\(=5\cdot\dfrac{x}{5y}=\dfrac{x}{y}\)

\(A=5x^2-3x-x^3+x^2+x^3-62x-10+3x\\ A=6x^2-62x-10\\ B=x^3+x^2+x-x^3-x^2-x+5=5\\ C=3x^2y-15xy^2+15xy^2-10y^3+10y^2-3x^2y-4=-4\)

b: Ta có: \(B=x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5\)

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

=5

a, \(\left(5x-4\right)\left(5x+4\right)-\left(5x-4\right)^2=\left(25x^2-16\right)-\left(25x^2-40x+16\right)=40x-32\)

b,\(\left(5x+3\right)^2-\left(4x-1\right)^2-\left(9x^2+8\right)=\left(x+4\right)\left(9x-2\right)-\left(9x^2+8\right)\)

\(=9x^2+34x-8-\left(9x^2+8\right)=34x\)

c,\(2\left(x-5y\right)\left(x+5y\right)+\left(x+5y\right)^2+\left(x-5y\right)^2=\left(2x\right)^2=4x^2\)

a: \(\dfrac{9x^3y^2-4xy^2+5x}{2x}=\dfrac{9}{2}x^2y^2-2y^2+\dfrac{5}{2}\)

b: \(\left(\dfrac{3}{4}x^3y^6+\dfrac{6}{5}x^4y^3-\dfrac{9}{10}x^5y\right):\dfrac{-3}{5}x^3y\)

\(=y^5\cdot\left(\dfrac{3}{4}:\dfrac{-3}{5}\right)-xy^2\cdot\left(\dfrac{6}{5}:\dfrac{3}{5}\right)+\dfrac{9}{10}:\dfrac{3}{5}\cdot x^2\)

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

\(4x=5y\Rightarrow\frac{y}{4}=\frac{x}{5}\Rightarrow\frac{y^2}{16}=\frac{x^2}{25}\)

Áp dụng dãy tỉ số bằng nhau:

\(\frac{y^2}{16}=\frac{x^2}{25}=\frac{x^2-y^2}{25-16}=\frac{1}{9}\)

=> \(y^2=\frac{16}{9}\Rightarrow y=\pm\frac{4}{3}\)

Với y=4/3=> x=5/3

Với x=-4/3=> y=-5/3

b) \(VT\ge0\Rightarrow VP\ge0\Rightarrow5x-10\ge0\Rightarrow5x\ge10\Rightarrow x\ge10:5\Rightarrow x\ge2\)

=> \(x+1>0,x-2\ge0,x+7>0\)

Với phương trinh tương đương với:

x+1+x-2+x+7=5x-10

<=> 3x+6=5x-10

<=> 2x=16

<=> x=8

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=10\\xy\left(x+y\right)+5\left(x+y\right)=32\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u^2-2v=10\\uv+5u=32\end{matrix}\right.\)

\(\Rightarrow u\left(\dfrac{u^2-10}{2}\right)+5u=32\)

\(\Leftrightarrow u^3=64\Rightarrow u=4\Rightarrow v=3\)

\(\Rightarrow\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)

A= x² + y² - 5x - 5y + 2xy + 2009

= (x² + 2xy + y²) - 5(x + y) + 2009

= (x + y)² - 5(x + y) + 2009

= 10² - 5.10 + 2009

= 2059

\(x^2+y^2-5x-5y+2xy+2009=\left(x^2+2xy+y^2\right)-5\left(x+y\right)+2009\)

\(=\left(x+y\right)^2-5\left(x+y\right)+2009\)

thay x + y = 10 đc: 

10² - 5*10 + 2009 = 2059

\(A=x^2+y^2-5x-5y+2xy+2009\)

\(=\left(x^2+2xy+y^2\right)-\left(5x+5y\right)+2009\)

\(=\left(x+y\right)^2-5\left(x+y\right)+2009\)

Thay x + y = 10 vào A, ta có:

\(10^2-5\times10+2009=100-50+2009=2059\)

Vậy x + y = 10, giá trị của biểu thức A là 2059

f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)

g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)

h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)

n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)

p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)

k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)

m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)

4x^2/5y^2 * 5y/6x * 3y/2x= 1/3

(x-2)(x+2)/3(x+4) * x+4/2(x-2)=x+2/6

5(x+2)/4(x-2)* -2(x-2)/x+2=-5/2