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

\(\Rightarrow ab+bc+ac=\frac{-\left(a^2+b^2+c^2\right)}{2}=-\frac{4}{2}=-2\)

\(\Leftrightarrow a^4+b^4+c^4=16-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

Mặt khác : \(\left(ab+bc+ac\right)^2=4\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=4\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=4\)

\(\Rightarrow a^4+b^4+c^4=16-2.4=8\)

b, Ta co: \(x^3+xy^2-x^2y-y^3+3\)

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

\(=\left(x-y\right)^3+3xy\left(x-y\right)-xy\left(x-y\right)+3\)

= 3 ( vì x-y = 0)

x^2=a;y^2=b(Đk:a,b không âm)
Từ giả thiết suy ra a+b=2
=>3x^4+5x^2y^2+2y^4+2y^2
=3a^2+5ab+2b^2+2b
=(3a^2+3ab)+(2ab+2b^2)+2b
=3a(a+b)+2b(a+b)+2b
=(a+b)(3a+2b)+2b
=2(3a+2b)+2b
=2(2a+2b)+2a+2b
=4.2+2*\.2=12

Đặt x^2=a;y^2=b(với Đk:a,b không âm) 
Từ giả thiết suy ra a+b=2 
=>3x^4+5x^2y^2+2y^4+2y^2 
=3a^2+5ab+2b^2+2b 
=(3a^2+3ab)+(2ab+2b^2)+2b 
=3a(a+b)+2b(a+b)+2b 
=(a+b)(3a+2b)+2b 
=2(3a+2b)+2b 
=2(2a+2b)+2a+2b 
=4.2+2.2=12

x^2+y^2=1 ma ban

Ta có a + b + c = 0 

<=> (a + b + c)² = 0

<=> a² + b² + c² + 2(ab + bc + ca) = 0 

<=> ab + bc + ca = \(-\frac{1}{2}\)

=> \(\left(ab+bc+ca\right)^2=\frac{1}{4}\)

<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2ab^2c+2a^2bc+2abc^2=\frac{1}{4}\)

<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=\frac{1}{4}\)

<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\frac{1}{4}\)

Lại có a² + b² + c² = 1

=> (a² + b² + c²)² = 1

<= > a⁴ + b⁴ + c⁴ + 2[(ab)² + (bc)² + (ca)²] = 1 

<=> \(a^4+b^4+c^4+2.\frac{1}{4}=1\)

<=> \(a^4+b^4+c^4=\frac{1}{2}\)

Từ a + b + c = 0 => ( a + b + c )² = 0 <=> a² + b² + c² + 2ab + 2bc + 2ca = 0

<=> ab + bc + ca = -1/2 => ( ab + bc + ca )² = 1/4

<=> a²b² + b²c² + c²a² + 2ab²c + 2bc²a + 2a²bc = 1/4

<=> a²b² + b²c² + c²a² + 2abc( a + b + c ) = 1/4

<=> a²b² + b²c² + c²a² = 1/4 ( vì a + b + c = 0 )

Từ a² + b² + c² = 1 => ( a² + b² + c² )² = 1 <=> a⁴ + b⁴ + c⁴ + 2a²b² + 2b²c² + 2c²a² = 1

<=> a⁴ + b⁴ + c⁴ + 2( a²b² + b²c² + c²a² ) = 1 

<=> a⁴ + b⁴ + c⁴ + 1/2 = 1 <=> a⁴ + b⁴ + c⁴ = 1/2

Vậy A = 1/2