a: \(A=\left(\dfrac{2\left(2x+1\right)}{2\left(2x+4\right)}-\dfrac{x}{3x-6}-\dfrac{2x^3}{3x^3-12x}\right):\dfrac{6x+13x^2}{24x-12x^2}\)

\(=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^3}{3x\left(x^2-4\right)}\right):\dfrac{x\left(13x+6\right)}{x\left(24-12x\right)}\)

\(=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^2}{3\left(x-2\right)\left(x+2\right)}\right):\dfrac{13x+6}{-12\left(x-2\right)}\)

\(=\dfrac{3\left(2x+1\right)\left(x-2\right)-2x\left(x+2\right)-4x^2}{6\left(x+2\right)\left(x-2\right)}\cdot\dfrac{-12\left(x-2\right)}{13x+6}\)

\(=\dfrac{3\left(2x^2-3x-2\right)-2x^2-4x-4x^2}{x-2}\cdot\dfrac{-2}{13x+6}\)

\(=\dfrac{6x^2-9x-6-6x^2-4x}{x-2}\cdot\dfrac{-2}{13x+6}\)

\(=\dfrac{-\left(13x+6\right)\cdot\left(-2\right)}{\left(13x+6\right)\left(x-2\right)}=\dfrac{2}{x-2}\)

b: Để A>0 thì x-2>0

hay x>2

Để A>-1 thì A+1>0

\(\Leftrightarrow\dfrac{2+x-2}{x-2}>0\)

=>x/x-2>0

=>x>2 hoặc x<0

a) \(4x^2-4x=-1\)

\(\Leftrightarrow4x\left(x-1\right)=-1\)

\(\Leftrightarrow4x=-1\) hoặc \(x-1=-1\)

\(\Leftrightarrow x=\dfrac{-1}{4}\) hoặc \(x=0\)

Vậy S={\(\dfrac{-1}{4};0\)}

\(\text{a) }4x^2-4x=-1\\ \Leftrightarrow4x^2-4x+1=0\\ \Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\\ \Leftrightarrow\left(2x-1\right)^2=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\\ \text{Vậy }x=\dfrac{1}{2}\\ \)

\(\text{ b) }8x^3+12x^2+6x+1=0\\ \Leftrightarrow\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3=0\\ \Leftrightarrow\left(2x+1\right)^3=0\\ \Leftrightarrow2x+1=0\\ \Leftrightarrow2x=-1\\ \Leftrightarrow x-\dfrac{1}{2}\\ \text{Vậy }x=-\dfrac{1}{2}\)

1 , 

\(b,x^2-2x=0\)

\(\Rightarrow x\left(x-2\right)=0\)

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

KL :..

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

\(b,4x^2-4x+1=\left(2x\right)^2-2.2x+1\)

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

a) ( 4x - 1 ) ( x - 2 ) = 0

\(\Leftrightarrow\orbr{\begin{cases}4x-1=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=2\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{4};2\right\}\)

b) 4x² - 12x = 0

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

\(\Leftrightarrow\orbr{\begin{cases}4x=0\\x-3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x=3\end{cases}}\)

Vậy \(x\in\left\{0;3\right\}\)

c) ( x - 5 )4 + 25 - x² = 0

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

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

( x - 5 ) ( 9 + x ) = 0

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\9+x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-9\end{cases}}\)

Vậy \(x\in\left\{-9;5\right\}\)

a)x=0,25,x=2

b)x=3,x=0

a) 4x²-12x=9

<=> 4x(x-3)=9

<=> 4x=9                hoặc          x-3=9

=> x=4/9                              => x=12

b) 3.(x²-4)-5x(x+2)=0

<=> 3(x-2)(x+2)-5x(x+2)=0

<=> (x+2)(3x-6-5x)=0

<=> (x+2)(-2x-6)=0

<=> x+2=0                   hoặc -2x-6=0

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

đạng phân tich đa thức thành nhân tử nha các bn

a) \(4x^2-12x=-9\)

\(\Leftrightarrow4x^2-12x+9=0\)

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

\(\Leftrightarrow2x-3=0\Leftrightarrow x=\frac{3}{2}\)

b) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(25-4x^2\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(5+2x\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7+5+2x\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(4x+12\right)=0\)

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

c)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

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

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=0\\x=2\end{array}\right.\)

d) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right]\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=-\frac{23}{17}\end{array}\right.\)

e, Ta có: \(\Delta\)'\(=\left(-6\right)^2-4.5=16>0\)

Suy ra \(\sqrt{\Delta'}=\sqrt{16}=4\)

Vậy phương trình đã cho có 2 nghiệm phân biệt

\(x_1=\dfrac{-b'-\sqrt{\Delta'}}{a}=\dfrac{6-4}{4}=\dfrac{1}{2}\)

\(x_2=\dfrac{-b'+\sqrt{\Delta'}}{a}=\dfrac{6+4}{4}=\dfrac{5}{2}\)

Vậy phương trình đã cho có 2 nghiệm là 1/2;5/2

f,Ta có : a+-b+c=2-5+3=0

Do đó phương trình đã cho có 2 nghiệm \(x_1\)=-1 hoặc \(x_2=\dfrac{-c}{a}=-\dfrac{3}{2}\)

g,Ta có: a+b+c=1+1-2=0

Do phương trình đã cho có 2 nghiệm \(x_1\)=1 hoặc \(x_2=\dfrac{c}{a}=-2\)

h,Ta có a+b+c=1-4+3=0

Do đó phương trình đã cho có 2 nghiệm \(x_1=1\) hoặc \(x_2=\dfrac{c}{a}=3\)

g, \(x^2+x-2=0\)

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

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

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

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

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

Vậy......

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

\(\Rightarrow x^2-3x-x+3=0\)

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

\(\Rightarrow x.\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right).\left(x-1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy......

Chúc bạn học tốt!!!

a)12x-9-4x²=0

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

\(\Leftrightarrow2x-3=0\\ \Leftrightarrow x=\dfrac{3}{2}\)

b) x+x²-x³-x⁴ =0

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

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

=> x=0 hoặc x=1 hoặc x=-1

c)

b) Ta có: \(x^3-7x+6=0\)

\(\Leftrightarrow x^3-6x-x+6=0\)

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

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

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

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

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

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

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

Vậy: x∈{1;-3;2}

c) Ta có: \(x^4-4x^3+12x-9=0\)

\(\Leftrightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\\x^2-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=\pm\sqrt{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{3;1;\pm\sqrt{3}\right\}\)

d) Ta có: \(x^5-5x^3+4x=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=0\\x=\pm2\end{matrix}\right.\)

Vậy: x∈{-2;-1;0;1;2}

e) Ta có: \(x^4-4x^3+3x^2+4x-4=0\)

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

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

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

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

Vậy: x∈{-1;1;2}