a là nghiệm nên \(\sqrt{2}a^2+a-1=0\Rightarrow\sqrt{2}a^2=1-a\)

\(\Rightarrow2a^4=\left(1-a\right)^2=a^2-2a+1\)

\(\Rightarrow2a^4-2a+3=a^2-4a+4=\left(a-2\right)^2\)

Mặt khác \(1-a=\sqrt{2}a^2>0\Rightarrow a< 1\)

\(\Rightarrow\sqrt{2\left(2a^4-2a+3\right)}+2a^2=\sqrt{2\left(a-2\right)^2}+2a^2=\sqrt{2}\left(2-a\right)+2a^2\)

\(=\sqrt{2}\left(\sqrt{2}a^2-a+2\right)=\sqrt{2}\left(1-a-a+2\right)=\sqrt{2}\left(3-2a\right)\)

\(\Rightarrow C=\dfrac{2a-3}{\sqrt{2}\left(3-2a\right)}=-\dfrac{\sqrt{2}}{2}\)

Sử dụng delta thôi!

Xét \(4x^2+\sqrt{2}x-\sqrt{2}=0\) có \(4\cdot\left(-\sqrt{2}\right)=-4\sqrt{2}< 0\) nên PT có 2 nghiệm phân biệt

Mà a là nghiệm nguyên dương của PT nên ta có: \(4a^2+\sqrt{2}a-\sqrt{2}=0\)

Vì a > 0 \(\Rightarrow4a^2=-\sqrt{2}a+\sqrt{2}\)

\(\Rightarrow a^2=\frac{\sqrt{2}-\sqrt{2}a}{4}=\frac{\left(1-a\right)\sqrt{2}}{4}=\frac{1-a}{2\sqrt{2}}\)

\(\Rightarrow a^4=\left(\frac{1-a}{2\sqrt{2}}\right)^2=\frac{1-2a+a^2}{8}\)

Thay vào ta được:

\(B=\frac{a+1}{\sqrt{a^4+a+1}-a^2}=\frac{\left(a+1\right)\left(\sqrt{a^4+a+1}+a^2\right)}{\left(\sqrt{a^4+a+1}\right)^2-a^4}\)

\(=\frac{\left(a+1\right)\left(\sqrt{a^4+a+1}+a^2\right)}{a^4+a+1-a^4}=\frac{\left(a+1\right)\left(\sqrt{a^4+a+1}+a^2\right)}{a+1}=\sqrt{a^4+a+1}+a^2\)

\(=\sqrt{\frac{1-2a+a^2}{8}+a+1}+\frac{1-a}{2\sqrt{2}}=\sqrt{\frac{a^2+6a+9}{8}}+\frac{1-a}{2\sqrt{2}}\)

\(=\frac{a+3}{2\sqrt{2}}+\frac{1-a}{2\sqrt{2}}=\frac{4}{2\sqrt{2}}=\sqrt{2}\)

Vậy \(B=\sqrt{2}\)

\(T=\dfrac{\left(x1\cdot\sqrt{x_2}+x_2\cdot\sqrt{x_1}\right)}{x1^2+x_2^2}\)

\(=\dfrac{\sqrt{x_1\cdot x_2}\left(\sqrt{x_1}+\sqrt{x_2}\right)}{\left(x_1+x_2\right)^2-2x_1x_2}\)

\(=\dfrac{4\cdot\sqrt{x_1+x_2+2\sqrt{x_1x_2}}}{9^2-2\cdot16}=\dfrac{4\cdot\sqrt{9+2\cdot4}}{81-32}\)

\(=\dfrac{4\sqrt{17}}{49}\)

Ta có: \(x-3\sqrt{x}+2=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)

=>x=4 hoặc x=1

Khi x=4 thì \(A=2\cdot\left(1-2\right)=-2\)

Khi x=1 thì \(A=1\cdot\left(1-1\right)=0\)