrightarrow\frac{-3\left(x-1\right)^2\left(x^2-2x+7\right)}{\left(x^2-2x+2\right)\left(x^2-2x+5\right)}=0\)

\(\Rightarrow-3\left(x-1\right)^2\left(x^2-2x+7\right)=0\)

(vì \(\hept{\begin{cases}x^2-2x+2\\x^2-2x+5\end{cases}}>0\forall x\))

\(\Leftrightarrow\left(x-1\right)^2=0\) \(\left(x^2-2x+7>0\forall x\right)\)

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Cảm ơn đã theo dõi, 1 h= ủng hộ, kb để gắn kết tình bạn ^^

Cảm ơn nhiều nha.

Đặt \(A=\frac{ab}{\left(b-c\right)\left(c-a\right)}+\frac{bc}{\left(c-a\right)\left(a-b\right)}+\frac{ca}{\left(b-c\right)\left(a-b\right)}=-1\)

\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)^2\ge0\)

\(\Leftrightarrow\left(\frac{a}{b-c}\right)^2+\left(\frac{b}{c-a}\right)^2+\left(\frac{c}{a-b}\right)^2+2A\ge0\)

\(\Leftrightarrow\left(\frac{a}{b-c}\right)^2+\left(\frac{b}{c-a}\right)^2+\left(\frac{c}{a-b}\right)^2\ge2\)

\(\ge2\)