Question

câu 2 bài 3 ạ

 

 

Answer 1

2:

a: Thay x=-2 vào (P), ta được:

\(y=\left(-2\right)^2=4\)

Thay x=-2 và y=4 vào (d), ta được:

2*(-2)-m+1=4

=>-4-m+1=4

=>-m-3=4

=>-m=7

=>m=-7

=>(d): y=2x-(-7)+1=2x+8

Phương trình hoành độ giao điểm là:

\(x^2=2x+8\)

=>\(x^2-2x-8=0\)

=>(x-4)(x+2)=0

=>x=4 hoặc x=-2

Khi x=4 thì \(y=4^2=16\)

Vậy: Tọa độ điểm còn lại là B(4;16)

b: Phương trình hoành độ giao điểm là:

\(x^2=2x-m+1\)

=>\(x^2-2x+m-1=0\)

\(\Delta=\left(-4\right)^2-4\cdot\left(m-1\right)=16-4m+4=-4m+20\)

Để (P) cắt (d) tại hai điểm phân biệt thì Δ>0

=>-4m+20>0

=>-4m>-20

=>m<5

Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=2;x_1x_2=\frac{c}{a}=m-1\)

\(\frac{x_1}{x_2^2+2x_1+1}+\frac{x_2}{x_1^2+2x_2+1}=\frac14\)

=>\(\frac{x_1\left(x_1^2+2x_2+1\right)+x_2\left(x_2^2+2x_1+1\right)}{\left(x_2^2+2x_1+1\right)\left(x_1^2+2x_2+1\right)}=\frac14\)

\(x_1\left(x_1^2+2x_2+1\right)+x_2\left(x_2^2+2x_1+1\right)\)

\(=\left(x_1^3+x_2^3\right)+4x_1x_2+\left(x_1+x_2\right)\)

\(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)+4x_1x_2+\left(x_1+x_2\right)\)

\(=2^3-3\cdot2\cdot\left(m-1\right)+4\cdot\left(m-1\right)+2\)

=8-6(m-1)+4(m-1)+2

=10-2(m-1)

=10-2m+2

=-2m+12

\(\left(x_2^2+2x_1+1\right)\left(x_1^2+2x_2+1\right)\)

\(=\left(x_1x_2\right)^2+2x_2^3+x_2^2+2\cdot x_1^3+4\cdot x_1x_2+2x_1+x_1^2+2x_2+1\)

\(=\left(x_1x_2\right)^2+2\left(x_1^3+x_2^3\right)+\left(x_1^2+x_2^2\right)+4x_1x_2+2\left(x_1+x_2\right)+1\)

\(=\left(m-1\right)^2+2\cdot\left\lbrack\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\right\rbrack+\left\lbrack\left(x_1+x_2\right)^2-2x_1x_2\right\rbrack\) +4(m-1)+2*2+1

=\(\left(m-1\right)^2+4\left(m-1\right)+4+2\cdot\left\lbrack2^3-3\cdot2\cdot\left(m-1\right)\right\rbrack+\left\lbrack2^2-2\left(m-1\right)\right\rbrack\) +1

\(=\left(m-1+2\right)^2+2\cdot\left\lbrack8-6m+6\right\rbrack+4-2m+2+1\)

\(=\left(m+1\right)^2+2\left(-6m+14\right)-2m+7\)

\(=m^2+2m+1-12m+28-2m+7=m^2-12m+36=\left(m-6\right)^2\)

\(\frac{x_1\left(x_1^2+2x_2+1\right)+x_2\left(x_2^2+2x_1+1\right)}{\left(x_2^2+2x_1+1\right)\left(x_1^2+2x_2+1\right)}=\frac14\)

=>\(\frac{-2\left(m-6\right)}{\left(m-6\right)^2}=\frac14\)

=>\(\frac{-2}{m-6}=\frac14=\frac{-2}{-8}\)

=>m-6=-8

=>m=-2(nhận)