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\(x^3y^2-x^2y^2-2x^2y+2xy+3x-3=0\)
\(\Leftrightarrow x^2y^2\left(x-1\right)-2xy\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2y^2-2xy+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(xy-1\right)^2+2=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow y^2-y-3m+1=0\) (1)
\(\Delta=1-4\left(-3m+1\right)=12m-3>0\Rightarrow m>\frac{1}{4}\)
Gọi \(y_1;y_2\) là 2 nghiệm của (1) \(\Rightarrow\left\{{}\begin{matrix}y_1+y_2=1\\y_1y_2=-3m+1\end{matrix}\right.\)
\(\left(1+y_2\right)\left(1+y_1\right)+3=0\)
\(\Leftrightarrow y_1y_2+y_1+y_2+4=0\)
\(\Leftrightarrow-3m+1+5=0\) \(\Rightarrow m=2\)
a/ ĐKXĐ: ...
\(3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}=2x^2-3x+10\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\ge0\\\sqrt{x^2-2x+4}=b>0\end{matrix}\right.\) \(\Rightarrow2a^2+b^2=2x^2-3x+10\)
Phương trình trở thành:
\(3ab=2a^2+b^2\)
\(\Leftrightarrow2a^2-3ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)=0\Rightarrow\left[{}\begin{matrix}a=b\\b=2a\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+2}=\sqrt{x^2-2x+4}\\\sqrt{x+2}=2\sqrt{x^2-2x+4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x^2-2x+4\\x+2=4x^2-8x+16\end{matrix}\right.\)
2/ \(\Leftrightarrow\left\{{}\begin{matrix}x^2+1+y^2+xy-4y=0\\\left(x^2+1\right)\left(x+y-2\right)=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+1+y\left(x+y-4\right)=0\\\left(x^2+1\right)\left(x+y-2\right)=y\end{matrix}\right.\)
Nhận thấy \(y=0\) không phải nghiệm, hệ tương đương:
\(\left\{{}\begin{matrix}\frac{x^2+1}{y}+x+y-2=2\\\left(\frac{x^2+1}{y}\right)\left(x+y-2\right)=1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\frac{x^2+1}{y}=a\\x+y-2=b\end{matrix}\right.\)
Theo Viet đảo, a và b là nghiệm của: \(t^2-2t+1=0\Rightarrow t=1\)
\(\Rightarrow a=b=1\Rightarrow\left\{{}\begin{matrix}\frac{x^2+1}{y}=1\\x+y-2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+1=y\\x-3=-y\end{matrix}\right.\) \(\Rightarrow x^2+x-2=0\)
Phương trình hoành độ giao điểm:
\(x^2-x+m=0\)
\(\Delta=1-4m>0\Rightarrow m< \frac{1}{4}\)
Theo định lý Viet: \(\left\{{}\begin{matrix}x_1+x_2=1\\x_1x_2=m\end{matrix}\right.\)
\(\left(x_2-x_1\right)^2=\left(x_1+x_2\right)^2-4x_1x_2=1-4m\Rightarrow\left(x_2-x_1\right)^4=\left(1-4m\right)^2\)
\(y_2-y_1=x_2-m-\left(x_1-m\right)=x_2-x_1\)
\(\Rightarrow\left(y_2-y_1\right)^4=\left(x_2-x_1\right)^4=\left(1-4m\right)^2\)
Thay vào bài toán:
\(2\left(1-4m\right)^2=18\)
\(\Rightarrow\left(1-4m\right)^2=9\)
Nhớ chỉ lấy nghiệm \(m< \frac{1}{4}\)
\(\left\{{}\begin{matrix}x^2+2xy-3y^2=-4\left(1\right)\\2x^2+xy+4y^2=5\left(2\right)\end{matrix}\right.\)\(với\)\(y=0\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}x^2=-4\\2x^2=5\end{matrix}\right.\)\(\left(loại\right)\)
\(y\ne0\) \(đặt:x=t.y\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}t^2y^2+2ty^2-3y^2=-4\left(3\right)\\2t^2y^2+ty^2+4y^2=5\left(4\right)\end{matrix}\right.\)
\(\Leftrightarrow5t^2y^2+10ty^2-15y^2=-8t^2y^2-4ty^2-16y^2\)
\(\Leftrightarrow13t^2y^2+14ty^2+y^2=0\)
\(\Leftrightarrow13t^2+14t+1=0\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{1}{13}\\t=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{13}y\left(5\right)\\x=-y\left(6\right)\end{matrix}\right.\)
\(thay\left(5\right)và\left(6\right)\) \(lên\left(1\right)hoặc\left(2\right)\Rightarrow\left(x;y\right)=\left\{\left(1;-1\right);\left(-1;1\right);\left(-\dfrac{1}{\sqrt{133}};\dfrac{13}{\sqrt{133}}\right)\right\}\)
\(pt:x^4-4x^3+x^2+6x+m+2=0\)
\(\Leftrightarrow x^4-4x^3+4x^2-3x^2+6x+m+2=0\)
\(\Leftrightarrow\left(x^2-2x\right)^2-3\left(x^2-2x\right)+m+2=0\left(1\right)\)
\(đặt:x^2-2x=t\ge-1\)
\(\Rightarrow\left(1\right)\Leftrightarrow t^2-3t=-m-2\)
\(xét:f\left(t\right)=t^2-3t\) \(trên[-1;+\text{∞})\) \(và:y=-m-2\)
\(\Rightarrow f\left(-1\right)=4\)
\(f\left(-\dfrac{b}{2a}\right)=-\dfrac{9}{4}\)
\(\left(1\right)\) \(có\) \(3\) \(ngo\) \(pb\Leftrightarrow-m-2=4\Leftrightarrow m=-6\)
\(2)mx^2-2\left(m-1\right)x+m-1=0\)
Để pt có nghiệm kép \(\Leftrightarrow\left\{{}\begin{matrix}a\ne0\\\Delta=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ne0\\\left[-2\left(m-1\right)\right]^2-4m\left(m-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow4\left(m^2-2m+1\right)-4m^2+4m=0\)
\(\Leftrightarrow4m^2-8m+4-4m^2+4m=0\)
\(\Leftrightarrow-4m+4=0\)
\(\Leftrightarrow m=1\)
Vậy để pt trên có nghiệm kép thì \(\left\{{}\begin{matrix}m\ne0\\m=1\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}x+my=3\\mx+4y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}mx+m^2y=3m\\mx+4y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2y-4y=3m-6\\mx+4y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y\left(m^2-4\right)=3m-6\\mx+4y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3m-6}{m^2-4}\\mx=6-4y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3\left(m-2\right)}{\left(m+2\right)\left(m-2\right)}=\dfrac{3}{m+2}\\mx=6-4\cdot\dfrac{3}{m+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{m+2}\\mx=6-\dfrac{12}{m+2}=\dfrac{6\left(m+2\right)-12}{m+2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{m+2}\\mx=\dfrac{6m+12-12}{m+2}=\dfrac{6m}{m+2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{6m}{m+2}:m=\dfrac{6m}{m+2}\cdot\dfrac{1}{m}=\dfrac{6}{m+2}\\y=\dfrac{3}{m+2}\end{matrix}\right.\)
Để phương trình có nghiệm x>1 và y>0 thì \(\left\{{}\begin{matrix}\dfrac{6}{m+2}>1\\\dfrac{3}{m+2}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{m+2}-1>0\\m+2>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{m+2}-\dfrac{m+2}{m+2}>0\\m>-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6-m-2}{m+2}>0\\m>-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4-m>0\\m>-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-m>-4\\m>-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< 4\\m>-2\end{matrix}\right.\Leftrightarrow-2< m< 4\)
Vậy: Để hệ phương trình có nghiệm x>1 và y>0 thì -2<m<4