Question

x^2+1-12=0

A=x1^2+x2^2 +x1^2. x2 + x1. x^2

Giúp gấp nhaa cam on mn

Answer 1

\(x^2+1-12=0\)

Theo Vi - ét , ta có :

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=0\\x_1x_2=\dfrac{c}{a}=-11\end{matrix}\right.\)

Ta có : 

\(A=x_1^2+x_2^2+x_1^2x_2+x_1x^2_2\)

\(=\left(x_1+x_2\right)^2-2x_1x_2+x_1x_2\left(x_1+x_2\right)\)

\(=0^2-2\left(-11\right)-11\left(0\right)\)

\(=22-11\)

\(=11\)

Vậy \(A=11\)

Answer 2

Bạn xem lại giúp mình pt \(x^2+1-12=0\) có thiếu \(x\) không vậy ?

Answer 3

\(x^2+1-12=0\)

\(\Leftrightarrow x^2-11=0\)

\(\Delta=b^2-4ac=0^2-4.1.\left(-11\right)=44\)

Vậy phương trình có hai nghiệm:

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-0+\sqrt{44}}{2.1}=\sqrt{11}\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-0-\sqrt{44}}{2.1}=-\sqrt{11}\end{matrix}\right.\)