Áp dụng hệ thức Vi-ét,ta có : \(\hept{\begin{cases}x_1+x_2=-2\left(m-1\right)\\x_1x_2=4m\end{cases}}\)

Ta có : \(4x_1^2\left(1+x_2\right)+4x_2\left(1+x_1\right)+x_1^2x_2^2=36\)

\(\Rightarrow4\left(x_1^2+x_2^2\right)+4x_1x_2\left(x_1+x_2\right)+x_1^2x_2^2=36\)

\(\Rightarrow4\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+4x_1x_2\left(x_1+x_2\right)+x_1^2x_2^2=36\)

thay vào rồi tìm m thôi 

Ta có: \(x^2-5x+3=0\)

Áp dụng định lí viet ta có: \(\hept{\begin{cases}x_1+x_2=5\\x_1x_2=3\end{cases}}\)

a) \(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=5^2-2.3=19\)

b) \(B=x_1^3+x_2^3=\left(x_1+x_2\right)^3-3\left(x_1+x_2\right)x_1x_2=5^3-3.5.3=80\)

c) \(C=\left|x_1-x_2\right|\)>0

=> \(C^2=x_1^2+x_2^2-2x_1x_2=19-2.3=13\)

=> C = căn 13

d) \(D=x_2+\frac{1}{x_1}+x_1+\frac{1}{x_2}=\left(x_1+x_2\right)+\frac{x_1+x_2}{x_1x_2}=5+\frac{5}{3}=5\frac{5}{3}\)

e) \(E=\frac{1}{x_1+3}+\frac{1}{x_2+3}=\frac{\left(x_1+x_2\right)+6}{x_1x_2+3\left(x_1+x_2\right)+9}=\frac{5+6}{3+3.5+9}=\frac{11}{27}\)

g) \(G=\frac{x_1-3}{x_1^2}+\frac{x_2-3}{x_2^2}=\left(\frac{1}{x_1}+\frac{1}{x_2}\right)-3\left(\frac{1}{x_1^2}+\frac{1}{x_2^2}\right)\)

\(=\frac{x_1+x_2}{x_1x_2}-3\frac{x_1^2+x_2^2}{x_1^2.x_2^2}=\frac{5}{3}-3.\frac{19}{3^2}=-\frac{14}{3}\)

\(\sqrt{32}+10\sqrt{7}+\sqrt{32}-10\sqrt{7}\)

\(=\left(\sqrt{32}+\sqrt{32}\right)+\left(10\sqrt{7}-10\sqrt{7}\right)\)

\(=\sqrt{16\times2}\times2\)

\(=\sqrt{\left(4\right)^2\times2}\times2\)

\(=4\sqrt{2}\times2\)

\(=8\sqrt{2}\)

https://www.youtube.com/watch?v=DzTRaA42rvY

Bạn ơi. Bạn giải được câu b này chưa mình xin với ạ