Ta có \(x^2-x-1=0\Rightarrow x^2-x=1\Rightarrow\left(x^2-x\right)^3=1\)
\(\Rightarrow x^6-3x^5+3x^4-x^3=1\)
Mặt khác \(x^2-x-1-0\Rightarrow x^2=x+1\)
\(\Rightarrow x^6=\left(x+1\right)^3=x^3+2=3x^2+3x+1\)
\(\Rightarrow P=\frac{1+2017}{1+2017}=1\)
Ta có \(x^2-x-1=0\Rightarrow x^2-x=1\Rightarrow\left(x^2-x\right)^3=1\)
\(\Rightarrow x^6-3x^5+3x^4-x^3=1\)
Mặt khác \(x^2-x-1-0\Rightarrow x^2=x+1\)
\(\Rightarrow x^6=\left(x+1\right)^3=x^3+2=3x^2+3x+1\)
\(\Rightarrow P=\frac{1+2017}{1+2017}=1\)
cho x2-x-1=0 tính Q=\(\frac{x^6-3^5+3x^4-x^3+2013}{x^6-x^3-3x^2-3x+2013}\)
Cho \(x^2-x-1=0\).Tính \(P=\frac{x^6-3x^5+3x^4-x^3+2015}{x^6-x^3-3x^2-3x+2015}\)
Bài 1: Cho \(f\left(x\right)=\frac{x^3}{1-3x+3x^2}\)
Tính GTBT: \(f\left(\frac{1}{2017}\right)+f\left(\frac{2}{2017}\right)+...+f\left(\frac{2016}{2017}\right).\)
Bài 2: Giải HPT sau: \(\hept{\begin{cases}2x^2-y^2+xy+3y=2\\x^2-y^2=3\end{cases}}\)
Bài 3: Tìm m để PT: \(x^4+x^3+\left(m-2\right)x^2-4mx-2m^2=0\)có 4 nghiệm thỏa \(x_1^2+x_2^2+x_3^2+x_4^2=5\)
Cho x=\(\frac{3-\sqrt{5}}{2}\). Tính P=\(x^{2018}-3x^{2017}+5x^2-15x+2017\)
Giải phương trình, x>0
\(\frac{\left(x^3+3x^2\sqrt{x^3-3x+6}\right)\left(3x-x^3-2\right)}{2+\sqrt{x^3-3x+6}}=4\left[2\sqrt{\left(x^3-3x+6\right)^3}-\left(x^3-3x+6\right)^2\right]\)
Giải phương trình, x>0
\(\frac{\left(x^3+3x^2\sqrt{x^3-3x+6}\right)\left(3x-x^3-2\right)}{2+\sqrt{x^3-3x+6}}=4\left[2\sqrt{\left(x^3-3x+6\right)^3}-\left(x^3-3x+6\right)^2\right]\)
a,\(\frac{3}{x}+\frac{1}{x+3}+\frac{3}{x+6}+\frac{1}{x+7}=\frac{1}{1-x}\)
b, \(\frac{1}{x-5}+\frac{1}{x-2}+\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+3}=\frac{3x-3}{4}\)
c,\(\frac{1}{x-3}+\frac{1}{3x+1}+\frac{10x-13}{4x-6}=\frac{1}{x+1}+\frac{1}{2x-1}+\frac{1}{3x+7}\)
d,\(\frac{x^2+x+1}{2x-1}\left(\frac{3x^2-x+5}{4x-2}-3\right)=8\)
e,\(\frac{2x^2-3}{3x-1}\left(2x-\frac{7+4x}{3x-1}\right)=2\)
f,\(\frac{x\left(3x-1\right)\left(3x^2+1\right)\left(6x^2-3x-1\right)}{\left(x+1\right)^3}=\frac{1}{2}\)
g, \(x\left(x^2+2\right)\left(x^2+2x+8+\frac{12}{x-2}\right)=3\left(x-2\right)\)
Giải các PT
a,:\(x^4-2x^3-x^2-2x+1=0\)
b,\(\frac{1}{x^2-3x+3}+\frac{2}{x^2-3x+4}=\frac{6}{x^2-3x+5}\)
c,\(x^2+\sqrt{x+72}=72\)
Không cần các bạn giải quyết hết, nhé! GẤP, GẤP, GẤP
Giải các phương trình sau:
a) x4 + 2x3 - 39x2 - 4x + 4 = 0
b) (x + 4)4 + (x + 2)2 = 34
c) \(\frac{2x}{3x^2-5x+2}+\frac{13x}{3x^2+x+2}=6\)
d) \(\frac{x^4+3x^2+1}{x^3+x^2-x}=3\)
e) \(\frac{1}{x^2}+\frac{1}{\left(x+1\right)^2}=15\)
f) \(\left(\frac{x+1}{x-2}\right)^2+\frac{x-1}{x+3}=12\left(\frac{x-2}{x+3}\right)^2\)
g) \(\frac{2\left(x+1\right)}{3x^2+x}+\frac{13\left(x+1\right)}{3x^2+7x+6}=6\)