Lời giải:
a)

$M=2x^3-4xy+6y^2+x^2+3xy-y^2=2x^3-xy+5y^2+x^2$

b)

$M=-(2x^2-4xy+y^2)=-2x^2+4xy-y^2$

c)

$2M=(2x^2-7xy+3y^2)-(4x^2-5xy+9y^2)=-2x^2-2xy-6y^2$

$\Rightarrow M=-x^2-xy-3y^2$

\(5xy\left(2x^3y^2-7xy+3y\right)=10x^4y^3-35x^2y^2+15xy^2\\ \left(-6x^6+15x^2-4x^4\right):3x^2=-2x^4+5-\dfrac{4}{3}x^2\\ \left(x^2-y^2-12x+36\right):\left(x+y-6\right)\\ =\left[\left(x-6\right)^2-y^2\right]:\left(x+y-6\right)\\ =\left(x-y-6\right)\left(x+y-6\right):\left(x+y-6\right)\\ =x-y-6\)

\((x+3y)^2\\=x^2+2\cdot x\cdot3y+(3y)^2\\=x^2+6xy+9y^2\\---\\(x-5xy)^2\\=x^2-2\cdot x\cdot5xy+(5xy)^2\\=x^2-10x^2y+25x^2y^2\)

\((5+9y)^3\\=5^3+3\cdot5^2\cdot9y+3\cdot5\cdot(9y)^2+(9y)^3\\=125+675y+1215y^2+729y^3\\---\\(6x-7xy)^3\\=(6x)^3-3\cdot(6x)^2\cdot7xy+3\cdot6x\cdot(7xy)^2-(7xy)^3\\=216x^3-756x^3y+882x^3y^2-343x^3y^3\)

a. \(2a^2+5ab-3b^2-7b-2\)

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

\(=2a\left(a+3b+1\right)-b\left(a+3b+1\right)-2\left(a+3b+1\right)\)

\(=\left(2a-b-2\right)\left(a+3b+1\right)\)

b. \(2x^2-7xy+x+3y^2-3y\)

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

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

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

\(=\left(x-3y\right)\left(2x-y+1\right)\)

c. \(6x^2-xy-2y^2+3x-2y\)

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

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

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

\(=\left(3x-2y\right)\left(2x+y+1\right)\)

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:

Với điều kiện xy\(\ne\)0;+ -3/2 y;x\(\ne\)-y các phân thức có nghĩa. Ta có

\(\frac{5x\left(2x-3y\right)^2}{3y\left(4x^2-9y^2\right)}:\frac{\left(2x^2+2xy\right)\left(2x-3y\right)}{2x^2y+5xy^2+3y^3}\)\(=\)\(\frac{5x\left(2x-3y\right)^2.y\left(2x^2+5xy+3y^2\right)}{3y\left(4x^2-9y^2\right).2x\left(x+y\right).\left(2x-3y\right)}\)

\(=\)\(\frac{10xy\left(2x-3y\right)^2.\left(2x^2+2xy+3xy+3y^2\right)}{6xy\left(2x-3y\right).\left(2x+3y\right)\left(x+y\right)\left(2x-3y\right)}\)\(=\)\(\frac{10xy\left(2x-3y\right)^2\left(x+y\right).\left(2x+3y\right)}{6xy\left(2x-3y\right)^2.\left(2x+3y\right).\left(x+y\right)}\)

\(=\)\(\frac{5}{3}\)

ĐK \(\hept{\begin{cases}xy\ne0\\2x-3y\ne0,2x+3y\ne0\\x\ne-y\end{cases}}\)

\(=\frac{5x\left(2x-3y\right)^2}{3y\left(2x+3y\right)\left(2x-3y\right)}:\frac{2x\left(x+y\right)\left(2x-3y\right)}{xy\left(2x+3y\right)+y^2\left(2x+3y\right)}\)

\(=\frac{5x\left(2x-3y\right)}{3y\left(2x+3y\right)}:\frac{2x\left(x+y\right)\left(2x-3y\right)}{\left(2x+3y\right)\left(xy+y^2\right)}\)

\(=\frac{5x\left(2x-3y\right)}{3y\left(2x+3y\right)}.\frac{y\left(x+y\right)\left(2x+3y\right)}{2x\left(x+y\right)\left(2x-3y\right)}=\frac{5}{6}\)

Vậy giá trị của biểu thức không phụ thuộc vào biến

a) \(=x^2+4xy+4y^2-y^2\)

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

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

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

b) \(=2x^2-4xy-xy+2y^2\)

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

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