Bài 5:
a: Theo Vi-et, ta có:
\(x_1+x_2=-\frac{b}{a}=\frac{-\left\lbrack-\left(m-8\right)\right\rbrack}{1}=m-8;x_1x_2=\frac{c}{a}=12m-1\)
=>Sai
b: \(x_1\cdot x_2=\frac{c}{a}=12m-1\)
=>Đúng
c: \(x_1+x_2+x_1x_2=m-8+12m-1=13m-9\)
=>Sai
d: \(2\left(x_1+x_2\right)+x_1x_2=2\left(m-8\right)+12m-1=2m-16+12m-1=14m-17\)
=>Đúng
Bài 6:
a: Thay x=1 vào phương trình, ta được:
\(2\cdot1^2-\left(m-1\right)\cdot1-m+3=0\)
=>2-m+1-m+3=0
=>-2m+6=0
=>-2m=-6
=>m=3
=>Đúng
b: Khi m=3 thì phương trình sẽ trở thành:
\(2x^2-\left(3-1\right)x-3+3=0\)
=>\(2x^2-2x=0\)
=>2x(x-1)=0
=>x=0 hoặc x=1
=>Sai
c: Theo Vi-et, ta có:
\(x_1+x_2=-\frac{b}{a}=\frac{m-1}{2}=\frac{-3-1}{2}=-\frac42=-2\)
=>Đúng
d: Theo Vi-et, ta có:
\(x_1x_2=\frac{c}{a}=\frac{-m+3}{2}=\frac{-2+3}{2}=\frac12\)
=>Sai
Bài 7:
a: Theo Vi-et, ta có:
\(x_1+x_2=-\frac{b}{a}=m=-2;x_1x_2=\frac{c}{a}=m+3=-2+3=1\)
=>Đúng
b: \(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=m^2-2\left(m+3\right)=m^2-2m-6\)
=>Đúng
c: \(x_1^3-x_2^3=\left(x_1-x_2\right)\left(x_1^2+x_1x_2+x_2^2\right)\)
\(=\left(x_1-x_2\right)\left(x_1^2+2x_1x_2+x_2^2-x_1x_2\right)\)
\(=\left(x_1-x_2\right)\left\lbrack\left(x_1+x_2\right)^2-x_1x_2\right\rbrack\)
\(=\left(x_1-x_2\right)\left(m^2-m-3\right)\)
\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2=m^2-4\left(m+3\right)=m^2-4m-12\)
=>\(x_1-x_2=\pm\sqrt{m^2-4m-12}\)
=>\(x_1^3-x_2^3=\pm\sqrt{m^2-4m-12}\cdot\left(m^2-m-3\right)\)
=>Sai
d: Khi x=-3 thì ta có:
\(\left(-3\right)^2-m\cdot\left(-3\right)+m+3=0\)
=>9+3m+m+3=0
=>4m+12=0
=>m=-3
\(x_1\cdot x_2=\frac{c}{a}=-3+3=0\)
=>\(x_2=0\)
=>Sai
Bài 8:
\(3x^2+2x-21=0\)
=>\(3x^2+9x-7x-21=0\)
=>3x(x+3)-7(x+3)=0
=>(x+3)(3x-7)=0
=>x=-3 hoặc x=7/3
=>NGhiệm còn lại là \(x_2=\frac73\)
Bài 9:
Thay \(x=\frac13\) vào phương trình, ta được:
\(3\cdot\left(\frac13\right)^2-2\left(m-3\right)\cdot\frac13+5=0\)
=>\(3\cdot\frac19-\frac23\left(m-3\right)+5=0\)
=>\(\frac13-\frac23m+2+5=0\)
=>\(\frac23m=\frac13+7=\frac{22}{3}\)
=>m=11
Theo Vi-et, ta có:
\(x_1\cdot x_2=\frac{c}{a}=\frac53\)
=>\(x_2=\frac53:\frac13=5\)
Bài 4:
\(\Delta=\left\lbrack-2\left(m+1\right)\right\rbrack^2-4\left(m^2+2\right)\)
\(=4\left(m+1\right)^2-4\left(m^2+2\right)\)
\(=4\left(m^2+2m+1-m^2-2\right)=4\left(2m-1\right)\)
Để phương trình có hai nghiệm thì 4(2m-1)>=0
=>2m-1>=0
=>m>=1/2
Theo Vi-et, ta có:
\(x_1+x_2=-\frac{b}{a}=2\left(m+1\right);x_1x_2=\frac{c}{a}=m^2+2\)
\(A=x_1x_2-2\left(x_1+x_2\right)-6\)
\(=m^2+2-2\left(2m+2\right)-6=m^2-4-4m-4=m^2-4m-8\)
\(=m^2-4m+4-12=\left(m-2\right)^2-12\ge-12\forall m\)
Dấu '=' xảy ra khi m-2=0
=>m=2
