Question

[....]

Gọi
x1,x2 là hai nghiệm của pt \(x^2-2x-1=0\) tính giá trị của các biểu thức:

A=\(x_1^2+x_2^2\)

B=\(x_1^3+x_2^3\)

C=\(x_1^4+x_2^4\)

D=\(x_1^2.x_2+x_2^2.x_1\)

E=\(\dfrac{x_1^2}{x_2}+\dfrac{x_2^2}{x_1}\)

F=\(\left|x_1-x_2\right|\)

G=\(\dfrac{x_1}{x_2+1}+\dfrac{x_2}{x_1+1}\)

H=\(\left(x_1+\dfrac{2}{x_2}\right)\left(x_2+\dfrac{2}{x_1}\right)\)

Answer 1

[missing you =]

,có \(ac< 0\)=>pt đã cho luôn có 2 nghiệm phân biệt

vi ét \(=>\left\{{}\begin{matrix}x1+x2=2\\x1x2=-1\end{matrix}\right.\)

a,\(A=\left(x1+x2\right)^2-2x1x2=.....\) thay số tính

b,\(B=\left(x1+x2\right)^3-3x1x2\left(x1+x2\right)=.......\)

c,\(C=x1^{2^2}+x2^{2^2}=\left(x1^2+x2^2\right)^2-2\left(x1x2\right)^2=\left[\left(x1+x2\right)^2-2x1x2\right]^2-2\left(x1x2\right)^2=....\)

\(D=x1x2\left(x1+x2\right)=.....\)

\(x1,x2\ne0=>E=\dfrac{\left(x1+x2\right)^3-3x1x2\left(x1+x2\right)}{x1x2}=...\)

\(F=\sqrt{\left(x1-x2\right)^2}=\sqrt{\left(x1+x2\right)^2-4x1x2}=....\)

\(x1,x2\ne-1=>G=\dfrac{\left(x1+x2\right)^2-2x1x2+x1x2}{x1x2+x1+X2+1}=...\)

\(x1,x2\ne0=>H=\left(\dfrac{x1x2+2}{x2}\right)\left(\dfrac{x1x2+2}{x1}\right)=\dfrac{\left(x1x2+2\right)^2}{x1x2}\)

\(=\dfrac{\left(x1x2\right)^2+4x1x2+4}{x1x2}=..\)