`a)` Ptr có:`\Delta'=(-2)^2-(-15)=19 > 0`
`b)` Áp dụng Viét có:`{(x_1+x_2=[-b]/a=4),(x_1.x_2=c/a=-15):}`
Ta có:`B=[x_1+1]/[x_2+3]+[x_2+1]/[x_1+3]`
`<=>B=[(x_1+1)(x_1+3)+(x_2+1)(x_2+3)]/[(x_2+3)(x_1+3)]`
`<=>B=[x_1 ^2+3x_1+x_1+3+x_2 ^2+3x_2+x_2+3]/[x_1.x_2+3x_2+3x_1+9]`
`<=>B=[(x_1+x_2)^2-2x_1.x_2+4(x_1+x_2)+6]/[x_1.x_2+3(x_1+x_2)+9]`
`<=>B=[(-4)^2-2.(-15)+4.(-4)+6]/[-15+3.(-4)+9]=-2`
a: a=1; b=-4; c=-15
Vì ac<0 nên phương trình có hai nghiệm phân biệt
b: Theo Vi-et, ta được: \(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=-15\end{matrix}\right.\)
\(B=\dfrac{\left(x_1+1\right)\left(x_1+3\right)+\left(x_2+1\right)\left(x_2+3\right)}{\left(x_1+3\right)\left(x_2+3\right)}\)
\(=\dfrac{x_1^2+4x_1+3+x_2^2+4x_2+3}{x_1x_2+3\left(x_1+x_2\right)+9}\)
\(=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2+4\cdot4+6}{-15+3\cdot4+9}=\dfrac{4^2-2\cdot\left(-15\right)+16+6}{-15+12+9}\)
\(=\dfrac{16+30+22}{-3+9}=\dfrac{68}{6}=\dfrac{34}{3}\)
