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CHÀO BẠN
Áp dụng Viét
- x1*x2=4m (1)
- x1+x2=2(m+1) (2)
(*) (x1+m)(x2+m)=3m^2+12
<=>x1*x2+m(x1+x2)=3m^2+12 (**)
thay (1);(2) vô (**) =>....
Mình bày hướng có chỗ nào sai tự sửa
\(x^2+2\left(m+1\right)+4m-4=0\)
Theo Vi - ét, ta có :
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-2\left(m+1\right)\\x_1x_2=\dfrac{c}{a}=4m-4\end{matrix}\right.\)
Ta có :
\(x_1^2+x_2^2+3x_1x_2=0\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+3x_1x_2=0\)
\(\Leftrightarrow\left(x_1+x_2\right)^2+x_1x_2=0\)
\(\Leftrightarrow\left[-2\left(m+1\right)\right]^2+\left(4m-4\right)=0\)
\(\Leftrightarrow4\left(m^2+2m+1\right)+4m-4=0\)
\(\Leftrightarrow4m^2+8m+4+4m-4=0\)
\(\Leftrightarrow4m^2+12m=0\)
\(\Leftrightarrow4m\left(m+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=0\\m=-3\end{matrix}\right.\)
\(\Delta^'=\left(-1\right)^2-\left(m-1\right)=2-m\)
Để PT có nghiệm thì: \(m\le2\)
Khi đó theo hệ thức viet ta có: \(\hept{\begin{cases}x_1+x_2=2\\x_1x_2=m-1\end{cases}}\)
Ta có: \(x_1^4-x_1^3=x_2^4-x_2^3\)
\(\Leftrightarrow\left(x_1^4-x_2^4\right)-\left(x_1^3-x_2^3\right)=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1+x_2\right)\left(x_1^2+x_2^2\right)-\left(x_1-x_2\right)\left(x_1^2+x_1x_2+x_2^2\right)=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left[2\left(x_1^2+x_2^2\right)-x_1^2-x_1x_2-x_2^2\right]=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1^2-x_1x_2+x_2^2\right)=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left[\left(x_1+x_2\right)^2-3x_1x_2\right]=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left[4-3\left(m-1\right)\right]=0\)
Nếu \(x_1-x_2=0\Rightarrow x_1=x_2=1\Rightarrow m=1\left(tm\right)\)
Nếu \(4-3\left(m-1\right)=0\Rightarrow m=\frac{7}{3}\left(ktm\right)\)
Vậy m = 1
Ta có: \(\Delta=\left\lbrack2\left(m-3\right)\right\rbrack^2-4\left(3m^2-8m+5\right)\)
\(=4\left(m^2-6m+9\right)-12m^2+32m-20\)
\(=4m^2-24m+36-12m^2+32m-20=-8m^2+8m+16\)
\(=-8\left(m^2-m-2\right)=-8\left(m-2\right)\left(m+1\right)\)
Để phương trình có hai nghiệm thì Δ>=0
=>-8(m-2)(m+1)>=0
=>(m-2)(m+1)<=0
=>-1<=m<=2
Theo Vi-et, ta có: \(\begin{cases}x_1+x_2=-\frac{b}{a}=2\left(m-3\right)\\ x_1x_2=\frac{c}{a}=3m^2-8m+5=\left(3m-5\right)\left(m-1\right)\end{cases}\)
\(x_1^2+2x_2^2-3x_1x_2=x_1-x_2\)
=>\(\left(x_1-x_2\right)\left(x_1-2x_2\right)-\left(x_1-x_2\right)=0\)
=>\(\left(x_1-x_2\right)\left(x_1-2x_2-1\right)=0\)
TH1: \(x_1-x_2=0\)
=>\(x_1=x_2\)
mà \(x_1+x_2=2\left(m-3\right)\)
nên \(x_1=x_2=\frac{2\left(m-3\right)}{2}=m-3\)
\(x_1x_2=3m^2-8m+5\)
=>\(3m^2-8m+5=\left(m-3\right)^2=m^2-6m+9\)
=>\(2m^2-2m-4=0\)
=>\(m^2-m-2=0\)
=>(m-2)(m+1)=0
=>\(\left[\begin{array}{l}m-2=0\\ m+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}m=2\left(nhận\right)\\ m=-1\left(nhận\right)\end{array}\right.\)
TH2: \(x_1-2x_2-1=0\)
=>\(x_1-2x_2=1\)
mà \(x_1+x_2=2\left(m-3\right)=2m-6\)
nên \(x_1-2x_2-x_1-x_2=1-2m+6=-2m+7\)
=>\(-3x_2=-2m+7\)
=>\(x_2=\frac{2m-7}{3}\)
\(x_1+x_2=2m-6\)
=>\(x_1=2m-6-\frac{2m-7}{3}=\frac{3\left(2m-6\right)-2m+7}{3}=\frac{4m-11}{3}\)
\(x_1x_2=3m^2-8m+5\)
=>\(\frac{\left(2m-7\right)\left(4m-11\right)}{9}=3m^2-8m+5\)
=>\(9\left(3m^2-8m+5\right)=\left(2m-7\right)\left(4m-11\right)\)
=>\(27m^2-72m+45=8m^2-50m+77\)
=>\(19m^2-22m-32=0\)
=>(19m+16)(m-2)=0
=>\(\left[\begin{array}{l}19m+16=0\\ m-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}m=-\frac{16}{19}\left(nhận\right)\\ m=2\left(nhận\right)\end{array}\right.\)
Để pt có 2 nghiệm \(x_1,x_2\) thì \(\Delta'=4\left(m-1\right)^2-3\left(m^2-4m+1\right)=m^2+4m+1\ge0\)
\(\Leftrightarrow\)\(\left(m^2+4m+4\right)-3\ge0\)\(\Leftrightarrow\)\(\left(m+2\right)^2-3\ge0\)
\(\Leftrightarrow\)\(\left(m+2-\sqrt{3}\right)\left(m+2+\sqrt{3}\right)\ge0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}m\ge\sqrt{3}-2\\m\le-\sqrt{3}-2\end{cases}}\)
Ta có : \(\left|x_1-x_2\right|=2\)
\(\Leftrightarrow\)\(\left(x_1-x_2\right)^2=4\)
\(\Leftrightarrow\)\(x_1^2+x_2^2-2x_1x_2=4\)
\(\Leftrightarrow\)\(\left(x_1+x_2\right)^2-4x_1x_2=4\) \(\left(1\right)\)
Theo định lý Vi-et ta có \(\hept{\begin{cases}x_1+x_2=\frac{4\left(1-m\right)}{3}\\x_1x_2=\frac{m^2-4m+1}{3}\end{cases}}\)
\(\left(1\right)\)\(\Leftrightarrow\)\(\left(\frac{4-4m}{3}\right)^2-4\left(\frac{m^2-4m+1}{3}\right)=4\)
\(\Leftrightarrow\)\(\frac{16-32m+16m^2}{9}-\frac{4m^2-16m+4}{3}-4=0\)
\(\Leftrightarrow\)\(\frac{16m^2-32m+16-12m^2+48m-12-36}{9}=0\)
\(\Leftrightarrow\)\(4m^2+16m-32=0\)
\(\Leftrightarrow\)\(\left(m^2+4m+4\right)-12=0\)
\(\Leftrightarrow\)\(\left(m+2\right)^2=12\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}m=2\sqrt{3}-2\left(tm\right)\\m=-2\sqrt{3}-2\left(tm\right)\end{cases}}\)
Vậy để pt có hai nghiệm \(x_1,x_2\) thoả mãn \(\left|x_1-x_2\right|=2\) thì \(\orbr{\begin{cases}m=2\sqrt{3}-2\\m=-2\sqrt{3}-2\end{cases}}\)
chả biết đúng ko nhưng xem thử nha -_-
a) Có: `\Delta'=(m-2)^2-(m^2-4m)=m^2-4m+4-m^2+4m=4>0 forall m`
`=>` PT luôn có 2 nghiệm phân biệt với mọi `m`.
b) Viet: `x_1+x_2=-2m+4`
`x_1x_2=m^2-4m`
`3/(x_1) + x_2=3/(x_2)+x_1`
`<=> 3x_2+x_1x_2^2=3x_1+x_1^2 x_2`
`<=> 3(x_1-x_2)+x_1x_2(x_1-x_2)=0`
`<=>(x_1-x_2).(3+x_1x_2)=0`
`<=> \sqrt((x_1+x_2)^2-4x_1x_2) .(3+x_1x_2)=0`
`<=> \sqrt((-2m+4)^2-4(m^2-4m)) .(3+m^2-4m)=0`
`<=> 4.(3+m^2-4m)=0`
`<=> m^2-4m+3=0`
`<=>` \(\left[{}\begin{matrix}m=3\\m=1\end{matrix}\right.\)
Vậy `m \in {1;3}`.