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

\(=\dfrac{1}{x-1}\)

\(\dfrac{8}{x^2+3}+\dfrac{7}{x}=\dfrac{8x}{\left(x^2+3\right).x}+\dfrac{7\left(x^2+3\right)}{\left(x^2+3\right).x}=\dfrac{7x^2+28x+21}{x\left(x^2+3\right)}\)

x⁴+x²+1

=(x²)²+2x²+1-2x²+x²

=(x²+1)²-2x²+x² 

= (x² + 1)² − x² 

= (x² + x+ 1 )(x² − x+ 1 )

\(x^4+x^2+1\)
\(=\left[\left(x^2\right)^2+2.x^2.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]-\left(\frac{1}{2}\right)^2+1\)
\(=\left(x^2+\frac{1}{2}\right)^2-\frac{1}{4}+\frac{4}{4}\)
\(=\left(x^2+\frac{1}{2}\right)^2+\frac{3}{4}\)

x+\(\dfrac{y}{2}\)+x+\(\dfrac{2}{2}\)x2+4
=2x+\(\dfrac{4+y}{2}\)+4

Ta có biến đổi sau :

\(\dfrac{x^2-2x-3}{x^2+x}=\dfrac{x^2+x-3x-3}{x\left(x+1\right)}=\dfrac{x\left(x+1\right)-3\left(x+1\right)}{x\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}=\dfrac{x-3}{x}\left(1\right)\)Tương tự , ta có :

\(\dfrac{x^2-4x+3}{x^2-x}=\dfrac{x^2-x-3x+3}{x\left(x-1\right)}=\dfrac{x\left(x-1\right)-3\left(x-1\right)}{x\left(x-1\right)}=\dfrac{\left(x-1\right)\left(x-3\right)}{x\left(x-1\right)}=\dfrac{x-3}{x}\left(2\right)\)Do đó , ba phân thức bằng nhau

Giải:

a) \(3x^2y-6xy^2\)

\(=3xy\left(x-2y\right)\)

Vậy ...

b) \(\left(2x-a\right)x^2-\left(2x-a\right)y\)

\(=\left(2x-a\right)\left(x^2-y\right)\)

\(=\left(2x-a\right)\left(x-\sqrt{y}\right)\left(x+\sqrt{y}\right)\)

Vậy ...

c) \(25a^2-c^2\)

\(=\left(5a-c\right)\left(5a+c\right)\)

Vậy ...

d) \(4-36x+81x^2\)

\(=2^2-2.2.9x+\left(9x\right)^2\)

\(=\left(2-9x\right)^2\)

Vậy ...

e) \(\left(x+7\right)2-\left(2x-9\right)2\)

\(=2\left[\left(x+7\right)-\left(2x-9\right)\right]\)

\(=2\left(x+7-2x+9\right)\)

\(=2\left(16-x\right)\)

Vậy ...

f) \(x^2-6x+8\)

\(=x^2-6x+9-1\)

\(=\left(x-3\right)^2-1\)

\(=\left(x-4\right)\left(x-2\right)\)

Vậy ...

\(x^2-4x+4-y^2\)

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)