a, 3x2 - 8x2 - 2x+3=0

2x(3-8) - 2x+3=0

2x5 - 2x+3=0

2x5 - 2x=0-3=

2x5 - 2x=-3

2x(5-x)=-3

5-x=-3/2

5-x=1,5

x=5-1,5

x=3,5

3,5 nha bn

chúc bn học tốt

happy new year

b: 4x^2-20x+25=(x-3)^2

=>(2x-5)^2=(x-3)^2

=>(2x-5)^2-(x-3)^2=0

=>(2x-5-x+3)(2x-5+x-3)=0

=>(3x-8)(x-2)=0

=>x=8/3 hoặc x=2

c: x+x^2-x^3-x^4=0

=>x(x+1)-x^3(x+1)=0

=>(x+1)(x-x^3)=0

=>(x^3-x)(x+1)=0

=>x(x-1)(x+1)^2=0

=>\(x\in\left\{0;1;-1\right\}\)

d: 2x^3+3x^2+2x+3=0

=>x^2(2x+3)+(2x+3)=0

=>(2x+3)(x^2+1)=0

=>2x+3=0

=>x=-3/2

a: =>x^2(5x-7)-3(5x-7)=0

=>(5x-7)(x^2-3)=0

=>\(x\in\left\{\dfrac{7}{5};\sqrt{3};-\sqrt{3}\right\}\)

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

\(\Leftrightarrow x^2\left(x^2-2\right)+\left(x^2-2\right)=0\Leftrightarrow\left(x^2+1>0\right)\left(x^2-2\right)=0\Leftrightarrow x=\pm\sqrt{2}\)

b, \(\Leftrightarrow x^2\left(x^2+2x+1\right)=0\Leftrightarrow x^2\left(x+1\right)^2=0\Leftrightarrow x=0;x=-1\)

c, \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1>0\right)=0\Leftrightarrow x=1\)

d, \(\Leftrightarrow6x^2-3x-4x+2=0\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\Leftrightarrow x=\dfrac{2}{3};x=\dfrac{1}{2}\)

a) 

/ \(x^4+x^2-2=0\)

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

2x⁵ - 7x⁴ + 5x³ + 5x² - 7x + 2 = 0

<=> 2x⁵-4x⁴-3x⁴+6x³-x³+2x²+3x²-6x-x+2=0

<=> 2x⁴(x-2)-3x³(x-2)-x²(x-2)+3x(x-2)-(x-2)=0

<=>(x-2)(2x⁴-3x³-x²+3x-1)=0

<=>(x-2)(2x⁴-x³-2x³+x²-2x²+x+2x-1)=0

<=>(x-2)[x³(2x-1)-x²(2x-1)-x(2x-1)+2x-1]=0

<=>(x-2)(2x-1)(x³-x²-x+1)=0

<=>(x-2)(2x-1)[x²(x-1)-(x-1)]=0

<=>(x-2)(2x-1)(x-1)(x²-1)=0

<=>(x-2)(2x-1)(x-1)²(x+1)=0

=> x-2=0 => x=2

hoặc 2x-1=0=>x=1/2

hoặc x-1=0=>x=1

hoặc x+1=0=>x=-1

Vậy...

\(2x^5-7x^4+5x^3+5x^2-7x+2=0\)

\(\Leftrightarrow\left(2x^5-4x^4+2x^3\right)-\left(3x^4-6x^3+3x^2\right)-\left(3x^3-6x^2+3x\right)+\left(2x^2-4x+2\right)=0\)

\(\Leftrightarrow2x^3\left(x^2-2x+1\right)-3x^2\left(x^2-2x+1\right)-3x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)=0\)

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

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

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

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

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

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

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

\(\Leftrightarrow\)\(x-1=0\)

hoặc  \(x+1=0\)

hoặc \(x-2=0\)

hoặc \(2x-1=0\)

\(\Leftrightarrow\)\(x=1\)

hoặc \(x=-1\)

hoặc \(x=2\)

hoặc \(x=\frac{1}{2}\)

Vậy tập nghiệm của phương trình là \(S=\left\{1;-1;2;\frac{1}{2}\right\}\)

a) Ta có: \(x^3+x^2+x+1=0\)

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

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

mà \(x^2+1>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}

b) Ta có: \(x^3-6x^2+11x-6=0\) 

\(\Leftrightarrow x^3-x^2-5x^2+5x+6x-6=0\)

\(\Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\)

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

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

Vậy: S={1;2;3}

c) Ta có: \(x^3-x^2-21x+45=0\)

\(\Leftrightarrow x^3-3x^2+2x^2-6x-15x+45=0\)

\(\Leftrightarrow x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+2x-15\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+5x-3x-15\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy: S={3;-5}

d) Ta có: \(x^4+2x^3-4x^2-5x-6=0\)

\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6=0\)

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

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

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

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

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

mà \(x^2+x+1>0\forall x\)

nên (x-2)(x+3)=0

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

Vậy: S={2;-3}

\(5x\left(x-1\right)=x-1\)

\(\Leftrightarrow5x^2-5x=x-1\)

\(\Leftrightarrow5x^2-5x-x+1=0\)

\(\Leftrightarrow5x^2-6x+1=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-\frac{1}{5}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-\frac{1}{5}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)

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

\(2\left(x-7\right)-x\left(x-7\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x-7\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)

(2/2 + 3/6 + 4/12 + 6/30) nhân 10 - x=0

(9/6+5/4) nhân 10 -x=0

11/4 nhân 10 - x= 0

2,75 nhân 10 -x=0

27.5 - x=0

x=27.5 - 0= 27.5

dùm mình nha...

1/5x6 sửa thành 1/4x5 nhé!

(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5)x10-x=0

(1-1/5)x10-x=0

4/5x10-x=0

8-x=0

x=8-0=8