(x - 1)^3 - x(x + 1)^2 = 5x(2 - x) - 11(x + 2)

<=> -5x^2 + 2x - 1 = -5x² - x - 22

<=> 2x - 1 = -5x² - x - 22 + 5x²

<=> 2x - 1 = -x - 22

<=> 2x - 1 + x = -22

<=> 3x - 1 = -22

<=> 3x = -22 + 1

<=> 3x = -21

<=> x = -7

Vậy: phương trình có nghiệm duy nhất là: S = {-7}

a, ĐK: \(x\ge1\)

Đặt \(\sqrt{5x-1}=a;\sqrt{x-1}=b\left(a,b\ge0\right)\)

\(pt\Leftrightarrow\left(a+b\right)\left(\dfrac{a^2+b^2}{2}-ab\right)=a^2-b^2\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2=2\left(a-b\right)\left(a+b\right)\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)\left(a-b-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=b+2\end{matrix}\right.\)

TH1: \(a=b\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}\Leftrightarrow x=0\left(l\right)\)

TH2: \(a=b+2\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}+2\)

\(\Leftrightarrow5x-1=x-1+4+4\sqrt{x-1}\)

\(\Leftrightarrow4x-4-4\sqrt{x-1}=0\)

\(\Leftrightarrow4x-4-4\sqrt{x-1}+1=1\)

\(\Leftrightarrow\left(2\sqrt{x-1}-1\right)^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x-1}-1=1\\2\sqrt{x-1}-1=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-1}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

a: =>(x^2+4x-5)(x^2+4x-21)=297

=>(x^2+4x)^2-26(x^2+4x)+105-297=0

=>x^2+4x=32 hoặc x^2+4x=-6(loại)

=>x^2+4x-32=0

=>(x+8)(x-4)=0

=>x=4 hoặc x=-8

b: =>(x^2-x-3)(x^2+x-4)=0

hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{17}}{2};\dfrac{-1-\sqrt{17}}{2}\right\}\)

c: =>(x-1)(x+2)(x^2-6x-2)=0

hay \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)

ĐKXĐ: bla bla bla

\(3x\left(x-2\right)\sqrt{3x-1}=2\left(x^3-5x^2+7x-2\right)\)

\(\Leftrightarrow3x\left(x-2\right)\sqrt{3x-1}=2\left(x-2\right)\left(x^2-3x+1\right)\)

TH1: \(x=2\)

TH2: \(3x\sqrt{3x-1}=2\left(x^2-3x+1\right)\)

Đặt \(\sqrt{3x-1}=t\ge0\)

\(\Rightarrow3tx=2\left(x^2-t^2\right)\)

\(\Leftrightarrow2x^2-3tx-2t^2=0\)

\(\Leftrightarrow\left(2x+t\right)\left(x-2t\right)=0\)

\(\Rightarrow x=2t\)

\(\Leftrightarrow x=2\sqrt{3x-1}\)

\(\Leftrightarrow x^2=4\left(3x-1\right)\)

\(\Leftrightarrow x^2-12x+4=0\)

a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)

Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)

Suy ra: \(9-3x+10x-2=4\)

\(\Leftrightarrow7x+7=4\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)

a/

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-4=x^2-4\\x^2-5x-4=4-x^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-5x=0\\2x^2-5x-8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{5\pm\sqrt{89}}{4}\\\end{matrix}\right.\)

b/ - Với \(x\ge3\) pt trở thành:

\(x-1+3\left(x-3\right)=6\Leftrightarrow4x=16\Rightarrow x=4\)

- Với \(x\le1\) pt trở thành:

\(1-x+3\left(3-x\right)=6\)

\(\Leftrightarrow x=1\)

- Với \(1< x< 3\) pt trở thành:

\(x-1+3\left(3-x\right)=6\)

\(\Leftrightarrow-2x=-2\Rightarrow x=1\) (loại)

c/ ĐKXĐ: \(x\ne\pm2\)

\(\left[{}\begin{matrix}\frac{x^2-6x-4}{x^2-4}=1\\\frac{x^2-6x-4}{x^2-4}=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x-4=x^2-4\\x^2-6x-4=4-x^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-6x=0\\2x^2-6x-8=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=4\end{matrix}\right.\)

d/ - Với \(x\ge2\) pt trở thành:

\(x-1-2\left(x-2\right)=x^2-x-3\)

\(\Leftrightarrow x^2=6\Rightarrow\left[{}\begin{matrix}x=\sqrt{6}\\x=-\sqrt{6}\left(l\right)\end{matrix}\right.\)

- Với \(x\le1\) pt trở thành:

\(1-x-2\left(2-x\right)=x^2-x-3\) làm tương tự

- Với \(1< x< 2\):

\(x-1-2\left(2-x\right)=x^2-x-3\)