\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

a) Tương đương      b) Không tương đương.

Ta có: P = x − x + 2 ( x + 1 ) ( x − 2 ) − x x ( x − 2 ) : 1 − x 2 − x = x − x + 2 − x ( x + 1 ) ( x + 1 ) ( x − 2 ) . 2 − x 1 − x = 2 − 2 x ( x + 1 ) ( x − 1 ) = 2 ( 1 − x ) ( x + 1 ) ( x − 1 ) = − 2 x + 1  

Answer:

\(\left(2x-3\right).\left(x+1\right)-x.\left(2x+3\right)-9=0\)

\(\Rightarrow\left(2x^2+2x-3x-3\right)-2x^2-3x-9=0\)

\(\Rightarrow\left(2x^2-x-3\right)-2x^2-3x-9=0\)

\(\Rightarrow2x^2-x-3-2x^2-3x-9=0\)

\(\Rightarrow\left(2x^2-2x^2\right)-\left(x+3x\right)-\left(3+9\right)=0\)

\(\Rightarrow-4x-12=0\)

\(\Rightarrow x+3=0\)

\(\Rightarrow x=-3\)

\(2x.\left(x-3\right)-x+3=0\) (Sửa đề)

\(\Rightarrow2x.\left(x-3\right)-\left(x-3\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\2x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}}\)

\(2x.\left(x^2-4\right)+6.\left(4-x^2\right)=0\)

\(\Rightarrow2x.\left(x^2-4\right)-6.\left(x^2-4\right)=0\)

\(\Rightarrow2.\left(x-3\right).\left(x+2\right).\left(x-2\right)=0\)

Trường hợp 1: \(x-3=0\Rightarrow x=3\)

Trường hợp 2: \(x+2=0\Rightarrow x=-2\)

Trường hợp 3: \(x-2=0\Rightarrow x=2\)

2: \(3x\left(x-4\right)+2x-8=0\)

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

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

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

3: 4x(x-3)+x²-9=0

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

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

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

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

4: \(x\left(x-1\right)-x^2+3x=0\)

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

=>2x=0

=>x=0

5: \(x\left(2x-1\right)-2x^2+5x=16\)

=>\(2x^2-x-2x^2+5x=16\)

=>4x=16

=>x=4

a)\(=>2x=-10=>x=-5\)

b)\(=>-2x=-5=>x=\dfrac{-5}{-2}=\dfrac{5}{2}\)

c)\(4-x=0=>x=4-0=4\)

d)\(=>2x=-1=>x=-\dfrac{1}{2}\)

e)\(=>x^2=-2\)=> x ko tồn tại

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

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

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

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

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

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

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

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

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

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

( 2x - 3 )( x + 1 ) - 2x² + 6x = 0

<=> 2x² - x - 3 - 2x² + 6x = 0

<=> 5x - 3 = 0

<=> 5x = 3

<=> x = 3/5

( x² - x + 1 )( x - 3 ) - x³ + 4x² = 0

<=> x³ - 4x² + 4x - 3 - x³ + 4x² = 0

<=> 4x - 3 = 0

<=> 4x = 3

<=> x = 3/4

( x² - 2 )( x² + 2 ) - x⁴ - 2x + 5 = 0

<=> ( x² )² - 4 - x⁴ - 2x + 5 = 0

<=> x⁴ + 1 - x⁴ - 2x = 0

<=> 1 - 2x = 0

<=> 2x = 1

<=> x = 1/2

( x - 3 )( x² - 3x + 2 ) - ( x² - 2x - 7 )( x - 2 ) + 2x² - 2x = 0

<=> x³ - 6x² + 11x - 6 - ( x³ - 4x² - 3x + 14 ) + 2x² - 2x = 0

<=> x³ - 6x² + 11x - 6 - x³ + 4x² + 3x - 14 + 2x² - 2x = 0

<=> 12x - 20 = 0

<=> 12x = 20

<=> x = 20/12 = 5/3

a, \(\left(2x-3\right)\left(x+1\right)-2x^2+6x=0\)

\(\Leftrightarrow2x^2+2x-3x-3-2x^2+6x=0\Leftrightarrow5x-3=0\Leftrightarrow x=\frac{3}{5}\)

b, \(\left(x^2-x+1\right)\left(x-3\right)-x^3+4x^2=0\)

\(\Leftrightarrow x^3-3x^2-x^2+3x+x-3-x^3+4x^2=0\Leftrightarrow4x-3=0\Leftrightarrow x=\frac{3}{4}\)

c ; d tương tự nhé ! 

\(2x^2-6x=0\)

\(\Rightarrow2x.\left(x-3\right)=0\)

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

Vậy \(x\in\left\{0;3\right\}.\)

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

\(\Rightarrow\left(x+2\right).\left(2x-3\right)=0\)

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

Vậy \(x\in\left\{-2;\frac{3}{2}\right\}.\)

\(x^3-16x=0\)

\(\Rightarrow x.\left(x^2-16\right)=0\)

\(\Rightarrow x.\left(x^2-4^2\right)=0\)

\(\Rightarrow x.\left(x-4\right).\left(x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=0+4\\x=0-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

Vậy \(x\in\left\{0;4;-4\right\}.\)

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