mk ms 2k8

2/5+3/5:(3/5+-2/3)-3-1/2

=2/5+3/5:(9/15+-10/15)-3-1/2

=2/5+3/5:(-1)/15-3-1/2

=2/5+3/5.15/(-1)-3-1/2

=2/5+(-9)-3-1/2

=(2/5-1/2)+(-9-3)

=(4/10-5/10)+(-12)

=-1/10-12

=-1/10-120/10

=(-121)/10

-121/10

\(\Delta=m^2+4>0;\forall m\Rightarrow\) phương trình luôn có 2 nghiệm pb

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-m\\x_1x_2=-1\end{matrix}\right.\)

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

\(\Leftrightarrow-m^3-3m=-4\)

\(\Leftrightarrow m^3+3m-4=0\)

\(\Leftrightarrow\left(m-1\right)\left(m^2+m+4\right)=0\)

\(\Leftrightarrow m=1\)

\(\Delta=m^2-4.1.\left(-1\right)=m^2+4>0\) suy ra pt luôn có 2 nghiệm phân biệt 

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=-m\\x_1x_2=-1\end{matrix}\right.\)

\(x^3_1+x^3_2=-4\\ \Leftrightarrow\left(x_1+x_2\right)\left(x^2_1-x_1x_2+x_2^2\right)=-4\\ \Leftrightarrow-m\left[\left(x_1+x_2\right)^2-3x_1x_2\right]=-4\\ \Leftrightarrow m\left[\left(-m\right)^2-3.\left(-1\right)\right]=4\\ \Leftrightarrow m\left(m+3\right)-4=0\\ \Leftrightarrow m^2+3m-4=0\\ \Leftrightarrow m^2+4m-m-4=0\\ \Leftrightarrow m\left(m+4\right)-\left(m+4\right)=0\\ \Leftrightarrow\left(m+4\right)\left(m-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=-4\\m=1\end{matrix}\right.\)

1: \(\Delta=2^2-4\cdot1\left(m-1\right)\)

\(=4-4m+4=-4m+8\)

Để phương trình có hai nghiệm phân biệt thì \(\Delta>0\)

=>-4m+8>0

=>-4m>-8

=>m<2

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-2\\x_1\cdot x_2=\dfrac{c}{a}=m-1\end{matrix}\right.\)

\(x_1^3+x_2^3-6x_1x_2=4\left(m-m^2\right)\)

=>\(\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)-6x_1x_2=4\left(m-m^2\right)\)

=>\(\left(-2\right)^3-3\cdot\left(-2\right)\left(m-1\right)-6\left(m-1\right)=4\left(m-m^2\right)\)

=>\(-8+6\left(m-1\right)-6\left(m-1\right)=4\left(m-m^2\right)\)

=>\(4\left(m^2-m\right)=8\)

=>\(m^2-m=2\)

=>\(m^2-m-2=0\)

=>(m-2)(m+1)=0

=>\(\left[{}\begin{matrix}m-2=0\\m+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=2\left(loại\right)\\m=-1\left(nhận\right)\end{matrix}\right.\)

2: \(x_1^2+2x_2+2x_1x_2+20=0\)

=>\(x_1^2-x_2\left(x_1+x_2\right)+2x_1x_2+20=0\)

=>\(x_1^2-x_2^2+x_1x_2+20=0\)

=>\(\left(x_1-x_2\right)\left(x_1+x_2\right)+m-1+20=0\)

=>\(-2\left(x_1-x_2\right)=-m-19\)

=>2(x1-x2)=m+19

=>\(x_1-x_2=\dfrac{1}{2}\left(m+19\right)\)

=>\(\left(x_1-x_2\right)^2=\dfrac{1}{4}\left(m+19\right)^2\)

=>\(\left(x_1+x_2\right)^2-4x_1x_2=\dfrac{1}{4}\left(m+19\right)^2\)

=>\(\left(-2\right)^2-4\left(m-1\right)=\dfrac{1}{4}\left(m+19\right)^2\)

=>\(4-4m+4=\dfrac{1}{4}\left(m+19\right)^2\)

=>\(\left(m+19\right)^2=4\left(-4m+8\right)=-16m+32\)

=>\(m^2+38m+361+16m-32=0\)

=>\(m^2+54m+329=0\)

=>\(\left[{}\begin{matrix}m=-7\left(nhận\right)\\m=-47\left(nhận\right)\end{matrix}\right.\)

Ta có: \(\frac{c}{a}=-\frac{2}{2}=-1< 0\)

=> Phương trình luôn có 2 ngiệm trái dấu \(x_1;x_2\)

Theo định lí viet: \(x_1x_2=-1;x_1+x_2=\frac{1-m}{2}\)

Ta có: \(\left(x_1+\frac{1}{2}x^2_1-x^3_1\right)\left(x_2+\frac{1}{2}x^2_2-x^3_2\right)=4\)

<=> \(x_1x_2\left(x_1^2-\frac{1}{2}x_1-1\right)\left(x_2^2-\frac{1}{2}x_2-x_2\right)=4\)

<=> \(\left(2x_1^2-x_1-2\right)\left(2x_2^2-x_2-2\right)=-16\)

<=> \(\left(2x_1x_2\right)^2-2x_1^2x_2-4x_1^2-2x_1x_2^2+x_1x_2+2x_2-4x_2^2+2x_2+4=-16\)

<=> \(4+2x_1-4x_1^2+2x_2-1+2x_2-4x_2^2+2x_2+4=-16\)

<=> \(4x_1^2+4x_2^2-4x_1-4x_2=23\)

<=> \(4\left(x_1+x_2\right)^2-4\left(x_1+x_2\right)=15\)

<=> \(\orbr{\begin{cases}x_1+x_2=\frac{5}{2}\\x_1+x_2=-\frac{3}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1-m}{2}=\frac{5}{2}\\\frac{1-m}{2}=-\frac{3}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}m=-4\\m=4\end{cases}}\)

Vậy:.... 

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