a/ Nhận thấy \(x=0\) không phải nghiệm, chia cả 2 vế của pt cho \(x^2\):

\(x^2+5x-10+\frac{10}{x}+\frac{4}{x^2}=0\)

\(\Leftrightarrow x^2+\frac{4}{x^2}+5\left(x+\frac{2}{x}\right)-10=0\)

Đặt \(x+\frac{2}{x}=a\Rightarrow x^2+4+\frac{4}{x^2}=a^2\Rightarrow x^2+\frac{4}{x^2}=a^2-4\)

Phương trình trở thành:

\(a^2-4+5a-10=0\)

\(\Leftrightarrow a^2+5a-14=0\) \(\Rightarrow\left[{}\begin{matrix}a=2\\a=-7\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{2}{x}=2\\x+\frac{2}{x}=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+2=0\left(vn\right)\\x^2+7x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-7+\sqrt{41}}{2}\\x=\frac{-7-\sqrt{41}}{2}\end{matrix}\right.\)

b/ \(x^4-8x^2+x+12=0\)

\(\Leftrightarrow x^4-8x^2+16+x-4=0\)

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

Đặt \(x^2-4=a\Rightarrow-4=a-x^2\)

Phương trình trở thành:

\(a^2+x+a-x^2=0\)

\(\Leftrightarrow\left(a-x\right)\left(a+x\right)+x+a=0\)

\(\Leftrightarrow\left(a-x+1\right)\left(x+a\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-3=0\\x^2+x-4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{1\pm\sqrt{13}}{2}\\x=\frac{-1\pm\sqrt{17}}{2}\end{matrix}\right.\)

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

\(a,x^4-2x^3+5x^2-10x=0\\ \Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(b,\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\\ \Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-7\end{matrix}\right.\)

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

\(d,x^2\left(x-1\right)-4x^2+8x-4=0\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) \(x^4-2x^3+5x^2-10x=0\\ \Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\\ \Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\\ \Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2+5=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\\x=2\end{matrix}\right.\)

Vậy \(x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}\)

b) \(\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+5=2x-2\\3x+5=-2x+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

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

vậy ...

d) \(x^2\left(x-1\right)-4x^2+8x-4=0\\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)

    \(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a: Ta có: \(x^4-2x^3+5x^2-10x=0\)

\(\Leftrightarrow x\left(x^3-2x^2+5x-10\right)=0\)

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

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

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

\(\Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\)

\(\Leftrightarrow\left(3x+5-2x+2\right)\left(3x+5+2x-2\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(5x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

bạn viết rõ đề câu a;b nhé 

c, \(2x\left(x-5\right)-\left(x-5\right)=0\Leftrightarrow\left(2x-1\right)\left(x-5\right)=0\Leftrightarrow x=\dfrac{1}{2};x=5\)

d, \(\left(x+3\right)\left(x+3-5+x\right)=0\Leftrightarrow\left(x+3\right)\left(2x-2\right)=0\Leftrightarrow x=-3;x=1\)

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

\(\Leftrightarrow\left(x+2\right)\left(-5x+1\right)=0\Leftrightarrow x=-2;x=\dfrac{1}{5}\)

làm tạm câu này vậy

a/\(\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)^2=5x^4\)

\(\Leftrightarrow\left(x^2-x+1\right)^4+4x^2\left(x^2-x+1\right)+4x^4=9x^4\)

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

\(\Leftrightarrow\left(x^2-x+1\right)^2+2x^2=3x^2\)(vì 2 vế đều không âm)

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

\(\Leftrightarrow\left|x\right|=x^2-x+1\)\(\left(x^2-x+1=\left(x-\frac{1}{4}\right)^2+\frac{3}{4}>0\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x=x^2-x+1\\-x=x^2-x+1\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\x^2+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x^2+1=0\left(vo.nghiem\right)\end{cases}}}\)

Vậy...

chuẩn

i cum back <(") câu e/ bạn xét x=0 không là nghiệm của pt, sau đó chia 2 vế cho \(x^2\), đặt ẩn phụ \(t=x+\frac{1}{x}\)rồi giải 

2: Ta có: \(x^4-4x^3-9x^2+8x+4=0\)

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

\(\Leftrightarrow x^3\left(x-1\right)-3x^2\left(x-1\right)-12x\left(x-1\right)-4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2-12x-4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2-5x^2-10x-2x-4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)-5x\left(x+2\right)-2\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-5x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x^2-5x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\dfrac{5-\sqrt{33}}{2}\\x=\dfrac{5+\sqrt{33}}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-2;\dfrac{5-\sqrt{33}}{2};\dfrac{5+\sqrt{33}}{2}\right\}\)

1: Ta có: \(x^4+5x^3+10x^2+15x+9=0\)

\(\Leftrightarrow x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0\)

\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)+6x\left(x+1\right)+9\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2+6x+9\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[x^3+3x^2+x^2+6x+9\right]=0\)

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

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

mà \(x^2+x+3>0\forall x\)

nên (x+1)(x+3)=0

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

Vậy: S={-1;-3}

x⁴ - 3x³ + 4x² - 3x + 1 = 0

<=> x⁴ - 2x² + x² - x³ + 2x² - x + x² - 2x + 1 = 0

<=> x²(x² - 2x + 1) - x(x² - 2x + 1) + (x² - 2x + 1) = 0

<=> (x² - 2x + 1)(x² - x + 1) = 0

<=> (x - 1)²(x² - x + 1) = 0

<=> x - 1 = 0 (vì x² - x + 1 \(\ge\) 0,75 > 0)

<=> x = 1

Vậy tập nghiệm của pt là S = {1}

`1)x^4 -10x^3 +26x^2 -10x+1=0`
`x=0=>VT=1=>x=0(l)`
Chia 2 vế cho `x^2>0` ta có
`x^2-10x+26-10/x+1/x^2=0`
`=>x^2+1/x^2+26-10(x+1/x)=0`
`=>(x+1/x)^2-10(x+1/x)+24=0`
Đặt `a=x+1/x`
`pt<=>a^2-10a+24=0`
`<=>` $\left[ \begin{array}{l}a=4\\a=6\end{array} \right.$
`a=4<=>x+1/x=4<=>x^2-4x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt3+2\\x=-\sqrt3+2\end{array} \right.$
`a=6<=>x+1/x=6<=>x^2-6x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt8+3\\x=-\sqrt8+3\end{array} \right.$
Vậy `S={\sqrt3+2,-\sqrt3+2,\sqrt8+3,-\sqrt8+3}`

2)Do hệ số chẵn bằng=hệ số lẻ
`=>x=-1`
`pt<=>x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0`
`<=>(x+1)(x^3+4x^2+6x+9)=0`
`<=>(x+1)(x^3+3x^2+x^2+6x+9)=0`
`<=>(x+1)[x^2(x+3)+(x+3)^2]=0`
`<=>(x+1)(x+3)(x^2+x+3)=0`
Do `x^2+x+3=(x+1/2)^2+11/4>0`
`=>` $\left[ \begin{array}{l}x=-3\\x=-1\end{array} \right.$
Vậy `S={-1,-3}`