a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)

b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)

c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)

d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)

e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)

f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)

\(a,\dfrac{21x^2y^3}{24x^3y^2}=\dfrac{7y}{8x}\)

\(b,\dfrac{15xy^3\left(x^2-y^2\right)}{20x^2y\left(x+y\right)^2}=\dfrac{15xy^3\left(x-y\right)\left(x+y\right)}{20x^2y\left(x+y\right)^2}=\dfrac{3y^2\left(x-y\right)}{4x\left(x+y\right)}=\dfrac{3xy^2-3y^3}{4x^2+4xy}\)

a) Ta có: \(\dfrac{21x^2y^3}{24x^3y^2}\)

\(=\dfrac{21x^2y^3:3x^2y^2}{24x^3y^2:3x^2y^2}\)

\(=\dfrac{7y}{8x}\)

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.

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

\(\frac{x^2-x-6}{x^2+7x+10}\)

\(=\frac{x^2-3x+2x-6}{x^2+5x+2x+10}=\frac{x.\left(x-3\right)+2.\left(x-3\right)}{x.\left(x+5\right)+2.\left(x+5\right)}\)

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

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: \(\dfrac{x^2+7x+12}{x^2+5x+6}\)

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

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

a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)

b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)

c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)

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

e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)

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

\(b,\dfrac{x^3y+xy^3+xy}{x^3+y^3+x^2y+xy^2+x+y}\)

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

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

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

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

\(c,\dfrac{\left(3x+2\right)^2-\left(x+2\right)^2}{x^3-x^2}\)

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

\(=\dfrac{2x.\left(4x+4\right)}{x\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{8\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{8}{x-1}\)

a) ĐKXĐ: \(x\ne2y,x\ne-y;x\ne-1\) 

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

\(B=\left[\dfrac{y-x}{x-2y}-\dfrac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right]:\dfrac{4x^4+4x^2y+y^2-4}{x\left(x+y\right)+\left(x+y\right)}\)

\(B=\left[\dfrac{\left(y-x\right)\left(x+y\right)}{\left(x-2y\right)\left(x+y\right)}-\dfrac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right]:\dfrac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)

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

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

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

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