refer

\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Leftrightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)=40\)

\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)

\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+5+3\right)=40\)

\(\Leftrightarrow p\left(p+3\right)=40\) (khi đặt \(\left(x^2+6x+5\right)=p\)

\(\Leftrightarrow p^2+3p=40\)

\(\Leftrightarrow p^2\cdot2\cdot p\cdot\frac{3}{2}+\left(\frac{3}{2}\right)^2=\frac{169}{4}\)

\(\Leftrightarrow\left(p+\frac{3}{2}\right)^2-\left(\frac{13}{2}\right)^2=0\)

\(\Leftrightarrow\left(p+\frac{3}{2}-\frac{13}{2}\right)\left(p+\frac{3}{2}+\frac{13}{2}\right)=0\)

\(\Leftrightarrow\left(p-5\right)\left(p+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}p=5\\p=-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+6x+5=5\\x^2+6x+5=-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+6x=0\\x^2+2\cdot x\cdot3+9-9+5=-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\left(x+6\right)=0\\\left(x+3\right)^2=-4\left(\text{vôlí}\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-6\end{cases}}\)

\(\left(x-2\right)\left(x^2+5x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x^2+5x-7=0\end{cases}}\)

Ta có: \(\Delta=25-4\cdot\left(-7\right)=25+28=53\)

\(\Rightarrow\Delta>0\)

\(\Rightarrow\text{pt có 2 nghiệm pb}\)

\(\Rightarrow\hept{\begin{cases}x_1=\frac{-5-\sqrt{53}}{2}\\x_2=\frac{-5+\sqrt{53}}{2}\end{cases}}\)

\(\text{Vậy pt trên có nghiệm là x=2; x=}\frac{-5\pm\sqrt{53}}{2}\)

\(x^3+3x^2-4x-12=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=4\\x=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

\(\text{Vậy pt có nghiệm là x=2;x=-3}\)

ĐKXĐ: \(x\ne2;x\ne-2\)

\(\Rightarrow x-2+5\left(x+2\right)=2x-12\Leftrightarrow x-2+5x+10=2x-12\Leftrightarrow6x+8-2x=-12\Leftrightarrow4x+8=-12\Leftrightarrow4x=-20\Leftrightarrow x=-5\left(TM\right)\)

ĐKXĐ: \(x\ne\pm2\)

\(\dfrac{1}{x+2}+\dfrac{5}{x-2}=\dfrac{2x-12}{x^2-4}\)

\(\Leftrightarrow\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}+\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x-12}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow x-2+5x+10=2x-12\)

\(\Leftrightarrow2x-12-x+2-5x-10=0\)

\(\Leftrightarrow-4x-20=0\)

\(\Leftrightarrow-4\left(x+5\right)=0\)

\(\Leftrightarrow x+5=0\)

\(\Leftrightarrow x=-5\)

Vậy...

      3 - 2x = 3(x + 1) - x - 2

⇔  3 - 2x = 3x + 3 - x - 2

⇔   4x = 2 

⇔    x = 1/2

ĐKXĐ: \(x\ne\pm1\)

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

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

\(\Leftrightarrow4x=x^3+3\)

\(\Leftrightarrow x^3-4x+3=0\)

\(\Leftrightarrow x^3-x^2+x^2-x-3x+3=0\)

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

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

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

\(\Rightarrow x=\dfrac{-1\pm\sqrt{13}}{2}\)

Lời giải:

Ta sẽ thử phân tích $x^4-3x^3-x^2+2x-4$ thành nhân tử

Đặt $x^4-3x^3-x^2+2x-4=(x^2+ax+b)(x^2+cx+d)$ với $a,b,c,d$ nguyên.

$\Leftrightarrow x^4-3x^3-x^2+2x-4=x^4+x^3(a+c)+x^2(ac+b+d)+x(ad+bc)+bd$

Đồng nhất hệ số:

\(\left\{\begin{matrix} a+c=-3\\ ac+b+d=-1\\ ad+bc=2\\ bd=-4\end{matrix}\right.\). Từ $bd=-4$ ta xét các TH nguyên của $b,d$ để thay vào tìm $a,c$

Ta tìm được $a=-2;b=-4; c=-1; d=1$

Do đó:

$x^4-3x^3-x^2+2x-4=0$

$\Leftrightarrow (x^2-2x-4)(x^2-x+1)=0$

$\Leftrightarrow x^2-2x-4=0$ (do $x^2-x+1\neq 0$)

$\Leftrightarrow x=1\pm \sqrt{5}$

\(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-x\)

\(\Leftrightarrow\dfrac{2x}{6}-\dfrac{2x+1}{6}=\dfrac{x}{6}-\dfrac{6x}{6}\)

\(\Leftrightarrow2x-2x+1=x-6x\)

\(\Leftrightarrow1=-5x\)

\(\Leftrightarrow x=\dfrac{-1}{5}\)