(2x²+x-2013)²+4 (x²-5x-2012)²= 4 (2x²+x-2013)(x²-5x-2012)

Dat \(\hept{\begin{cases}a=2x^2+x-2013\\b=x^2-5x-2012\end{cases}}\)ta co phuong trinh 

(2x²+x-2013)²+4 (x²-5x-2012)²= 4 (2x²+x-2013)(x²-5x-2012)

<=>\(a^2+4b^2=4ab\)

<=>\(a^2+4b^2-4ab=0\) 

<=>\(\left(a-2b\right)^2=0\)

<=>\(a=2b\)

=>\(2x^2+x-2013=2x^2-10x-4024\)

<=>\(11x=2011\)

<=>x=\(\frac{2011}{11}\)

\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right).\)

\(\Rightarrow\left(2x^2+x-2013\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)+4\left(x^2-5x-2012\right)^2=0\)

\(\Leftrightarrow\left[\left(2x^2+x-2013\right)-2\left(x^2-5x-2012\right)\right]^2=0\)(Hằng đẳng thức)

\(\Leftrightarrow2x^2+x-2013-2x^2+10x+4024=0\)

\(\Leftrightarrow11x=-2011\)

\(\Leftrightarrow x=\frac{-2011}{11}\)

Đặt 2x^2 + x +2013 = a, x^2-5x+2012 = b

Ta có: a^2 + 4b^2 = 4ab

          a^2 - 4ab + 4b^2 = 0

          (a-2b)^2 = 0

Do đó: a = 2b

Hay: 2x^2 + x -2013 = 2(x^2 -5x -2012)     

        2x^2 + x -2013 = 2x^2 -10x -4024

        x-2013 = -10x -4024

        x+10x = -4024+2013

        11x = -2011

         x = -2011/11

Bạn hỏi nhiều câu hay đấy. Chúc bạn học tốt.   

\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2\)

\(=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)

Đặt  \(\hept{\begin{cases}2x^2+x-2013=a\\x^2-5x-2012=b\end{cases}}\) thì ta có :

\(a^2+4b^2=4ab\Rightarrow a^2+b^2-4ab=0\)

\(\Rightarrow\left(a-2b\right)^2=0\Rightarrow a-2b\Rightarrow a=2b\)

Tức là :

\(2x^2+x-2013=2\left(x^2-5x-2012\right)\)

\(\Leftrightarrow2x^2+x-2013=2x^2-10x-4024\)

\(\Leftrightarrow11x+2011=0\Leftrightarrow11x=-2011\Rightarrow x=-\frac{2011}{11}\)

Chúc bạn học tốt !!!

1: ĐKXĐ: x<>0

\(\sqrt{x^2+x+2} + \frac{1}{x} = \frac{13-7x}{2}\)

\(\Leftrightarrow \sqrt{x^2+x+2} = \frac{13}{2} - \frac{7x}{2} - \frac{1}{x} = \frac{13x - 7x^2 - 2}{2x}\)

=>\(\left(\sqrt{x^2+x+2} - 2\right) + \left(\frac{1}{x} - 1\right) + \frac{7x - 7}{2} = 0\)

=>\(\frac{x^2+x-2}{\sqrt{x^2+x+2} + 2}+\frac{1-x}{x}+\frac{7(x-1)}{2}=0\)

=>\((x-1)\left[\frac{x+2}{\sqrt{x^2+x+2} + 2}-\frac{1}{x}+\frac{7}{2}\right]=0\)

=>x-1=0

=>x=1(nhận)

7: ĐKXĐ: \(16-x^2>0\)

=>\(x^2<16\)

=>-4<x<4

\(\frac{x^3}{\sqrt{16-x^2}} + x^2 - 16 = 0\)

\(\Leftrightarrow \frac{x^3}{\sqrt{16-x^2}} - \left(\sqrt{16-x^2}\right)^2 = 0\)

\(\Leftrightarrow x^3 - \left(\sqrt{16-x^2}\right)^3 = 0\)

\(\Leftrightarrow x^3 = \left(\sqrt{16-x^2}\right)^3\)

\(\Leftrightarrow x = \sqrt{16-x^2} \quad (x > 0)\)

\(\Leftrightarrow x^2 = 16 - x^2\) và x>0

=>\(2x^2=16\) và x>0

=>\(x^2=8\) và 0<x<4

=>\(x=2\sqrt2\)

Sửa tí nha kết quả cuối sai dâu phải là \(x=\dfrac{-2011}{11}\)

\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\\ \Leftrightarrow\left(2x^2+x-2013\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)+4\left(x^2-5x-2012\right)^2=0\\ \Leftrightarrow\left[2x^2+x-2013-2\left(x^2-5x-2012\right)\right]^2=0\\ \Leftrightarrow\left(11x+2011\right)^2=0\\ \Leftrightarrow11x+2011=0\\ \Leftrightarrow x=\dfrac{2011}{11}\)