x₁₌\(-\frac{1}{3}\)         x₂=\(\frac{1}{2}\)            x₃=\(\frac{1}{4}\)

cach lam the nao

c: =>\(\dfrac{2x-1}{\left(x+5\right)\left(x-1\right)}+\dfrac{x-2}{\left(x-1\right)\left(x-9\right)}=\dfrac{3x-12}{\left(x-9\right)\left(x+5\right)}\)

=>(2x-1)(x-9)+(x-2)(x+5)=(3x-12)(x-1)

=>2x^2-19x+9+x^2+3x-10=3x^2-15x+12

=>-16x-1=-15x+12

=>-x=13

=>x=-13

=> 24x³ - 4x² - 4x - 6x² + x + 1 = 0 

=> 4x.(6x² - x - 1) - (6x² - x - 1) = 0

=> (6x² - x - 1)(4x - 1) = 0

=> (6x² - 3x + 2x - 1) (4x - 1) = 0

=> [ 3x.(2x - 1) + (2x - 1) ] . (4x - 1) = 0

=> (2x - 1)(3x + 1).(4x - 1) = 0

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

hoặc 3x + 1 = 0 => x = -1/3

hoặc 4x - 1 = 0 => x = 1/4

Vậy x = 1/2 , x = -1/3 , x = 1/4

`a)16x^2-24x+9=25`

`<=>(4x-3)^2=25`

`+)4x-3=5`

`<=>4x=8<=>x=2`

`+)4x-3=-5`

`<=>4x=-2`

`<=>x=-1/2`

`b)x^2+10x+9=0`

`<=>x^2+x+9x+9=0`

`<=>x(x+1)+9(x+1)=0`

`<=>(x+1)(x+9)=0`

`<=>` \(\left[ \begin{array}{l}x=-9\\x=-1\end{array} \right.\) 

`c)x^2-4x-12=0`

`<=>x^2+2x-6x-12=0`

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

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

`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\) 

`d)x^2-5x-6=0`

`<=>x^2+x-6x-6=0`

`<=>x(x+1)-6(x+1)=0`

`<=>(x+1)(x-6)=0`

`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\) 

`e)4x^2-3x-1=0`

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

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

`<=>` \(\left[ \begin{array}{l}x=1\\x=-\dfrac14\end{array} \right.\) 

`f)x^4+4x^2-5=0`

`<=>x^4-x^2+5x^2-5=0`

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

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

Vì `x^2+5>=5>0`

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

`<=>` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\) 

\(A=\dfrac{4\left(x^2-4x+4\right)+\left(x^2-8x+16\right)}{x^2-4x+4}=4+\left(\dfrac{x-4}{x-2}\right)^2\ge4\)

\(A_{min}=4\) khi \(x=4\) (A max ko tồn tại)

\(B=\dfrac{6\left(x^2+2x+1\right)+\left(4x^2+12x+9\right)}{x^2+2x+1}=6+\left(\dfrac{2x+3}{x+1}\right)^2\ge6\)

\(B_{min}=6\) khi \(x=-\dfrac{3}{2}\) 

B max ko tồn tại

a/ \(x^2-4y^2-3x+6y=\left(x^2-4y^2\right)-\left(3x-6y\right)=\left(x-2y\right)\left(x+2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x+2y-3\right)\)

b/ \(a^2+2ab+b^2-ac-bc=\left(a^2+2ab+b^2\right)-\left(ac+bc\right)=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b\right)\left(a+b-c\right)\)

c/ \(25x^2-10x-3x=25x^2-13x=x\left(25x-13\right)\)

d/ \(16x^2+24x-7=16x^2-4x+28x-7=4x\left(4x-1\right)+7\left(4x-1\right)=\left(4x-1\right)\left(4x+7\right)\)

1) \(P=-2x^2-12x=-2\left(x^2+6x+9\right)+18=-2\left(x+3\right)^2+18\le18\)

\(maxP=18\Leftrightarrow x=-3\)

2) \(Q=-5x^2+10x=-5\left(x^2-2x+1\right)+5=-5\left(x-1\right)^2+5\le5\)

\(maxQ=5\Leftrightarrow x=1\)

3) \(A=-3x^2+12x-6=-3\left(x^2-4x+4\right)+6=-3\left(x-2\right)^2+6\le6\)

\(maxA=6\Leftrightarrow x=2\)

4) \(B=-2x^2-24x+12=-2\left(x^2+12x+36\right)+84=-2\left(x+6\right)^2+84\le84\)

\(maxB=84\Leftrightarrow x=-6\)

3: \(15x^3+29x^2-8x-12\)

\(=15x^3+30x^2-x^2-2x-6x-12\)

\(=\left(x+2\right)\left(15x^2-x-6\right)\)

\(=\left(x+2\right)\left(15x^2-10x+9x-6\right)\)

\(=\left(x+2\right)\left(3x-2\right)\left(3x+5\right)\)

5: \(x^3+9x^2+26x+24\)

\(=x^3+4x^2+5x^2+20x+6x+24\)

\(=\left(x+4\right)\left(x^2+5x+6\right)\)

\(=\left(x+4\right)\left(x+2\right)\left(x+3\right)\)