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

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

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

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

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

Vì x^2 + 2 > 0  \(\forall x\)

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

Vậy ...

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

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

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

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+2\right)=0\text{ mà }x^2+2>0\text{ nên:}x-1=0\text{ hoặc:}x+2=0\)

x=1 hoặc x=-2

bài 1: 

2(x^2-9).4(x^2-1)

=(2x^2-18)(4x^2-4)

=8x^4-8x^2-72x^2+72

=8x^4-80x^2+72

\(Bai1:2\left(x-3\right)\left(x+3\right)+4\left(x-1\right)\left(x+1\right)\)

\(=2\left(x^2-9\right)+4\left(x^2-1\right)\)

\(=2x^2-18+4x^2-4\)

\(=6x^2-22\)

\(Bai2:-\left(6x-1\right)\left(3-2x\right)+\left(3x-2\right)\left(4x-3\right)=17\)

\(\Leftrightarrow-\left(18x-12x^2-3+2x\right)+12x^2-9x-8x+6=17\)

\(\Leftrightarrow-18x+12x^2+3-2x+12x^2-9x-8x+6=17\)

\(\Leftrightarrow24x^2-37x+9-17=0\)

\(\Leftrightarrow24x^2-37x-8=0\)

Đề sai??

a: \(\Leftrightarrow4\left(2x+1\right)-3\left(6x-1\right)=2x+1\)

=>8x+4-18x+3=2x+1

=>-10x+7=2x+1

=>-12x=-6

hay x=1/2

b: \(\Leftrightarrow4x^2-12x+7x-21-x^2=3x^2+6x\)

=>5x-21=6x

=>-x=21

hay x=-21

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

Mình khuyên bạn thế này : 

Bạn nên tách những câu hỏi ra 

Như vậy các bạn sẽ dễ giúp

Và cũng có nhiều bạn giúp hơn !

Bài 1.

a) ( x - 3 )( x + 7 ) = 0

<=> x - 3 = 0 hoặc x + 7 = 0

<=> x = 3 hoặc x = -7

Vậy S = { 3 ; -7 }

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

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

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

<=> x - 2 = 0 hoặc 2x - 5 = 0

<=> x = 2 hoặc x = 5/2

Vậy S = { 2 ; 5/2 }

c) x² - 5x + 6 = 0

<=> x² - 2x - 3x + 6 = 0

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

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

<=> x - 2 = 0 hoặc x - 3 = 0

<=> x = 2 hoặc x = 3

Bài 2.

a) \(\frac{x}{x+1}-1=\frac{3}{2}x\)

ĐKXĐ : x khác -1

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

<=> \(\frac{-1}{x+1}=\frac{3x}{2}\)

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

<=> 3x² + 3x + 2 = 0

Vi 3x² + 3x + 2 = 3( x² + x + 1/4 ) + 5/4 = 3( x + 1/2 )² + 5/4 ≥ 5/4 > 0 ∀ x

=> phương trình vô nghiệm

b) \(\frac{4x}{x-2}-\frac{7}{x}=4\)

ĐKXĐ : x khác 0 ; x khác 2

<=> \(\frac{4x^2}{x\left(x-2\right)}-\frac{7x-14}{x\left(x-2\right)}=\frac{4x^2-8x}{x\left(x-2\right)}\)

=> 4x² - 7x + 14 = 4x² - 8x

<=> 4x² - 7x - 4x² + 8x = -14

<=> x = -14 ( tm )

Vậy phương trình có nghiệm x = -14

\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

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

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

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

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

ĐK \(x\ge-3\)

PT <=> \(x^3+5x^2+6x+2=4\sqrt{x+3}+2\sqrt{2x+7}\)

<=> \(2\left(x+3-2\sqrt{x+3}\right)+\left(x+5-2\sqrt{2x+7}\right)+x^3+5x^2+3x-9=0\)

+  Với x=-3 =>thỏa mãn 

+Với \(x>-3\) ta liên hợp

\(2.\frac{x^2+2x-3}{x+3+2\sqrt{x+3}}+\frac{x^2+2x-3}{x+5+2\sqrt{2x+7}}+\left(x+3\right)\left(x^2+2x-3\right)=0\)

<=> \(\left(x^2+2x-3\right)\left(\frac{2}{x+3+2\sqrt{x+3}}+\frac{1}{x+5+2\sqrt{2x+7}}+x+3\right)=0\)

Do \(x>-3\)=> \(\frac{2}{x+3+2\sqrt{x+3}}+\frac{1}{x+5+2\sqrt{2x+7}}+x+3>0\)

=> \(x=1\)(TMĐKXĐ)

Vậy \(x=1;x=-3\)

giúp mình bài ni với :3x^2(x+1)-5x(x+1)^2+4(x+1)

a) \(\dfrac{x+1}{4}-\dfrac{5+2x}{8}=\dfrac{3-4x}{2}\)

⇔\(\dfrac{2\left(x+1\right)}{8}-\dfrac{5+2x}{8}=\dfrac{4\left(3-4x\right)}{8}\) 

⇔ 2x + 2 - 5 - 2x = 12 -16x

⇔ 16x = 15 

⇔ x = 15/16

b) \(\dfrac{4-3x}{5}-\dfrac{4-x}{10}=\dfrac{x+2}{2}\)

⇔\(\dfrac{2\left(4-3x\right)}{10}-\dfrac{4-x}{10}=\dfrac{5\left(x+2\right)}{10}\)

⇔ 8 - 6x - 4 + x = 5x + 10

⇔ 10x = -6

⇔ x = -6/10

Câu 1:

x + 1/4 - 5 + 2x/8 = 3 - 4x/2

<=> 2x + 2/8 - 5 + 2x/8 = 12 - 16x/8

<=> 2x + 2 - 5 - 2x = 12 - 16x

<=> -3 = 12 - 16x <=> 15 = 16x <=> x = 15/16

Câu 2:

4 - 3x/5 - 4 - x/10 = x + 2/2

<=> 8 - 6x/10 - 4 - x/10 = 5x + 10/10

<=> 8 - 6x - 4 + x = 5x + 10

<=> 4 - 5x = 5x + 10

<=> 4 = 10x + 10 <=> 10x = -6 <=> x = -3/5