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3 tháng 2 2019

\(2x^3+7x^2+7x+2=0\)

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

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

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

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

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

.......................................................................................

\(x^3-8x^2-8x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-8x\left(x+1\right)=0\)

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11 tháng 2 2019

cảm ơn nha 

12 tháng 8 2021

1/ x2-3x+2=0

⇒ (x2-2x)-(x-2)=0

⇒ x(x-2)-(x-2)=0

⇒ (x-1)(x-2)=0

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

2) x2-6x+5=0

⇒x2-6x+9-4=0

⇒(x2-6x+9)-22=0

⇒(x-3)2-22=0

⇒(x-3-2)(x-3+2)=0

⇒(x-5)(x-1)=0

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

3) 2x2+5x+3=0

⇒ (2x2+2x)+(3x+3)=0

⇒ 2x(x+1)+3(x+1)=0

⇒ (x+1)(2x+3)=0

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

4) x2-8x+15=0

⇒ (x2-8x+16)-1=0

⇒ (x-4)2-12=0

⇒ (x-4-1)(x-4+1)=0

⇒ (x-5)(x-3)=0

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

5) x2-x-12=0

⇒ (x2-4x)+(3x-12)=0

⇒ x(x-4)+3(x-4)=0

⇒ (x-4)(x+3)=0

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

1: Ta có: \(x^2-3x+2=0\)

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

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

2: Ta có: \(x^2-6x+5=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

3: Ta có: \(2x^2+5x+3=0\)

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

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

4: Ta có: \(x^2-8x+15=0\)

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

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

5: Ta có: \(x^2-x-12=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

1: Ta có: \(2x\left(x+3\right)-6\left(x-3\right)=0\)

\(\Leftrightarrow2x^2+6x-6x+18=0\)

\(\Leftrightarrow2x^2+18=0\left(loại\right)\)

2: Ta có: \(2x^2\left(2x+3\right)+\left(2x+3\right)=0\)

\(\Leftrightarrow2x+3=0\)

hay \(x=-\dfrac{3}{2}\)

3: Ta có: \(\left(x-2\right)\left(x+1\right)-4x\left(x-2\right)=0\)

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

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

4: Ta có: \(2x\left(x-5\right)-3x+15=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)

5: Ta có: \(3x\left(x+4\right)-2x-8=0\)

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

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

6: Ta có: \(x^2\left(2x-6\right)+2x-6=0\)

\(\Leftrightarrow2x-6=0\)

hay x=3

1: Ta có: \(x^2+7x+6=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)

2: Ta có: \(x^2+7x+12=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)

3: Ta có: \(x^2+8x+15=0\)

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

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

4: Ta có: \(x^2+5x+4=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\end{matrix}\right.\)

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

14 tháng 7 2023

`@` `\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}.`

12 tháng 8 2021

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

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

2/ x (2x-1) (3x+15) =0

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\3x+15=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-5\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)

4/ (2x-10)(x2+1)=0

\(\Rightarrow\left[{}\begin{matrix}2x-10=0\\x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x^2=-1\left(loại\right)\end{matrix}\right.\)

5/ (x2+3) (2x-1) =0

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

6/ (3x-1) (2x2 +1)=0

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

 

1: Ta có: \(\left(x-1\right)\left(2x+1\right)=0\)

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

2: Ta có: \(x\left(2x-1\right)\left(3x+15\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\3x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-5\end{matrix}\right.\)

3: Ta có: \(\left(2x-6\right)\left(3x+4\right)x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)

26 tháng 1 2021

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}

30 tháng 7 2021

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

\(\Rightarrow x=32\)

30 tháng 7 2021

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