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: \(=\dfrac{27a^6b^3\cdot a^2b^6}{a^8b^8}=27b\)

b: \(=3y^2-5x^2y^3-2y^2+3x^2y^3\)

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

c: \(=6x-y+2x^2+3y-2x^2+x\)

\(=7x+2y\)

d: \(=x-y+2y^2-6xy+\dfrac{10x^2}{y}\)

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

b: \(=6x-y+2x^2+3y^2-2x^2+x=7x-y+3y^2\)

c: \(=x-y+4y^2-6xy+\dfrac{10x^2}{y}\)

\(a.\left(9x^2y^3-15x^4y^4\right):3x^2y-\left(2-3x^2y\right)y^2\)

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

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

\(b.\left(6x^2-xy\right):x+\left(2x^3y+3xy^2\right):xy-\left(2x-1\right)x\)

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

\(=2x^2+7x+2y-2\)

\(c.\left(x^2-xy\right):x+\left(6x^2y^5-9x^3y^4+15x^4y^3\right):\dfrac{3}{2}x^2y^3\)

\(=x-y+4y^2-6xy+10x^2\)

b: Đặt \(x^2-6x-2=a\)

Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)

=>(a+2)(a+7)=0

\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)

=>x(x-6)(x-1)(x-5)=0

hay \(x\in\left\{0;1;6;5\right\}\)

c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)

\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)

\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)

\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)

=>26x=-3

hay x=-3/26

1.

a)\(\left(\dfrac{1}{2}\cdot\left(-2\right)\cdot\dfrac{-1}{3}\right)\cdot\left(x^2\cdot x^2\cdot x^2\right)\cdot\left(y^2\cdot y^3\right)\cdot z\)

\(\dfrac{1}{3}x^6y^5z\)

Deg=12

Mấy câu kia tương tự nha cố gắng lên!

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Thick thể hiện à

haizzzz

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 = \(\dfrac{17}{19}\)