a)\(x^3y+3x^2y\)

b)\(-24x^2+4xy\)

m) \(\dfrac{1}{4}x^2-4x^2=\left(\dfrac{1}{2}x-2x\right)\left(\dfrac{1}{2}x+2x\right)\)

n) \(\dfrac{4}{49}-4x^2=\left(\dfrac{2}{7}-2x\right)\left(\dfrac{2}{7}+2x\right)\)

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

Mik cần gấp 

(x+1)+(x+2)+(x+3)=4x

x+1+x+2+x+3=4x

(x+x+x)+(1+2+3)=4x

x*3+6=4x

6=1*x(bớt cả hai vế đi 3*x)

x=6/1(Tìm thừa số)

x=6

a,<=>   x²-4x+2²+y²-8y+4²-14

<=> (x²-2x2+2²)+(y²-2x4+4²)-14

<=> (x-2)²+(y-4)²-14 

Vì (x-2)²+(y-4)²>= 0

=> F >= -14 => MIn F = -14 <=> x=2, y=4

b, <=> (x²+5²+(2y)²-4xy+10x-20y) +(y²-2y+1)+2

<=> (x+5-2y )²+(y-1)²+2 

Vì (x+5-2y) ²+(y-1)² >= 0

=> G >= 2 => Min =2 <=> y=1, x= -3

\(F=x^2-4x+y^2-8y+6\)

\(F=\left(x^2-2.2x+2^2\right)+\left(y^2-2.4.y+4^2\right)-14\)

\(F=\left(x-2\right)^2+\left(y-4\right)^2-14\)

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\left(y-4\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\forall x\)

\(F=-14\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=4\end{cases}}\)

Vậy \(F_{min}=-14\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)

a,\(A=\left(\frac{2x-x^2}{2\left(x^2+4\right)}-\frac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\left(\frac{2x+x^2\left(1-x\right)}{x^3}\right)\left(ĐKXĐ:x\ne2;x\ne0\right)\)

\(A=\frac{\left(2x-x^2\right)\left(x-2\right)-4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{-x^3+x^2+2x}{x^3}\)

\(=\frac{-x^3-4x}{2\left(x^2+4\right)\left(x-2\right)}.\frac{x^2-x-2}{-x^2}\)

\(=\frac{-x\left(x^2+4\right)}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{-x^2}=\frac{x+1}{2x}\)

b, \(A=x\Leftrightarrow\frac{x+1}{2x}=x\Rightarrow2x^2=x+1\Leftrightarrow2x^2-x-1=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)(thỏa mãn điều kiện)

c, \(A\in Z\Leftrightarrow\frac{x+1}{2x}\in Z\Leftrightarrow x+1⋮\left(2x\right)\)

\(\Leftrightarrow2x+2⋮2x\Leftrightarrow2⋮2x\Leftrightarrow1⋮x\Leftrightarrow x=\pm1\) (thỏa mãn ĐKXĐ)

1. -4x( x + 3 )( x - 4 ) - 3x( x² - x + 1 )

= -4x( x² - x - 12 ) - 3x( x² - x + 1 )

= -4x³ + 4x² + 48x - 3x³ + 3x² - 3x

= -7x³ + 7x² + 45x

2. a) 4x( x - 5 ) - ( x - 1 )( 4x - 3 ) = 5

<=> 4x² - 20x - ( 4x² - 7x + 3 ) = 5

<=> 4x² - 20x - 4x² + 7x - 3 = 5

<=> -13x - 3 = 5

<=> -13x = 8

<=> x = -8/13

b) 6( x - 3 )( x - 4 ) - 6x( x - 2 ) = 4

<=> 6( x² - 7x + 12 ) - 6x² + 12x = 4

<=> 6x² - 42x + 72 - 6x² + 12x = 4

<=> -30x + 72 = 4

<=> -30x = -68

<=> x = 34/15

Bài 1 : 

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

\(=-7x^3+7x^2+45x\)

Bài 2 : 

a, \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)

\(\Leftrightarrow4x^2-20x-\left[4x^2-7x+3\right]=5\)

\(\Leftrightarrow4x^2-20x-4x^2+7x-3=5\)

\(\Leftrightarrow-13x-8=0\Leftrightarrow x=-\frac{8}{13}\)

b, \(6\left(x-3\right)\left(x-4\right)-6x\left(x-2\right)=4\)

\(\Leftrightarrow6x^2-42x+72-6x^2+12x=4\)

\(\Leftrightarrow-30x+68=0\Leftrightarrow x=\frac{34}{15}\)