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\(x^2+\left(16-x\sqrt{3}\right)^2=4\left(12-x\right)^2\)

\(\Leftrightarrow x^2+256-32\sqrt{3}x+3x^2=4\left(144-24x+x^2\right)\)

\(\Leftrightarrow4x^2-32\sqrt{3}x+256=576-96x+4x^2\)

\(\Leftrightarrow4x^2-4x^2-32\sqrt{3}x+96x+256-576=0\)

\(\Leftrightarrow\left(96-32\sqrt{3}\right)x-320=0\)

\(\Leftrightarrow\left(96-32\sqrt{3}\right)x=320\)

\(\Leftrightarrow x=\frac{320}{96-32\sqrt{3}}=\frac{15+5\sqrt{3}}{3}\)

Lời giải:

PT \(\Leftrightarrow [(x-5)(x-8)][(x-4)(x-10)]=72x^2\)

\(\Leftrightarrow (x^2-13x+40)(x^2-14x+40)=72x^2\)

Đặt \(x^2-13x+40=a\) thì pt trở thành:

\(a(a-x)=72x^2\)

\(\Leftrightarrow a^2-ax-72x^2=0\)

\(\Leftrightarrow a^2-9ax+8ax-72x^2=0\)

\(\Leftrightarrow a(a-9x)+8x(a-9x)=0\)

\(\Leftrightarrow (a-9x)(a+8x)=0\)

Nếu $a-9x=0$

\(\Leftrightarrow x^2-13x+40-9x=0\)

\(\Leftrightarrow x^2-22x+40=0\)

\(\Leftrightarrow (x-2)(x-20)=0\Rightarrow \left[\begin{matrix} x=2\\ x=20\end{matrix}\right.\)

Nếu $a+8x=0$

\(\Leftrightarrow x^2-13x+40+8x=0\)

\(\Leftrightarrow x^2-5x+40=0\Leftrightarrow (x-\frac{5}{2})^2=-\frac{135}{4}\) (vô lý)

Vậy........

ĐKXĐ:...

\(x^2+\frac{36}{x^2}-4\left(x-\frac{6}{x}\right)-17=0\)

Đặt \(x-\frac{6}{x}=a\Rightarrow a^2=x^2+\frac{36}{x^2}-12\Rightarrow x^2+\frac{36}{x^2}=a^2+12\)

\(a^2+12-4a-17=0\)

\(\Leftrightarrow a^2-4a-5=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{6}{x}=-1\\x-\frac{6}{x}=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-6=0\\x^2-5x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x+5\right)=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+5=5\\x^2+5x+5=-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+5x=0\\x^2+5x+10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+5\right)=0\\\left(x+\frac{5}{2}\right)^2=-\frac{15}{4}\left(VL\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\) ( TM )

PT ⇒ \(2\left(x^2-4x+5\right)-3\sqrt{x^2-4x+5}=22\)

Đặt \(\sqrt{x^2-4x+5}=y>0\), ta có:

\(2y^2-3y-22=0\) \(\Rightarrow y=\frac{3\pm\sqrt{185}}{4}\)

Số xấu quá, ko muốn giải nữa :D

Có vẻ phương trình có 4 nghiệm

a)\(\left(x-2\right)\left(5-x\right)=7x-\left(x-1\right)\left(3-2x\right)\Leftrightarrow5x-x^2-10+2x=7x-3x+2x^2+3-2x\Leftrightarrow-3x^2+5x-13=0\)\(\Delta=b^2-4ac=25-4.\left(-3\right).\left(-13\right)=-131< 0\)

\(\Rightarrow\)phương trình vô nghiệm

\(\frac{2x}{x+1}=\frac{x^2-x+8}{\left(x+1\right)\left(x+1\right)}\) (ĐKXĐ: x \(\ne\) 1)

\(\Leftrightarrow\) \(\frac{2x\left(x+1\right)}{\left(x+1\right)\left(x+1\right)}=\frac{x^2-x+8}{\left(x+1\right)\left(x+1\right)}\)

\(\Leftrightarrow\) 2x(x + 1) = x² - x + 8

\(\Leftrightarrow\) 2x² + 2x - x² + x - 8 = 0

\(\Leftrightarrow\) x² + 3x - 8 = 0

\(\Leftrightarrow\) (x + \(\frac{3}{2}\))² - \(\frac{41}{4}\) = 0

\(\Leftrightarrow\) (x + \(\frac{3}{2}\) - \(\frac{\sqrt{41}}{2}\))(x + \(\frac{3}{2}\) + \(\frac{\sqrt{41}}{2}\)) = 0

\(\Leftrightarrow\) (x + \(\frac{3-\sqrt{41}}{2}\))(x + \(\frac{3+\sqrt{41}}{2}\)) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3-\sqrt{41}}{2}=0\\x+\frac{3+\sqrt{41}}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{41}-3}{2}\\x=\frac{-3-\sqrt{41}}{2}\end{matrix}\right.\)

Vậy S = {\(\frac{\sqrt{41}-3}{2}\); \(\frac{-3-\sqrt{41}}{2}\)}

Chúc bn học tốt!!

\(\frac{2x}{x+1}=\frac{x^2-x+8}{\left(x+1\right)\cdot\left(x-4\right)}\\ \Leftrightarrow\frac{2x^2-8x}{\left(x+1\right)\cdot\left(x-4\right)}-\frac{x^2-x+8}{\left(x+1\right)\cdot\left(x-4\right)}=0\\ \Leftrightarrow\frac{2x^2-8x-x^2+x-8}{\left(x+1\right)\cdot\left(x-4\right)}=0\\ \Leftrightarrow\frac{x^2-7x-8}{\left(x+1\right)\cdot\left(x-4\right)}=0\\ \Leftrightarrow x^2-7x-8=0\\ \Rightarrow\left[{}\begin{matrix}x=\frac{7+\sqrt{17}}{2}\\x=\frac{7-\sqrt{17}}{2}\end{matrix}\right.\)