Bài 1:

\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)

\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)

Bài 2:

\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)

Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9

Bài 4:

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

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

a) 4x²(x² - 5x + 2)

= 4x².x² - 4x².5x + 4x².2

= 4x⁴ - 20x³ + 8x²

b) (2x²  - 5x + 3) : (2x - 3)

= (2x² - 3x - 2x + 3) : (2x - 3)

= [(2x² - 3x) - (2x - 3)] : (2x - 3)

= [x(2x - 3) - (2x - 3)] : (2x - 3)

= (2x - 3)(x - 1) : (2x - 3)

= x - 1

a, \(4x^2\left(x^2-5x+2\right)\\ =4x^4-20x^3+8x^2\)

b, \(\left(2x^2-5x+3\right):\left(2x-3\right)\\ =x-1\)

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

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

Bài 4:

$(x-1)(x^2+x+1)-x^3-6x=11$

Dùng $(x-1)(x^2+x+1)=x^3-1$:

$x^3-1-x^3-6x=11$

$-6x-1=11$

$-6x=12$

$x=-2$

Vậy $x=-2$.

$16x^2-(3x-4)^2=0$

$(4x)^2-(3x-4)^2=0$

$(4x-3x+4)(4x+3x-4)=0$

$(x+4)(7x-4)=0$

$x=-4$ hoặc $x=\dfrac47$

Vậy $x=-4,\dfrac47$.

$x^3-x^2+3-3x=0$

$=x^2(x-1)-3(x-1)=0$

$=(x-1)(x^2-3)=0$

$x-1=0$ hoặc $x^2-3=0$

$x=1$ hoặc $x=\pm\sqrt3$

Vậy $x=1,\sqrt3,-\sqrt3$.

$\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}$

Điều kiện: $x\ne-2,-1$.

$(x-1)(x+1)=(x+2)^2$

$x^2-1=x^2+4x+4$

$-4x=5$

$x=-\dfrac54$

Vậy $x=-\dfrac54$.

$\dfrac1{x+2}+\dfrac2{x+1}=0$

Điều kiện: $x\ne-2,-1$.

$\dfrac{x+1+2(x+2)}{(x+2)(x+1)}=0$

$x+1+2x+4=0$

$3x+5=0$

$x=-\dfrac53$

Vậy $x=-\dfrac53$.

$\dfrac{9-x^2}{x}:(x-3)=1$

Điều kiện: $x\ne0,3$.

$\dfrac{9-x^2}{x(x-3)}=1$

$9-x^2=x(x-3)$

$9-x^2=x^2-3x$

$2x^2-3x-9=0$

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

$x=-\dfrac32$ hoặc $x=3$

Nhưng $x=3$ không thỏa điều kiện.

Vậy $x=-\dfrac32$.

\(\dfrac{10x^2\left(x+2\right)^3-8x^3\left(x+2\right)^2+4x\left(x+2\right)^3}{2x\left(x+2\right)^2}\)

\(=\dfrac{10x^2\left(x+2\right)^3}{2x\left(x+2\right)^2}-\dfrac{8x^3\left(x+2\right)^2}{2x\left(x+2\right)^2}+\dfrac{4x\left(x+2\right)^3}{2x\left(x+2\right)^2}\)

\(=\dfrac{2x\cdot\left(x+2\right)^2\cdot5x\cdot\left(x+2\right)}{2x\left(x+2\right)^2}-\dfrac{2x\cdot\left(x+2\right)^2\cdot4x^2}{2x\left(x+2\right)^2}+\dfrac{2x\left(x+2\right)^2\cdot2\cdot\left(x+2\right)}{2x\left(x+2\right)^2}\)

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

\(=5x^2+10x-4x^2+2x+4\)

\(=x^2+12x+4\)