a) `x^4-2x^3+10x^2-20x=0`

`<=>x(x^3-2x^2+10x-20)=0`

`<=>x[x^2(x-2)+10(x-2)]=0`

`<=>x(x-2)(x^2+10)=0`

`<=> x in {0;2}`

b) `(2x-3)^2=(x+5)^2`

`<=>(2x-3)^2-(x+5)^2=0`

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

`<=>(3x+2)(x-8)=0`

`<=> x in {-2/3;8}`

c) `x^2(x-1)-4x^2+8x-4=0`

`<=>x^2(x-1)-4(x^2-2x+1)=0`

`<=>x^2(x-1)-4(x-1)^2=0`

`<=>(x-1)(x^2-4x+4)=0`

`<=>(x-1)(x-2)^2=0`

`<=> x in {1;2}`

 

$ a )x^{4}-$ $2x^{3}+$ $10x^{2}-20x=0$

$\text{ <=> }$ $x^{3}( x-2)+10x(x-2)=0$

$\text{ <=> x ( }$ $x^{}-2)($$x^{2}+10)=0$

$\text{ <=> x ( x - 2 ) (}$ $x^{2}+10)=0$

$<=>\left[\begin{array}{ccc}x=0\\x-2=0\\x^{2}+10=0 (ktm)\end{array}\right]<=>$ \(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)

$\text{ x ∈ (0 ; }$ $2 ^{} )$

$\text{b ) ( 2x - 3 )}$$^{2}=(x+5)$$^{2}$

$\text{ <=>}$ $(2x - 3 )^{2}-(x+5)$$^{2}=0$

$\text{ <=> ( 2x - 3 + x + 5 ) ( 2x - 3 - x - 5 ) = 0}$

$\text{ <=> ( 3x + 2 ) ( x - 8 ) = 0}$

<=>\(\left[ \begin{array}{l} 3x + 2 =0\\ x - 8 =0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=- \dfrac{2}{3}\\x=8\end{array} \right.\)

$\text{ x ∈ (} $ $-\dfrac{2}{3};8 )$

$c ) x^{2}(x-1)-$ $4x^{2}+8x-4=0$

$<=> x^{2}(x-1)-($ $4x^{2}-8x+4)=0$

$<=> x^{2}(x-1)-($ $4x^{2}-4x-4x+4)=0$

$<=> x^{2}(x-1)-$$\text{ [ 4x ( x - 1) - 4 ( x - 1) ]=0}$

$<=> x^{2}(x-1)- (4x-4)(x-1)=0$

$<=>(x^{2}-4x+4)(x-1)=0$

$<=>(x^{2}-4x+4)(x-1)=0$

$<=>(x-2)^{2}(x-1)=0$

<=>\(\left[ \begin{array}{l}x-2=0\\x-1=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\)

$\text{ x ∈ (1 ; }$ $2 ^{})$

sr bn nhiều lắm 

mik đang bận hok bài cho nên mới lm được 

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