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15 tháng 11 2016

Phân tích đa thức thành nhân tử , ta đươc :

\(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x^2+x+6\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x_1=-2\\x_2=1\end{array}\right.;x^2+x+6=\left(x+\frac{1}{2}\right)^2+5\frac{3}{4}\ne0\forall x.\)

Vậy pt đã cho các nghiệm : \(x_1=-2;x_2=1.\)

24 tháng 1 2021

(4x - 3)2 - (2x + 1)2 = 0

\(\Leftrightarrow\) (4x - 3 - 2x - 1)(4x - 3 + 2x + 1) = 0

\(\Leftrightarrow\) (2x - 4)(6x - 2) = 0

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

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\)

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

Vậy ...

3x - 12 - 5x(x - 4) = 0

\(\Leftrightarrow\) 3x - 12 - 5x2 + 20x = 0

\(\Leftrightarrow\) -5x2 + 23x - 12 = 0

\(\Leftrightarrow\) 5x2 - 23x + 12 = 0

\(\Leftrightarrow\) 5x2 - 20x - 3x + 12 = 0

\(\Leftrightarrow\) 5x(x - 4) - 3(x - 4) = 0

\(\Leftrightarrow\) (x - 4)(5x - 3) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-4=0\\5x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy ...

(8x + 2)(x2 + 5)(x2 - 4) = 0

\(\Leftrightarrow\) (8x + 2)(x2 + 5)(x - 2)(x + 2) = 0

Vì x2 \(\ge\) 0 \(\forall\) x nên x2 + 5 > 0 \(\forall\) x

\(\Rightarrow\) (8x + 2)(x - 2)(x + 2) = 0

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

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=2\\x=-2\end{matrix}\right.\)

Vậy ...

Chúc bn học tốt!

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

\(\Leftrightarrow\left(4x-3-2x-1\right)\left(4x-3+2x+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{2;\dfrac{1}{3}\right\}\)

b) Ta có: \(3x-12-5x\left(x-4\right)=0\)

\(\Leftrightarrow3\left(x-4\right)-5x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{4;\dfrac{3}{5}\right\}\)

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

\(\Leftrightarrow2\left(4x+1\right)\left(x^2+5\right)\left(x-2\right)\left(x+2\right)=0\)

mà \(2>0\)

và \(x^2+5>0\forall x\)

nên \(\left(4x+1\right)\left(x-2\right)\left(x+2\right)=0\)

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

Vậy: \(S=\left\{-\dfrac{1}{4};2;-2\right\}\)

16 tháng 4 2020

Hình như đề của bạn sai nên mình sửa lại nhé

x4 + 2x3 +5x2 +4x-12=0

⇔x4-x3+3x3-3x2+8x2-8x+12x-12=0

⇔x3(x-1)+3x2(x-1)+8x(x-1)+12(x-1)=0

⇔(x-1)(x3+3x2+8x+12)=0

⇔(x-1)(x+2)(x2+x+6)=0

ta có x2+x+6 >0 ∀x

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

Vậy...

27 tháng 3 2020

Đề sai không bạn

3 tháng 2 2018

x^4 + 2x^3 + 5x^2 + 4x-12 = 0 
<=> (x^4 - x^3) + (3x^3-3x^2) + (8x^2 - 8x) + (12x-12) = 0 
<=> (x-1).(x^3 + 3x^2 + 8x+12) = 0 
<=> (x-1).[(x^3+2x^2)+(x^2+2x)+(6x+12)] = 0 
<=>(x-1).(x+2).(x^2+x+6) = 0 
<=> x= 1 hoặc x = -2 

Chúc học tốt ( hên xui đó nha )

3 tháng 2 2018

\(x^4+2x^3+5x^2+4x-12=0.\)

\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}}\)

\(\text{Vì }x^2+x+6=\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\ge\frac{23}{4}\left(\text{nên vô No}\right)\)

1) Ta có: 3x-12=5x(x-4)

\(\Leftrightarrow3x-12-5x\left(x-4\right)=0\)

\(\Leftrightarrow3x-12-5x^2+20x=0\)

\(\Leftrightarrow-5x^2+23x-12=0\)

\(\Leftrightarrow-5x^2+20x+3x-12=0\)

\(\Leftrightarrow\left(-5x^2+20x\right)+\left(3x-12\right)=0\)

\(\Leftrightarrow5x\left(-x+4\right)+3\left(x-4\right)=0\)

\(\Leftrightarrow5x\left(4-x\right)-3\left(4-x\right)=0\)

\(\Leftrightarrow\left(4-x\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\5x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{3}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{4;\frac{3}{5}\right\}\)

2) Ta có: 3x-15=2x(x-5)

\(\Leftrightarrow3x-15-2x\left(x-5\right)=0\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

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

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

Vậy: \(x\in\left\{5;\frac{3}{2}\right\}\)

3) Ta có: 3x(2x-3)+2(2x-3)=0

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

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

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

4) Ta có: (4x-6)(3-3x)=0

\(\Leftrightarrow\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=6\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{6}{4}=\frac{3}{2}\\x=1\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{3}{2};1\right\}\)

10 tháng 2 2020

4) (4x - 6 ) ( 3 - 3x ) = 0

<=> \(\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}4x=6\\3x=3\end{matrix}\right.\)

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

30 tháng 1 2018

a) 2x2-4x-x+2=0

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

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

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

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

b) 3x2-12x+5x-20=0

=> 3x(x-4)+5.(x-4)=0

=> (x-4)(3x+5)=0

=> \(\left[{}\begin{matrix}x-4=0\\3x+5=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=4\\x=-\dfrac{5}{3}\end{matrix}\right.\)

c)x3+2x2-x2-2x+2x+4=0

=> x2(x+2)-x(x+2)+2(x+2)=0

=>(x2-x+2)(x+2)=0

=> x=-2( vi x2-x+2>0)

d) x3-x2-4x2+4x+4x-4=0

=> x2(x-1)-4x(x-1)+4(x-1)=0

=>(x-1)(x2-4x+4)=0

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

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

30 tháng 1 2018

2x2-5x+2=0

⇔2x2-x-4x+2=0

⇔x(2x-1)-2(2x-1)=0

⇔(x-2)(2x-1)=0

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

sậy S=\(\left\{2;\dfrac{1}{2}\right\}\)

x3+x2+4=0

⇔x3+2x2-x2-2x+2x+4=0

⇔(x3+2x2)-(x2+2x)+(2x+4)=0

⇔x2(x+2)-x(x+2)+2(x+2)=0

⇔(x+2)(x2-x+2)=0

⇔x+2=0 và x2-x+2=0

⇔x=-2 và \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\)(vô lý)

vậy S={-2}

4 tháng 5 2018

A, 2x-7=5x+20

<=>3x=-27

<=>x=-9

B, x^3-4x=0

<=>x(x2​-4)=0

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

<=>x=0, 2,-2

C, 

4 tháng 5 2018

Y c sao

a) \(x^4+2x^3-12x^2-13x+42=0\)

\(\Leftrightarrow x^4+3x^3-x^3-3x^2-9x^2-27x+14x+42=0\)

\(\Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)-9x\left(x+3\right)+14\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^3-x^2-9x+14\right)=0\)

8 tháng 4 2018

\(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x^2+12x-12=0\)

\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)

Ta có:

\(x^2+x+6=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{23}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0\)

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

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

Vậy...........