a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)

\(=2x^2-x-2x+\dfrac{5}{2}+4x^2\)

\(=6x^2-3x+\dfrac{5}{2}\)

b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)

\(=3x-y-y-x+2x^2-2x\)

\(=2x^2-2y\)

a) 5x² ( 3x² -7x+2)-15x(x-3)

=15x⁴-35x³+10x²-15x²+45x

=15x⁴-35x³-5x²+45x

c) (x+3)(x-3)(x-2)(x+1)

=(x²-9)(x²+x-2x-2)

=(x²-9)(x²-x-2)

=x⁴-x³-2x²-9x²+9x+18

=x⁴-x³-11x²+9x+18

d)(2x+1)²+(4x-1)²+2(2x+1)(4x+1)

=2x²+4x+1-16x²-8x+1

=2x²+4x+1-16x²-8x+1+16x²-4x+8x-2

=2x²+7

e) (2x²-3x)(5x²-2x+1)-10x²(x+3)

=10x⁴ -4x³+2x²-15x³+6x²-3 -10x²-30x

=10x⁴-19x³-2x²-30x-3

thanks bn nka

Bài 1:

a: ĐKXĐ: \(x+4\ne0\)

=>\(x\ne-4\)

b: ĐKXĐ: \(2x-1\ne0\)

=>\(2x\ne1\)

=>\(x\ne\dfrac{1}{2}\)

c: ĐKXĐ: \(x\left(y-3\right)\ne0\)

=>\(\left\{{}\begin{matrix}x\ne0\\y-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\y\ne3\end{matrix}\right.\)

d: ĐKXĐ: \(x^2-4y^2\ne0\)

=>\(\left(x-2y\right)\left(x+2y\right)\ne0\)

=>\(x\ne\pm2y\)

e: ĐKXĐ: \(\left(5-x\right)\left(y+2\right)\ne0\)

=>\(\left\{{}\begin{matrix}x\ne5\\y\ne-2\end{matrix}\right.\)

 Bài 2:

a: \(\dfrac{-12x^3y^2}{-20x^2y^2}=\dfrac{12x^3y^2}{20x^2y^2}=\dfrac{12x^3y^2:4x^2y^2}{20x^2y^2:4x^2y^2}=\dfrac{3x}{5}\)

b: \(\dfrac{x^2+xy-x-y}{x^2-xy-x+y}\)

\(=\dfrac{\left(x^2+xy\right)-\left(x+y\right)}{\left(x^2-xy\right)-\left(x-y\right)}\)

\(=\dfrac{x\left(x+y\right)-\left(x+y\right)}{x\left(x-y\right)-\left(x-y\right)}=\dfrac{\left(x+y\right)\left(x-1\right)}{\left(x-y\right)\left(x-1\right)}\)

\(=\dfrac{x+y}{x-y}\)

c: \(\dfrac{7x^2-7xy}{y^2-x^2}\)

\(=\dfrac{7x\left(x-y\right)}{\left(y-x\right)\left(y+x\right)}\)

\(=\dfrac{-7x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{-7x}{x+y}\)
d: \(\dfrac{7x^2+14x+7}{3x^2+3x}\)

\(=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)

\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)

e: \(\dfrac{3y-2-3xy+2x}{1-3x-x^3+3x^2}\)

\(=\dfrac{3y-2-x\left(3y-2\right)}{1-3x+3x^2-x^3}\)

\(=\dfrac{\left(3y-2\right)\left(1-x\right)}{\left(1-x\right)^3}=\dfrac{3y-2}{\left(1-x\right)^2}\)

g: \(\dfrac{x^2+7x+12}{x^2+5x+6}\)

\(=\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x+3\right)\left(x+2\right)}\)

\(=\dfrac{x+4}{x+2}\)

1) 30x-30x^2-31

2)6x^4-2x^3-15x^2+23x-6

a) 2x².(5x³-4x²y-7xy +1) =10x⁵-8x⁴y-14x³y+2x² b) (5x -2y)(x² -xy +1) =5x³-5x²y+5x-2x²y+2xy²-2y =5x³-7x²y+2xy²+5x-2y c) (\(\dfrac{1}{2}\)x -1)(2x -3) =x²-\(\dfrac{3}{2}\)x-2x+3 =x²-\(\dfrac{7}{2}\)x+3 d) (x +3y)² =x²+6xy+9y² e) (3x -2y)² =9x²-12xy+4y² g) (\(\dfrac{1}{4}\)x - 3y)(\(\dfrac{1}{4}\)x +3y) =\(\dfrac{1}{16}\)x²-9y² f) (2x +3)³ =8x³+36x²+54x+27 h) (3 -2y)³ =27-54y+36y²-8y³

a: Ta có: \(x^2-4x\left(3x-4\right)+7x-5\)

\(=x^2-12x^2+16x+7x-5\)

\(=-11x^2+23x-5\)

b: Ta có: \(7x\left(x^2-5\right)-3x^2y\left(xy-6y^2\right)\)

\(=7x^3-35x-3x^3y^2+18x^2y^3\)

c: Ta có: \(\left(5x+4\right)\left(2x-7\right)\)

\(=10x^2-35x+8x-28\)

\(=10x^2-27x-28\)