a) 2x⁴ - x³ -2x² -x +2=0

=> (2x⁴- 2x³) +(x³-x²) -(x² -x) -(2x-2)=0

=>(x-1)(2x³+x²-x-2)=0

=>(x-1)²( 2x²+3x+2)=0 ( vì 2x²+3x+2>0)

=> x-1=0 => x =1

chia cho x² , rồi đặt ẩn

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x-2=0\\x^2+2x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\frac{3}{2}\right)^2=\frac{17}{4}\\\left(x+1\right)^2=3\end{matrix}\right.\)

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

a) Dễ thấy x = 0 không là nghỉ=ệm của pt đã cho

Chia cả 2 vế của pt cho \(x^2\ne0\) ta đc :

\(2x^2-21x+74-\frac{105}{x}+\frac{50}{x^2}=0\)

\(\Leftrightarrow2\left(x^2+\frac{25}{x^2}+10\right)-21\left(x+\frac{5}{x}\right)+54=0\)

\(\Leftrightarrow2\left(x+\frac{5}{x}\right)^2-21\left(x+\frac{5}{x}\right)+54=0\)

\(\Leftrightarrow2t^2-21t+54=0\) ( với \(t=x+\frac{5}{x}\) )

\(\Leftrightarrow\left(2t-9\right)\left(t-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{9}{2}\\t=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+\frac{5}{x}=\frac{9}{2}\\x+\frac{5}{x}=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-\frac{9}{2}x+5=0\\x^2-6x+5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\frac{9}{4}\right)^2=\frac{1}{16}\\\left(x-1\right)\left(x-5\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{9}{4}=\frac{1}{4}\\x-\frac{9}{4}=-\frac{1}{4}\\x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=2\\x=1\\x=5\end{matrix}\right.\) ( TM )

Vậy tập nghiệm của pt đã cho là \(S=\left\{\frac{5}{2};2;1;5\right\}\)

2)  2x⁴-21x³+74x²-105x+50=0

<=>(2x⁴-2x³)+(-19x³+19x²)+(55x²-55x)+(-50x+50)=0

<=>2x³.(x-1)-19x².(x-1)+55x.(x-1)-50.(x-1)=0

<=>(x-1)(2x³-19x²+55x-50)=0

<=>(x-1)[(2x³-20x²+50x)+(x²+5x-50)]=0

<=>(x-1)[2x.(x-5)²+(x²-5x+10x-50)]=0

<=>(x-1){2x.(x-5)²+[x.(x-5)+10.(x-5)]}=0

<=>(x-1)[2x.(x-5)²+(x-5)(x+10)]=0

<=>(x-1)(x-5)(2x²-10x+x+10)=0

<=>(x-1)(x-5)(2x²-5x-4x+10)=0

<=>(x-1)(x-5)[x.(2x-5)-2.(2x-5)]=0

<=>(x-1)(x-5)(x-2)(2x-5)=0

<=>x=1 hoặc x=5 hoặc x=2 hoặc x=5/2

\(2x^4-21x^3+74x^2-105x+50=0\)

\(< =>2x^4-10x^3-11x^3+55x^2+19x^2-95x^2-10x+50=0\)

\(< =>2x^3\left(x-5\right)-11x^2\left(x-5\right)+19x\left(x-5\right)-10\left(x-5\right)=0\)

\(< =>\left(x-5\right).\left(2x^3-11x^2+19x-10\right)=0\)

\(< =>\left(x-5\right).\left(2x^3-2x^2-9x^2+9x+10x-10\right)=0\)

\(< =>\left(x-5\right).\left(x-1\right).\left(2x^2-9x+10\right)=0\)

\(2x^2-9x+10\ge0\)

\(< =>x=5\)hoặc \(x=1\)

Vậy S = 1 hoặc 5

Vì x = 0 ko là nghiệm của phương trình

Chia 2 vế cho x² ≠ 0 ta đc \(2\left(x^2+\frac{25}{x}\right)-21\left(x+\frac{5}{x}\right)+74=0\)

Đặt \(t=x+\frac{5}{x}\) thì \(t^2=x^2+\frac{25}{x^2}+10\)

Phương trình trở thành: \(2\left(t^2-10\right)-21t+74=0\Leftrightarrow2t^2-21t+54=0\Leftrightarrow t=6,t=\frac{9}{2}\)

Khi \(t=6\) ta có phương trình \(x+\frac{5}{x}=6\Leftrightarrow x^2-6x+5=0\Leftrightarrow x=1\) hoặc \(x=5\)

Khi \(t=\frac{9}{2}\) ta có phương trình \(x+\frac{5}{x}=\frac{9}{2}\Leftrightarrow2x^2-9x+10=0\Leftrightarrow x=2\) hoặc \(x=\frac{5}{2}\)

Vậy...