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24 tháng 9 2018

(x^2-9)^2=12x-1 
<=>x^4-18x^2-12x+80=0 
<=>x^4-2x^3+2x^3-4x^2-14x^2+28x-40x+80... 
<=>(x-2)(x^3+2x^2-14x-40)=0 
<=>(x-2)(x-4)(x^2+6x+10)=0 
Ta thấy x^2+6x+10=(x+3)^2+1>0 
=>x=2 hhoặc x=4

8 tháng 2 2018

\(\left(x^2-9\right)^2=12x+1\)

\(\Leftrightarrow x^4-18x^2+81=12x+1\)

\(\Leftrightarrow x^4-18x^2+81-12x-1=0\)

\(\Leftrightarrow x^4-2x^3+2x^3-4x^2-14x^2+28x-40x+80=0\)

\(\Leftrightarrow x^3\left(x-2\right)+2x^2\left(x-2\right)-14x\left(x-2\right)-40\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2-14x-40\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2+6x+10\right)=0\)

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

24 tháng 4 2016

(x^2-9)^2=12x-1 

<=>x^4-18x^2-12x+80=0 

<=>x^4-2x^3+2x^3-4x^2-14x^2+28x-40x+80... 

<=>(x-2)(x^3+2x^2-14x-40)=0 

<=>(x-2)(x-4)(x^2+6x+10)=0 

Ta thấy x^2+6x+10=(x+3)^2+1>0 

=>x=2 hhoặc x=4

24 tháng 4 2016

(x^2-9)^2=12x-1 

<=>x^4-18x^2-12x+80=0 

<=>x^4-2x^3+2x^3-4x^2-14x^2+28x-40x+80... 

<=>(x-2)(x^3+2x^2-14x-40)=0 

<=>(x-2)(x-4)(x^2+6x+10)=0 

Ta thấy x^2+6x+10=(x+3)^2+1>0 

=>x=2 hhoặc x=4

3 tháng 7 2019

Phương trình đưa được về dạng ax + b = 0

26 tháng 3 2020

DKXD : x khac -1

\(\frac{-x}{x+1}\)+ 3 =\(\frac{2x+3}{x+1}\)

<=> \(\frac{-x}{x+1}\)+\(\frac{3\left(x+1\right)}{x+1}\)\(\frac{2x+3}{x+1}\)

=>   -x + 3x +3 = 2x +3

<=> 2x -2x =3-3

<=>  0x=0

<=> x=0(TMDK)

12 tháng 3 2018

\(\Leftrightarrow\)4x-18-12x-1=0

\(\Leftrightarrow\)-8x=19

\(\Leftrightarrow\)x=\(\dfrac{-19}{8}\)

18 tháng 12 2020

5x( x - y ) + 12x - 12y 

= 5x( x - y ) + 12( x - y )

= ( x - y )( 5x + 12 )

x2 + 2xy - 9 + y2

= ( x2 + 2xy + y2 ) - 9

= ( x + y )2 - 32

= ( x + y - 3 )( x + y + 3 )

18 tháng 12 2020

\(5x\left(x-y\right)+12x-12y\)

\(=5x\left(x-y\right)+12\left(x-y\right)\)

\(=\left(x-y\right)\left(5x+12\right)\)

\(x^2+2xy-9+y^2\)

\(=\left(x^2+2xy+y^2\right)-9\)

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

\(=\left(x+y-3\right)\left(x+y+3\right)\)

22 tháng 10 2023

2:

a: \(x^2-12x+20\)

\(=x^2-2x-10x+20\)

=x(x-2)-10(x-2)

=(x-2)(x-10)

b: \(2x^2-x-15\)

=2x^2-6x+5x-15

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

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

c: \(x^3-x^2+x-1\)

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

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

d: \(2x^3-5x-6\)

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

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

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

e: \(4y^4+1\)

\(=4y^4+4y^2+1-4y^2\)

\(=\left(2y^2+1\right)^2-\left(2y\right)^2\)

\(=\left(2y^2+1-2y\right)\left(2y^2+1+2y\right)\)

f; \(x^7+x^5+x^3\)

\(=x^3\left(x^4+x^2+1\right)\)

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

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

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

g: \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)

\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-3\left(x^2+x\right)+6\)

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

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

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

h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)

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

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

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

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

\(=\left(x+1\right)^2\left[\left(x+1\right)^2-4\right]\)

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

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

i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)

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

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

\(=\left(x+2y\right)\left(x+2y-1\right)-3\left(x+2y-1\right)\)

\(=\left(x+2y-1\right)\left(x+2y-3\right)\)

j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)

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

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

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

6 tháng 2 2022

\(x^3-7x^2=3x^2-12x\)

\(\Leftrightarrow x^3-10x^2+12x=0\Leftrightarrow x\left(x^2-10x+12\right)=0\)

\(\Leftrightarrow x\left(x-5-\sqrt{13}\right)\left(x-5+\sqrt{13}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5+\sqrt{13}\\x=5-\sqrt{13}\end{matrix}\right.\)

6 tháng 2 2022

<=>\(x^3-10x^2+12x=0\)

<=>\(x\left(x^2-10x+12\right)=0\)

<=>\(x\left(x-5-\sqrt{13}\right)\left(x-5+\sqrt{13}\right)=0\)

<=>\(\left[{}\begin{matrix}x=0\\x-5-\sqrt{13}=0\\x-5+\sqrt{13}=0\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x=0\\x=5+\sqrt{13}\\x=5-\sqrt{13}\end{matrix}\right.\)