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: \(=\dfrac{x-z}{2}\)

b: \(=\dfrac{3x}{4y^3}\)

Bài 2:

a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)

b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)

\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)

\(=x^4-22x^3+108x^2-45x\)

c: \(=12x^5-18x^4+30x^3-24x^2\)

d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)

a, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)

⇔ \(6x^2-5x+3=2x-9x+6x^2\)

⇔ \(6x^2-5x+3-6x^2+9x-2x=0\)

⇔ \(2x+3=0\)

⇔ \(2x=-3\)

⇔ \(x=-\dfrac{3}{2}\)

b, \(\dfrac{2\left(x-4\right)}{4}-\dfrac{3+2x}{10}=x+\dfrac{1-x}{5}\)

⇔ \(\dfrac{20\left(x-4\right)}{4.10}-\dfrac{4\left(3+2x\right)}{4.10}=\dfrac{5x}{5}+\dfrac{1-x}{5}\)

⇔ \(\dfrac{20x-80}{40}-\dfrac{12+8x}{40}=\dfrac{5x+1-x}{5}\)

⇔ \(\dfrac{20x-80-12-8x}{40}=\dfrac{4x+1}{5}\)

⇔ \(\dfrac{12x-92}{40}-\dfrac{4x+1}{5}=0\)

⇔ \(\dfrac{12x-92}{40}-\dfrac{8\left(4x+1\right)}{40}=0\)

⇔ \(12x-92-8\left(4x+1\right)=0\)

⇔ 12x - 92 - 32x - 8 = 0

⇔ -100 - 20x = 0

⇔ 20x = -100

⇔ x = -100 : 20

⇔ x = -5

c, \(\dfrac{2x}{3}+\dfrac{3x-5}{4}=\dfrac{3\left(2x-1\right)}{2}-\dfrac{7}{6}\)

⇔ \(\dfrac{8x}{12}+\dfrac{9x-15}{12}=\dfrac{18x-9}{6}-\dfrac{7}{6}\)

⇔ \(\dfrac{17x-15}{12}=\dfrac{18x-16}{6}\)

⇔ \(\dfrac{17x-15}{12}-\dfrac{18x-16}{6}=0\)

⇔ \(\dfrac{17x-15}{12}-\dfrac{36x-32}{12}=0\)

⇔ 17x - 15 - 36 + 32 = 0

⇔ 17 - 19x = 0

⇔ 19x = 17

⇔ x = \(\dfrac{17}{19}\)

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

câu d

\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)