p bt cách giải bài này k

Ta có:

Vì n là tổng của 2 số chính phương

=> đặt n = a² + b²

=> 2n = (a² + b²) + (a² + b²)

=> 2n = (a² + a²) + (b² + b²)

=> 2n = 2a² + 2b² là tổng của 2 số chính phương (ĐPCM)
Vậy...

\(\left(3x-7\right)^2-4\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x-7\right)^2-\left(2x+2\right)^2=0\\ \Leftrightarrow\left(5x-5\right)\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)

\(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\\ \Rightarrow2x^3+3x^2-8x-12=x^3-x^2-4x+4\\ \Rightarrow x^3+4x^2-4x-16=0\\ \Rightarrow\left(x^3-4x\right)+\left(4x^2-16\right)=0\\ \Leftrightarrow x\left(x^2-4\right)+4\left(x^2-4\right)=0\\ \Leftrightarrow\left(x^2-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\pm2\\x=-4\end{matrix}\right.\)

{1,2,3,4,6,8,12,24}
 

\(x^4+2010x^2+2009x+2010\\ =\left(x^4-x\right)+\left(2010x^2+2010x+2010\right)\\ =x\left(x^3-1\right)+2010\left(x^2+x+1\right)\\ =x\left(x-1\right)\left(x^2+x+1\right)+2010\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^2-x+2010\right)\)

= (-53). (-240)- 241.53

= 12720 - 12773

= -53

\(x^2-y^2+2x-4y-10=0\\ \Rightarrow\left(x^2+2x+1\right)-\left(y^2+4y+4\right)-7=0\\ \Rightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)

Đề có sao k

\(\dfrac{y+3}{y-6}=\dfrac{y-4}{y+1}\\ \Rightarrow\left(y+3\right)\left(y+1\right)=\left(y-4\right)\left(y-6\right)\\ \Rightarrow y^2+4y+3=y^2-10y+24\\ \Rightarrow4y+3=-10y+24\\ \Rightarrow14y=21\\ \Rightarrow y=\dfrac{3}{2}\)