a) Thay m=-2 vào phương trình, ta được:

\(x^2+4x+3=0\)

a=1; b=4; c=3

Vì a-b+c=0 nên phương trình có hai nghiệm phân biệt là:

\(x_1=-1;x_2=\dfrac{-c}{a}=-3\)

PT có 2 nghiệm phân biệt`<=> \Delta' >0`

`<=> m^2-1>0`

`<=> m<-1 ; 1 <m`

Viet: `x_1+x_2=2m`

`x_1x_2=1`

Theo đề: `x_1^2+x_2^2=8`

`<=> (x_1+x_2)^2-2x_1x_2=8`

`<=> 4m^2-2=8`

`<=> 4m^2 - 10=0`

`<=>` \(\left[{}\begin{matrix}m=\dfrac{\sqrt{10}}{2}\\m=-\dfrac{\sqrt{10}}{2}\end{matrix}\right.\)

Vậy `m=\pm \sqrt10/2`.

nhanh đi đang gấp lắm

`x_1^2+x_2^2 = (x_1^2+2x_1x_2+x_2^2)-2x_1x_2 = (x_1+x_2)^2-2x_1x_2`

Bài 1:

a, Thay m=-1 vào (1) ta có:
\(x^2-2\left(-1+1\right)x+\left(-1\right)^2+7=0\\ \Leftrightarrow x^2+1+7=0\\ \Leftrightarrow x^2+8=0\left(vô.lí\right)\)

Thay m=3 vào (1) ta có:

\(x^2-2\left(3+1\right)x+3^2+7=0\\ \Leftrightarrow x^2-2.4x+9+7=0\\ \Leftrightarrow x^2-8x+16=0\\ \Leftrightarrow\left(x-4\right)^2=0\\ \Leftrightarrow x-4=0\\ \Leftrightarrow x=4\)

b, Thay x=4 vào (1) ta có:

\(4^2-2\left(m+1\right).4+m^2+7=0\\ \Leftrightarrow16-8\left(m+1\right)+m^2+7=0\\ \Leftrightarrow m^2+23-8m-8=0\\ \Leftrightarrow m^2-8m+15=0\\ \Leftrightarrow\left(m^2-3m\right)-\left(5m-15\right)=0\\ \Leftrightarrow m\left(m-3\right)-5\left(m-3\right)=0\\ \Leftrightarrow\left(m-3\right)\left(m-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=3\\m=5\end{matrix}\right.\)

c, \(\Delta'=\left[-\left(m+1\right)\right]^2-\left(m^2+7\right)=m^2+2m+1-m^2-7=2m-6\)

Để pt có 2 nghiệm thì \(\Delta'\ge0\Leftrightarrow2m-6\ge0\Leftrightarrow m\ge3\)

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=m^2+7\end{matrix}\right.\)

\(x_1^2+x_2^2=0\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=0\\ \Leftrightarrow\left(2m+2\right)^2-2\left(m^2+7\right)=0\\ \Leftrightarrow4m^2+8m+4-2m^2-14=0\\ \Leftrightarrow2m^2+8m-10=0\\ \Leftrightarrow\left[{}\begin{matrix}m=1\left(ktm\right)\\m=-5\left(ktm\right)\end{matrix}\right.\)

\(x_1-x_2=0\\ \Leftrightarrow\left(x_1-x_2\right)^2=0\\ \Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=0\\ \Leftrightarrow\left(2m+2\right)^2-4\left(m^2+7\right)=0\\ \Leftrightarrow4m^2+8m+4-4m^2-28=0\\ \Leftrightarrow8m=28=0\\ \Leftrightarrow m=\dfrac{7}{2}\left(tm\right)\)

Bài 2:

a,Thay m=-2 vào (1) ta có:

\(x^2-2x-\left(-2\right)^2-4=0\\ \Leftrightarrow x^2-2x-4-4=0\\ \Leftrightarrow x^2-2x-8=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

b, \(\Delta'=\left(-m\right)^2-\left(-m^2-4\right)\ge0=m^2+m^2+4=2m^2+4>0\)

Suy ra pt luôn có 2 nghiệm phân biệt

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=-m^2-4\end{matrix}\right.\)

\(x_1^2+x_2^2=20\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=20\\ \Leftrightarrow2^2-2\left(-m^2-4\right)=20\\ \Leftrightarrow4+2m^2+8-20=0\\ \Leftrightarrow2m^2-8=0\\ \Leftrightarrow m=\pm2\)

\(x_1^3+x_2^3=56\\ \Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=56\\ \Leftrightarrow2^3-3\left(-m^2-4\right).2=56\\ \Leftrightarrow8-6\left(-m^2-4\right)-56\\ =0\\ \Leftrightarrow8+6m^2+24-56=0\\ \Leftrightarrow6m^2-24=0\\ \Leftrightarrow m=\pm2\)

\(x_1-x_2=10\\ \Leftrightarrow\left(x_1-x_2\right)^2=100\\ \Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2-100=0\\ \Leftrightarrow2^2-4\left(-m^2-4\right)-100=0\\ \Leftrightarrow4+4m^2+16-100=0\\ \Leftrightarrow4m^2-80=0\\ \Leftrightarrow m=\pm2\sqrt{5}\)

x1^2+x2^2=(x1+x2)^2-2x1x2

=((m+1)/2)^2-2*(-6/2)

=1/4(m^2+2m+1)+6

=>x1^2=1/4m^2+1/2m+25/4-x2^2

x1^2+x2=-2

=>1/4m^2+1/2m+25/4-x2^2+x2=-2

=>-x2^2+x2+1/4m^2+1/2m+33/4=0

=>x2^2-x2-1/4m^2-1/2m-33/4=0

Δ=(-1)^2-4*1*(-1/4m^2-1/2m-33/4)

=1+m^2+2m+33

=(m+1)^2+33>=33

=>Phương trình luôn có m thỏa mãn

2 nghiệp pt phải:

 (2m - 1)²-4(m² - 1)

Vì x₁ là nghiệm nên

x21−(2m−1)x1+m2−1=0

<=> x12−(2m−1)x1+m2−1=0

<=>x12−2mx1+m2=x1+1

<=> 9m2=0 <=>m=0

#YQ

học tốt nhé !

Ptr có nghiệm `<=>\Delta' > 0`

   `<=>(-m)^2-2m+1 > 0`

  `<=>(m-1)^2 > 0<=>m-1 ne 0<=>m ne 1`

`=>` Áp dụng Viét có: `{(x_1+x_2=-b/a=2m),(x_1.x_2=c/a=2m-1):}`

Ta có: `(x_1 ^2-2mx_1 +3)(x_2 ^2-2mx_2 -2)=50`

`<=>[x_1 ^2-(x_1+x_2)x_1+3][x_2 ^2-(x_1+x_2)x_2 -2]=50`

`<=>(-x_1.x_2+3)(-x_1.x_2-2)=50`

`<=>(1-2m+3)(1-2m-2)=50`

`<=>(4-2m)(-1-2m)=50`

`<=>-4-8m+2m+4m^2=50`

`<=>4m^2-6m-54=0`

`<=>4m^2+12m-18m-54=0`

`<=>(m+3)(4m-18)=0<=>[(m=-3),(m=9/2):}`  (t/m)

ê phải n.nam 9c ko

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}`.