Ta có:

 \(\begin{array}{l}S = 2{x^2}{y^3} - \dfrac{3}{5}{x^2}{y^3} - 14{x^2}{y^3} + \dfrac{8}{5}{x^2}{y^3}\\ = \left( {2 - \dfrac{3}{5} - 14 + \dfrac{8}{5}} \right){x^2}{y^3}\\ =  - 11{x^2}{y^3}.\end{array}\)

Quy đồng mẫu thức các phân thức sau :

a) 

\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)

\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)

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

\(=6x^2y\)

\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)

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

\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)

1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy

2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3

=6x^2y

3: =(x+y-x+y)^2=(2y)^2=4y^2

4: =(2x+3-2x-5)^2=(-2)^2=4

5: =18^8-18^8+1=1

\(3x^2y^3-5x^2y^3+x^2y^3=-x^2y^3\)

chọn B 

Bài 1.

a)

\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)

b)

\(x(x+3y+1)-2y(x-1)-(y+x+1)x\\=x^2+3xy+x-2xy+2y-xy-x^2-x\\=(x^2-x^2)+(3xy-2xy-xy)+(x-x)+2y\\=2y\)

Bài 2.

a)

\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)

b)

\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)

Ta có: \(P=3x^2y+\frac12x^2y^3+\frac13x^2y^3+\left(-\frac12x^2y^3\right)\)

\(=3x^2y+\frac13x^2y^3+\left(\frac12x^2y^3-\frac12x^2y^3\right)\)

\(=3x^2y+\frac13x^2y^3\)

Thay x=-3 và y=2 vào P, ta được:

\(P=3\cdot\left(-3\right)^2\cdot2+\frac13\cdot\left(-3\right)^2\cdot2^3\)

=54+1/3*9*8

=54+3*8

=54+24

=78

a: \(y^3+2xy^2+y^2-4x^2\)

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

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

\(\frac{8x^3+y^3}{y^3+2xy^2+y^2-4x^2}\)

\(=\frac{\left(2x+y\right)\left(4x^2-2xy+y^2\right)}{\left(2x+y\right)\left(y^2+y-2x\right)}=\frac{4x^2-2xy+y^2}{y^2+y-2x}\)

b: \(\frac{x^2-2x-8}{2x^2+9x+10}\)

\(=\frac{x^2-4x+2x-8}{2x^2+4x+5x+10}\)

\(=\frac{\left(x-4\right)\cdot\left(x+2\right)}{\left(x+2\right)\left(2x+5\right)}=\frac{x-4}{2x+5}\)

c: \(\frac{6x-x^2-5}{5x^6-x^7}\)

\(=\frac{x^2-6x+5}{x^7-5x^6}\)

\(=\frac{\left(x-5\right)\left(x-1\right)}{x^6\cdot\left(x-5\right)}=\frac{x-1}{x^6}\)

d: \(\frac{x^3+64}{2x^3-8x^2+32x}=\frac{\left(x+4\right)\left(x^2-4x+16\right)}{2x\left(x^2-4x+16\right)}=\frac{x+4}{2x}\)

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

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

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