1) \(8x^3-12x^2+6x-1=0\)

\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)

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

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

2) \(x^3-6x^2+12x-8=27\)

\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)

\(\Leftrightarrow\left(x-2\right)^3=27\)

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

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=3+2\)

\(\Leftrightarrow x=5\)

3) \(x^2-8x+16=5\left(4-x\right)^3\)

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

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

\(\Leftrightarrow5\left(4-x\right)=1\)

\(\Leftrightarrow4-x=\dfrac{1}{5}\)

\(\Leftrightarrow x=4-\dfrac{1}{5}\)

\(\Leftrightarrow x=\dfrac{19}{5}\)

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

\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)

\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)

\(\Leftrightarrow8-x^3=0\)

\(\Leftrightarrow x^3=8\)

\(\Leftrightarrow x^3=2^3\)

\(\Leftrightarrow x=2\)

5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)

\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)

\(\Leftrightarrow12x-4=-10\)

\(\Leftrightarrow12x=-10+4\)

\(\Leftrightarrow12x=-6\)

\(\Leftrightarrow x=\dfrac{-6}{12}\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)

\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)

\(\Leftrightarrow-54x-2x^3=36x^2-54x\)

\(\Leftrightarrow-2x^3=36x^2\)

\(\Leftrightarrow-2x^3-36x^2=0\)

\(\Leftrightarrow-2x^2\left(x+18\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)

`1,(4x^3+3x^3):x^3+(15x^2+6x):(-3x)=0`

`<=> 4 + 3 + (-5x) + (-2)=0`

`<=> -5x+5=0`

`<=>-5x=-5`

`<=>x=1`

`2,(25x^2-10x):5x +3(x-2)=4`

`<=> 5x - 2 + 3x-6=4`

`<=> 8x -8=4`

`<=> 8x=12`

`<=>x=12/8`

`<=>x=3/2`

`3,(3x+1)^2-(2x+1/2)^2=0`

`<=> [(3x+1)-(2x+1/2)][(3x+1)+(2x+1/2)]=0`

`<=>( 3x+1-2x-1/2)(3x+1+2x+1/2)=0`

`<=>( x+1/2) (5x+3/2)=0`

`@ TH1`

`x+1/2=0`

`<=>x=0-1/2`

`<=>x=-1/2`

` @TH2`

`5x+3/2=0`

`<=> 5x=-3/2`

`<=>x=-3/2 : 5`

`<=>x=-15/2`

`4, x^2+8x+16=0`

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

`<=>x+4=0`

`<=>x=-4`

`5, 25-10x+x^2=0`

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

`<=>5-x=0`

`<=>x=5`

\(x^2+8x+16=x^2+2.x.4+4^2=\left(x+4\right)^2\)

\(25-10x+x^2=5^2-2.5.x+x^2=\left(5-x\right)^2\)

1. x(x + 1) - x² + 1 = 0

<=> x(x + 1) - (x² - 1) = 0

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

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

<=> x + 1 = 0\

<=> x = -1

2. 4x(x - 2) - 6 + 3x = 0

<=> 4x(x - 2) - (3x - 6) = 0

<=> 4x(x - 2) - 3(x - 2) = 0

<=> (4x - 3)(x - 2) = 0

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

<=> \(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=2\end{matrix}\right.\)

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

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

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

<=> \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Bài 1:

a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)

\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)

b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)

c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

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

e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Bài 2:

a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)

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

b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

1)
\(4x^2-4x+1-4x^2-16x-16=9\)
\(-20x-15=9\)
-20x=24
x=-1,2

3)
(2x+1)²=5²
2x+1=5
2x=4
x=2
 

\(1,\Rightarrow4x^2-4x+1-4x^2-16x-16=9\\ \Rightarrow-20x=23\Rightarrow x=-\dfrac{23}{20}\\ 2,\Rightarrow9x^2-6x+1+2x+6+11-11x^2=15\\ \Rightarrow2x^2+4x-3=0\\ \Rightarrow2\left(x^2+2x+1\right)-5=0\\ \Rightarrow2\left(x+1\right)^2-5=0\\ \Rightarrow\left[\sqrt{2}\left(x+1\right)-\sqrt{5}\right]\left[\sqrt{2}\left(x+1\right)+\sqrt{5}\right]=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{2}\left(x+1\right)=\sqrt{5}\\\sqrt{2}\left(x+1\right)=-\sqrt{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{10}}{2}\\x+1=-\sqrt{\dfrac{5}{2}}=\dfrac{-\sqrt{10}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-2}{2}\\x=\dfrac{-\sqrt{10}-2}{2}\end{matrix}\right.\)

\(3,\Rightarrow\left(2x+1\right)^2-25=0\Rightarrow\left(2x+1-5\right)\left(2x+1+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

\(4,\Rightarrow x^3+3x^2+3x+1-x^3-2x^2-2x+1-x^2=15\\ \Rightarrow x+2=15\Rightarrow x=13\)

\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

x²+4x+4=0
(x+2)²=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)²+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0           Th2: x+1=0
            x=-3                      x=-1
vậy ...

1)(x²-4x+16)(x+4)-x(x+1)(x+2)+3x²=0

\(\Rightarrow\)(x³+64)-x(x²+2x+x+2)+3x²=0

\(\Rightarrow\)x³+64-x³-2x²-x²-2x+3x²=0

\(\Rightarrow\)-2x+64=0

\(\Rightarrow\)-2x=-64

\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)

\(\Rightarrow x=32\)

2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50

\(\Rightarrow\)8x-24x²+2-6x+24x²-60x-4x+10=50

\(\Rightarrow\)-62x+12=50

\(\Rightarrow\)-62x=50-12

\(\Rightarrow\)-62x=38

\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)

5: =>4x^2-1/9=0

=>(2x-1/3)(2x+1/3)=0

=>x=1/6 hoặc x=-1/6

6: =>x-1=2

=>x=3

7:=>(2x-1)^3=-27

=>2x-1=-3

=>2x=-2

=>x=-1

8: =>1/8(x-1)^3=-125

=>(x-1)^3=-1000

=>x-1=-10

=>x=-9

3: =>(5x-5)^2-4=0

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

=>x=3/5 hoặc x=7/5

4: =>(5x-1)^2=0

=>5x-1=0

=>x=1/5

1: =>(3x-1)(2x-1)=0

=>x=1/3 hoặc x=1/2

2: =>x^2(2x-3)-4(2x-3)=0

=>(2x-3)(x^2-4)=0

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

=>x=3/2;x=2;x=-2

`@` `\text {Answer}`

`\downarrow`

`1,`

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

`<=> 2x(3x - 1) - 3x + 1 = 0`

`<=> 2x(3x - 1) - (3x - 1) = 0`

`<=> (2x - 1)(3x-1) = 0`

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

`<=>`\(\left[{}\begin{matrix}2x=1\\3x=1\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy,  `S = {1/2; 1/3}`

`2,`

\(x^2\left(2x-3\right)+12-8x=0\)

`<=> x^2(2x - 3) - 8x + 12 =0`

`<=> x^2(2x - 3) - (8x - 12) = 0`

`<=> x^2(2x - 3) - 4(2x - 3) = 0`

`<=> (x^2 - 4)(2x - 3) = 0`

`<=>`\(\left[{}\begin{matrix}x^2-4=0\\2x-3=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x^2=4\\2x=3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x^2=\left(\pm2\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\pm2\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy, `S = {+-2; 3/2}`

`3,`

\(25\left(x-1\right)^2-4=0\)

`<=> 25(x-1)(x-1) - 4 = 0`

`<=> 25(x^2 - 2x + 1) - 4 = 0`

`<=> 25x^2 - 50x + 25 - 4 = 0`

`<=> 25x^2 - 15x - 35x + 21 = 0`

`<=> (25x^2 - 15x) - (35x - 21) = 0`

`<=> 5x(5x - 3) - 7(5x - 3) = 0`

`<=> (5x - 7)(5x - 3) = 0`

`<=>`\(\left[{}\begin{matrix}5x-7=0\\5x-3=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}5x=7\\5x=3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy, `S = {7/5; 3/5}`

`4,`

\(25x^2-10x+1=0\)

`<=> 25x^2 - 5x - 5x + 1 = 0`

`<=> (25x^2 - 5x) - (5x - 1) = 0`

`<=> 5x(5x - 1) - (5x - 1) = 0`

`<=> (5x - 1)(5x-1)=0`

`<=> (5x-1)^2 = 0`

`<=> 5x - 1 = 0`

`<=> 5x = 1`

`<=> x = 1/5`

Vậy,` S = {1/5}.`

1)    x^2-x-(3x-3)=0

⇔   X^2-x-3x+3=0

⇔  x^2-4x+3     =0

⇔x^2-3x-x+3    =0

⇔ x(x-3)-(x-3)   =0

⇔(x-1)(x-3)       =0

⇔  x-1=0       -> x=1

      x-3=0       ->  x=3

Vậy tập nghiệm S={ 1;3}

\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)