MÌNH KHÔNG BIẾT ^_^

\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)

Đặt \(a=x^2+x\)

\(\Leftrightarrow a^2+4a=12\)

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

\(\Leftrightarrow a^2+6a-2a-12=0\)

\(\Leftrightarrow a\left(a+6\right)-2\left(a+6\right)=0\)

\(\Leftrightarrow\left(a+6\right)\left(a-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-6\\a=2\end{cases}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{23}{4}=0\\x^2+2x-x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\\\left(x+2\right)\left(x-1\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)

Vậy....

\(\left(x^2+6x+10\right)^2+\left(x+3\right)\left(3x^2+20x+36\right)=0\)

( rút gọn phá ngoặc tất cả )

\(\Leftrightarrow x^4+15x^3+85x^2+216x+208=0\)

\(\Leftrightarrow x^4+4x^3+11x^3+44x^2+41x^2+164x+52x+208=0\)

\(\Leftrightarrow x^3\left(x+4\right)+11x^2\left(x+4\right)+41x\left(x+4\right)+52\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^3+11x^2+41x+52\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^3+4x^2+7x^2+28x+13x+52\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left[x^2\left(x+4\right)+7x\left(x+4\right)+13\left(x+4\right)\right]=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+4\right)\left(x^2+7x+13\right)=0\)

\(\Leftrightarrow\left(x+4\right)^2\left(x^2+2\cdot x\cdot\frac{7}{2}+\frac{49}{4}+\frac{3}{4}\right)=0\)

\(\Leftrightarrow\left(x+4\right)^2\left[\left(x+\frac{7}{2}\right)^2+\frac{3}{4}\right]=0\)

\(\Leftrightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy....

ĐKXĐ:...

pt\(\Leftrightarrow4\left(x^2-2x\right)+16\sqrt{x^2-2x-3}-21=0\)

Đặt \(\sqrt{x^2-2x-3}=t\left(t\ge0\right)\Rightarrow t^2=x^2-2x-3\Leftrightarrow t^2+3=x^2-2x\)

\(\Rightarrow4\left(t^2+3\right)+16t-21=0\)

\(\Leftrightarrow4t^2+12+16t-21=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{1}{2}\\t=-\frac{9}{2}\left(l\right)\end{matrix}\right.\Rightarrow t=\frac{1}{2}\)

\(\Rightarrow x^2-2x-3=\frac{1}{4}\Leftrightarrow\left[{}\begin{matrix}x=\frac{2+\sqrt{17}}{2}\\x=\frac{2-\sqrt{17}}{2}\left(l\right)\end{matrix}\right.\)

Vậy \(x=\frac{2+\sqrt{17}}{2}\)

\(\left(x^2+7x+12\right).\left(4x-16\right)-\left(x+3\right)\left(x^2-5x+4\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(x^2+3x+4x+12\right).4.\left(x-4\right)-\left(x+3\right)\left(x^2-x-4x+4\right)\left(x-4\right)=0\)

\(\Leftrightarrow4\left(x+4\right)\left(x+3\right)\left(x-4\right)-\left(x+3\right)\left(x-4\right)\left(x+4\right)\left(x-4\right)=0\)

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

\(\Leftrightarrow\left(x+4\right)\left(x-4\right)\left(x+3\right)\left(8-x\right)=0\)

\(\Leftrightarrow\frac{\orbr{\begin{cases}x+4=0\\x-4=0\end{cases}}}{\orbr{\begin{cases}x+3=0\\8-x=0\end{cases}}}\Leftrightarrow\frac{\orbr{\begin{cases}x=-4\\x=4\end{cases}}}{\orbr{\begin{cases}x=-3\\x=8\end{cases}}}\)

\(\left|x-1\right|+2\left|x-2\right|+3\left|x-3\right|+4\left|x-4\right|+5\left|x-5\right|+20x=0\left(1\right)\)

TH1: x<1

(1) trở thành 1-x+2(2-x)+3(3-x)+4(4-x)+5(5-x)+20x=0

=>\(1-x+4-2x+9-3x+16-4x+25-5x+20x=0\)

=>\(5x+55=0\)

=>x=-11(nhận)

TH2: 1<=x<2

Phương trình (1) sẽ trở thành:

\(x-1+2\left(2-x\right)+3\left(3-x\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)

=>\(x-1+4-2x+9-3x+16-4x+25-5x+20x=0\)

=>\(7x+53=0\)

=>\(x=-\dfrac{53}{7}\left(loại\right)\)

TH3: 2<=x<3

Phương trình (1) sẽ trở thành:

\(x-1+2\left(x-2\right)+3\left(3-x\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)

=>\(x-1+2x-4+9-3x+16-4x+25-5x+20x=0\)

=>\(11x+45=0\)

=>\(x=-\dfrac{45}{11}\left(loại\right)\)

TH4: 3<=x<4

Phương trình (1) sẽ trở thành:

\(x-1+2\left(x-2\right)+3\left(x-3\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)

=>\(x-1+2x-4+3x-9+16-4x+25-5x+20x=0\)

=>\(-3x+27=0\)

=>x=9(loại)

TH5: 4<=x<5

Phương trình (1) sẽ trở thành:

\(\left(x-1\right)+2\left(x-2\right)+3\left(x-3\right)+4\left(x-4\right)+5\left(5-x\right)+20x=0\)

=>\(x-1+2x-4+3x-9+4x-16+25-5x+20x=0\)

=>\(25x-5=0\)

=>x=1/5(loại)

TH6: x>=5

Phương trình (1) sẽ trở thành:

\(\left(x-1\right)+2\left(x-2\right)+3\left(x-3\right)+4\left(x-4\right)+5\left(x-5\right)+20x=0\)

=>\(x-1+2x-4+3x-9+4x-16+5x-25+20x=0\)

=>35x-55=0

=>x=55/35(loại)

\(48x\left(x+1\right)\left(x^3-4\right)=\left(x^4+8x+12\right)^2\)

\(\Leftrightarrow4\left(12x+12\right)\left(x^4-4x\right)=\left(x^4+8x+12\right)^2\)

Đặt \(\left\{{}\begin{matrix}x^4-4x=a\\12x+12=b\end{matrix}\right.\)

\(\Rightarrow4ab=\left(a+b\right)^2\)

\(\Leftrightarrow4ab=a^2+a^2+2ab\)

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

\(\Leftrightarrow a-b=0\)

\(\Leftrightarrow x^4-16x-12=0\)

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

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

\(\Rightarrow x=1\pm\sqrt{3}\)

b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)

\(\Leftrightarrow x^2+7x+6=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)