1, \(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=15^2-2.6^2=153\)

2, chú ý: \(n^2-\left(n+1\right)^2=-\left(2n+1\right)\)

\(M=\left(1^2-2^2\right)+\left(3^2-4^2\right)+...+\left(2015^2-2016^2\right)+2017^2\)

\(=-3-7-11-...-4031+2017^2\)

\(=-1008.4034+2017^2=2017^2-2017.2016=\)\(2017\left(2017-2016\right)=2017\)

Từ x²+y²= 15 và xy=6 ta có hệ pt

\(\hept{\begin{cases}^{x^2+y^2=15}\\x=\frac{6}{y}\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(\frac{6}{y}\right)^2+y^2=15\Leftrightarrow36+y^4-15y^2=0\left(1\right)\\x=\frac{6}{y}\end{cases}}\)

giải pt (1)\(y^4-15y^2+36=y^4-3y^2-12y^2+36=y^2\left(y^2-3\right)-12\left(y^2-3\right)\)

tiếp \(\left(y^2-3\right)\left(y^2-12\right)=0\Leftrightarrow\orbr{\begin{cases}y^2=3\Rightarrow x^2=\frac{36}{3}=12\\y^2=12\Rightarrow x^2=\frac{36}{12}=3\end{cases}}\)

Không mất tính tổng quát nên x⁴+y⁴=(x²)²+(y²)²=12²+3²=153

Ta có:

\(\frac{x}{x^2+x+1}=-\frac{1}{4}\Rightarrow x^2+x+1=-4x\)

\(\Rightarrow x^2+5x+1=0\Rightarrow x^2=5x+1\)

Với x²=5x+1 ta được:

\(P=\frac{2x\left(5x+1\right)^2+10\left(5x+1\right)^2+2x\left(5x+1\right)-7\left(5x+1\right)-35x+2009}{2029+60x+11\left(5x+1\right)-5x\left(5x+1\right)-\left(5x+1\right)^2}\)

\(P=\frac{2x\left(25x^2+10x+1\right)+10\left(25x^2+10x+1\right)+10x^2+2x-35x-7-35x+2009}{2029+60x+55x+11-25x^2-5x-\left(25x^2+10x+1\right)}\)

\(P=\frac{50x^3+20x^2+2x+250x^2+100x+10+10x^2+2x-35x-7-35x+2009}{2029+60x+55x+11-25x^2-5x-25x^2-10x-1}\)

\(P=\frac{50x^3+280x^2+34x+2012}{2039+100x-50x^2}\)

\(P=\frac{50x\left(5x+1\right)+280\left(5x+1\right)+34x+2012}{2039+100x-50\left(5x+1\right)}\)

\(P=\frac{250x^2+50x+1400x+280+34x+2012}{2039+100x-250x-50}\)

\(P=\frac{250\left(5x+1\right)+50x+1400x+280+34x+2012}{1989-150x}\)

\(P=\frac{1250x+250+50x+1400x+280+34x+2012}{1989-150x}\)

bài trên sai rồi

Bài 1:

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

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

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

\(=3x^2-3x+8\)

\(=3\left(x^2-x\right)+8=3\cdot3+8=17\)