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Bài 1.
\(\left\{{}\begin{matrix}x-3y=5-2m\\2x+y=3\left(m+1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3y=5-2m\\6x+3y=9m+9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=7m+14\\x-3y=5-2m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\m+2-3y=5-2m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\-3y=-3m+3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\y=m-1\end{matrix}\right.\)
\(x_0^2+y_0^2=9m\)
\(\Leftrightarrow\left(m+2\right)^2+\left(m-1\right)^2=9m\)
\(\Leftrightarrow m^2+4m+4+m^2-2m+1-9m=0\)
\(\Leftrightarrow2m^2-7m+5=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}m=1\\m=\dfrac{5}{2}\end{matrix}\right.\) ( Vi-ét )
\(\left\{{}\begin{matrix}x^2+2xy-3y^2=-4\left(1\right)\\2x^2+xy+4y^2=5\left(2\right)\end{matrix}\right.\)\(với\)\(y=0\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}x^2=-4\\2x^2=5\end{matrix}\right.\)\(\left(loại\right)\)
\(y\ne0\) \(đặt:x=t.y\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}t^2y^2+2ty^2-3y^2=-4\left(3\right)\\2t^2y^2+ty^2+4y^2=5\left(4\right)\end{matrix}\right.\)
\(\Leftrightarrow5t^2y^2+10ty^2-15y^2=-8t^2y^2-4ty^2-16y^2\)
\(\Leftrightarrow13t^2y^2+14ty^2+y^2=0\)
\(\Leftrightarrow13t^2+14t+1=0\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{1}{13}\\t=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{13}y\left(5\right)\\x=-y\left(6\right)\end{matrix}\right.\)
\(thay\left(5\right)và\left(6\right)\) \(lên\left(1\right)hoặc\left(2\right)\Rightarrow\left(x;y\right)=\left\{\left(1;-1\right);\left(-1;1\right);\left(-\dfrac{1}{\sqrt{133}};\dfrac{13}{\sqrt{133}}\right)\right\}\)
\(pt:x^4-4x^3+x^2+6x+m+2=0\)
\(\Leftrightarrow x^4-4x^3+4x^2-3x^2+6x+m+2=0\)
\(\Leftrightarrow\left(x^2-2x\right)^2-3\left(x^2-2x\right)+m+2=0\left(1\right)\)
\(đặt:x^2-2x=t\ge-1\)
\(\Rightarrow\left(1\right)\Leftrightarrow t^2-3t=-m-2\)
\(xét:f\left(t\right)=t^2-3t\) \(trên[-1;+\text{∞})\) \(và:y=-m-2\)
\(\Rightarrow f\left(-1\right)=4\)
\(f\left(-\dfrac{b}{2a}\right)=-\dfrac{9}{4}\)
\(\left(1\right)\) \(có\) \(3\) \(ngo\) \(pb\Leftrightarrow-m-2=4\Leftrightarrow m=-6\)
1)
\(\left\{{}\begin{matrix}x+y=4\\2x+3y=m\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x+3y=12\\2x+3y=m\end{matrix}\right.\)
trừ 2 vế của pt cho nhau ta tìm được
\(\left\{{}\begin{matrix}x=12-m\\y=m-8\end{matrix}\right.\)
để \(\left\{{}\begin{matrix}x>0\\y< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m< 12\\m< 8\end{matrix}\right.\Rightarrow}m< 8}\)
Lời giải:
Cộng 2 pt theo vế có:
$3x=3m+3\Rightarrow x=m+1$
$y=x-(2m+1)=m+1-(2m+1)=-m$
Khi đó:
$(x+1)(y-3)<0$
$\Leftrightarrow (m+1+1)(-m-3)<0$
$\Leftrightarrow (m+2)(m+3)>0$
$\Leftrightarrow m>-2$ hoặc $m<-3$
=>x=3m-my và m(3m-my)-y=m^2-2
=>x=3m-my và 3m^2-m^2y-y=m^2-2
=>x=3m-my và 3m^2-y(m^2+1)=m^2-2
=>x=3m-my và y(m^2+1)=3m^2-m^2+2=2m^2+2
=>y=2 và x=3m-2m=m
x^2-y=2x+1
=>m^2-2=2m+1
=>m^2-2m-3=0
=>m=3 hoặc m=-1
a/ \(\Delta'=\left(m+2\right)^2-\left(3m+2\right)=m^2+m+2>0\) \(\forall m\)
Pt đã cho luôn có 2 nghiệm pb
Kết hợp Viet và đề bài ta được: \(\left\{{}\begin{matrix}x_1+x_2=2m+4\\-2x_1+x_2=3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x_1=2m+1\\x_2=2x_1+3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{2m+1}{3}\\x_2=\frac{4m+11}{3}\end{matrix}\right.\)
Cũng theo Viet:
\(x_1x_2=3m+2\Leftrightarrow\left(\frac{2m+1}{3}\right)\left(\frac{4m+11}{3}\right)=3m+2\)
\(\Leftrightarrow8m^2+26m+11=27m+18\)
\(\Leftrightarrow8m^2-m-7=0\Rightarrow\left[{}\begin{matrix}m=1\\m=-\frac{7}{8}\end{matrix}\right.\)
Câu 2:
\(2x^2+xy-y^2+3y-2=0\)
\(\Leftrightarrow2x^2+2xy-2x-xy-y^2+y+2x+2y-2=0\)
\(\Leftrightarrow2x\left(x+y-1\right)-y\left(x+y-1\right)+2\left(x+y-1\right)=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(2x-y+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1-x\\y=2x+2\end{matrix}\right.\)
Thay xuống dưới:
\(\Rightarrow\left[{}\begin{matrix}x^2-\left(1-x\right)^2=3\\x^2-\left(2x+2\right)^2=3\end{matrix}\right.\)
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