Lời giải:

 $\frac{x}{y}=\frac{2}{3}\Rightarrow \frac{x}{2}=\frac{y}{3}$. Đặt $\frac{x}{2}=\frac{y}{3}=k$ thì:

$x=2k; y=3k$

Khi đó: $3x-2y=3.2k-3.2k=0$. Mẫu số không thể bằng $0$ nên $A$ không xác định. Bạn xem lại.

$B=\frac{2(2k)^2-2k.3k+3(3k)^2}{3(2k)^2+2.2k.3k+(3k)^2}=\frac{29k^2}{33k^2}=\frac{29}{33}$

b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)

\(=\dfrac{2y^2+8y+12}{y-1}\)

\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

bài 1: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{x}{x+2}-\dfrac{x}{x-2}\)

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

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

Bài 2:

1: \(x^2y^2-8-1\)

\(=x^2y^2-9\)

\(=\left(xy-3\right)\left(xy+3\right)\)

2: \(x^3y-2x^2y+xy-xy^3\)

\(=xy\cdot x^2-xy\cdot2x+xy\cdot1-xy\cdot y^2\)

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

\(=xy\left[\left(x-1\right)^2-y^2\right]\)

\(=xy\left(x-1-y\right)\left(x-1+y\right)\)

3: \(x^3-2x^2y+xy^2\)

\(=x\cdot x^2-x\cdot2xy+x\cdot y^2\)

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

4: \(x^2+2x-y^2+1\)

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

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

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

5: \(x^2+2x-4y^2+1\)

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

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

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

6: \(x^2-6x-y^2+9\)

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

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

a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2

b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y

=>A-B=12xy^2-14x^2y

c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2

=>A-B=-5x^2y^3-x^3y^2

d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2

Thay \(x=\dfrac{1}{2};y=-1\) vào B, ta được:

\(B=\left[\left(\dfrac{1}{2}\right)^3-4\cdot\left(\dfrac{1}{2}\right)^2\cdot\left(-1\right)+3\cdot\left(-1\right)^2-4\right]:\left[3\cdot\left(\dfrac{1}{2}\right)^3-3\cdot\left(-1\right)^2-3\cdot\left(-1\right)\right]\)

\(=\left(\dfrac{1}{8}+4\cdot\dfrac{1}{4}+3\cdot1-4\right):\left(3\cdot\dfrac{1}{8}-3\cdot1+3\right)\)

\(=\left(\dfrac{1}{8}+1+3-4\right):\left(\dfrac{3}{8}-3+3\right)\)

\(=\dfrac{1}{8}\cdot\dfrac{8}{3}=\dfrac{1}{3}\)

\(B=\dfrac{3}{4}xy^2-\dfrac{1}{3}x^2y-\dfrac{5}{6}xy^2+2x^2y=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y\)

Bậc:3

Thay x=-1, y=1 vào B ta có:

\(B=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y=-\dfrac{1}{12}.\left(-1\right).1^2+\dfrac{5}{3}.\left(-1\right)^2.1=\dfrac{1}{12}+\dfrac{5}{3}=\dfrac{7}{4}\)

2: Thay \(x=\dfrac{1}{2}\) và y=2 vào M, ta được:

\(M=\dfrac{2\cdot\left(\dfrac{1}{2}\right)^2\cdot2-1.2\cdot\left(3\cdot\dfrac{1}{2}-2\cdot2\right)}{\dfrac{1}{2}\cdot2}\)

\(=4\cdot\dfrac{1}{4}-1.2\left(\dfrac{3}{2}-4\right)\)

\(=1-1.8+4.8\)

\(=4\)

1: Ta có: \(\left(-\dfrac{2}{3}x^3y^2\right)z\cdot5xy^2z^2\)

\(=\left(-\dfrac{2}{3}\cdot5\right)\cdot\left(x^3\cdot x\right)\cdot\left(y^2\cdot y^2\right)\cdot\left(z\cdot z^2\right)\)

\(=\dfrac{-10}{3}x^4y^4z^3\)

a: \(=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}\)

b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)

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

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

c: \(=\dfrac{x+3}{x+2}\cdot\dfrac{2x-1}{3\left(x+3\right)}\cdot\dfrac{2\left(x+2\right)}{2\left(2x-1\right)}\)

=1/3

d: \(=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)\)

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