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

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

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

Ta có: \(\Delta'=32>0\)

\(\Rightarrow\) Phương trình có 2 nghiệm phân biệt

Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=12\\x_1x_2=4\end{matrix}\right.\)

Mặt khác: \(T=\dfrac{x_1^2+x^2_2}{\sqrt{x_1}+\sqrt{x_2}}\)

\(\Rightarrow T^2=\dfrac{x_1^4+x^4_2+2x_1^2x_2^2}{x_1+x_2+2\sqrt{x_1x_2}}=\dfrac{\left(x_1^2+x_1^2\right)^2}{x_1+x_2+2\sqrt{x_1x_2}}\) \(=\dfrac{\left[\left(x_1+x_2\right)^2-2x_1x_2\right]^2}{x_1+x_2+2\sqrt{x_1x_2}}=\dfrac{\left(12^2-2\cdot4\right)^2}{12+2\sqrt{4}}=1156\)

Mà ta thấy \(T>0\) \(\Rightarrow T=\sqrt{1156}=34\) 

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