pt: \(x^2-5x+3m+1=0\left(1\right)\)
Để pt (1) có 2 nghiệm phân biệt \(x_1,x_2\) thỏa mãn:
\(\Delta>0\)
\(\Leftrightarrow\left(-5\right)^2-4\left(3m+1\right)>0\)
\(\Leftrightarrow m< \dfrac{7}{4}\)
Áp dụng hệ thức Vi-ét ta có:
\(x_1+x_2=5\)
\(x_1.x_2=3m+1\)
=> \(\left|x_1^2-x^2_2\right|=15\)
<=> \(\sqrt{\left(x_1-x_2\right)^2.\left(x_1+x_2\right)^2}=15\)
<=>\(\sqrt{\left(x_1-x_2\right)^2}=3\)
<=>\(\sqrt{\left(x_1+x_2\right)^2-4x_1.x_2}=3\)
<=> \(\sqrt{25-4x_1.x_2}=3\)
<=> \(-4\left(3m+1\right)=9-25\)
<=> \(m=1\)(TM)
Vậy............
ta có a=1 b=-5 c=3m+1 để pt có 2 no pb => \(\Delta\)>0 => (-5)2-4.(3m+1)>0
<=> 25-12m-4>0
<=> 21-12m>0
<=> m<21/12
theo viet ta có x1+x2=5
x1.x2=3m+1
/x12-x22/=15 => /(x1-x2)(x1+x2)/=15
<=> /5(x1-x2)/= 15
th1: (x1-x2)= -3
KH vs x1+x2=5 => x1=1; x2=4 => 3m+1=4 <=> m=1
th1: (x1-x2)=3
KH vs x1+x2=5 => x1=4; x2=1=> 3m+1=4 <=> m=1
vậy vs m=1 thì/x12-x22/=15