\(\Leftrightarrow x^4-x^3+x^3-x^2-8x^2+8x+16x-16=0\\ \Leftrightarrow\left(x-1\right)\left(x^3+x^2-8x+16\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3+4x^2-3x^2-12x+4x+16\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+4\right)\left(x^2-3x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\\\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{4}=0\left(vô.n_o\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

nhận thấy x = 0 không là nghiệm của phương trình

Chia 2 vế phương trình cho x², ta được : 

\(x^2-9x+24-\frac{27}{x}+\frac{9}{x^2}=0\)  ( 1 )

đặt \(t=x+\frac{3}{x}\)

( 1 ) \(\Leftrightarrow\left(x+\frac{3}{x}\right)^2-9\left(x+\frac{3}{x}\right)+18=0\)

\(\Leftrightarrow t^2-9t+18=0\Leftrightarrow\left(t-6\right)\left(t-3\right)=0\Leftrightarrow\orbr{\begin{cases}t=6\\t=3\end{cases}}\)

Khi đó : \(\orbr{\begin{cases}x+\frac{3}{x}=6\Leftrightarrow x=3\pm\sqrt{6}\\x+\frac{3}{x}=3\Leftrightarrow x\in\varnothing\end{cases}}\)

a. \(x^4-10x^3+25x^2-36=0\)

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

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

=>\(\left(x-3\right)\left[x^2\left(x-2\right)-5x\left(x-2\right)-6\left(x-2\right)\right]=0\)=> \(\left(x-3\right)\left(x-2\right)\left(x^2-5x-6\right)=0\)

=> \(\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(x-6\right)=0\)

=>\(\left[\begin{matrix}x=3\\x=2\\x=-1\\x=6\end{matrix}\right.\)

b) \(x^4\) - \(^{9x^2}\) - 24x - 16 = 0

=> \(x^3\left(x-4\right)+4x^2\left(x-4\right)+7x\left(x-4\right)+4\left(x-4\right)=0\)=>\(\left(x-4\right)\left(x^3+4x^2+7x+4\right)=0\)

=> \(\left(x-4\right)\left[x^2\left(x+1\right)+3x\left(x+1\right)+4\left(x+1\right)\right]=0\)=>\(\left(x-4\right)\left(x+1\right)\left(x^2+3x+4\right)=0\)

=> \(\left(x-4\right)\left(x+1\right)=0\) (vì x^2 + 3x + 4> 0)

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

a,pt\(\Leftrightarrow\left(x^4-10x^3+25x\right)-36=0\)\(\Leftrightarrow\left(x^2-5x\right)^2-36=0\)

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

\(\Leftrightarrow\left[\begin{matrix}\left(x+1\right)\left(x-6\right)=0\\\left(x-2\right)\left(x-3\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=-1,x=6\\x=2,x=3\end{matrix}\right.\)

vậy pt có 4 nghiệm x=(-1,6,2,3)

pt\(\Leftrightarrow x^4-\left(9x^2+24x+16\right)=0\)\(\Leftrightarrow\left(x^2\right)^2-\left(3x+4\right)^2=0\)\(\Leftrightarrow\left(x^2-3x-4\right)\left(x^2+3x+4\right)=0\)\(\Leftrightarrow\left[\begin{matrix}x^2-3x-4=0\\x^2+3X+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}\left(x-4\right)\left(x+1\right)=0\\x^2+3x+4>0\end{matrix}\right.\)\(\Leftrightarrow x=4,x=-1\)

vậy pt có nghiệm x=(4,-1)

\(x^4-9x^3+24x^2-27x+9=0\)

<=> \(x^4-3x^3+3x^2-6x^3+18x^2-18x+3x^2-9x+9=0\)

<=>\(x^2\left(x^2-3x+3\right)-6x\left(x^2-3x+3\right)+3\left(x^2-3x+3\right)=0\)

<=>\(\left(x^2-6x+3\right)\left(x^2-3x+3\right)=0\)

<=> \(\left[{}\begin{matrix}x^2-6x+3=0\\x^2-3x+3=0\end{matrix}\right.\)

Giải nốt :))

tương tự như phần vừa nãy nha bạn tự giải được kết quả x=-1 và x=4 là đúng

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

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

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

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

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

\(\left(\sqrt{2x+5}-\left(x+1\right)\right)^2+\left(\sqrt{3\left(x+1\right)}-\sqrt{x+7}\right)^2=0.\\ \)
Đến đây chắc biết phải làm gì =))