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\(\left\{{}\begin{matrix}3x+y=m+1\\x-2y=5m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+2y=2m+2\\x-2y=5m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\x-2y=5m-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=m\\m-2y=5m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=1-2m\end{matrix}\right.\\ 4x^2-y^2=10\Leftrightarrow4m^2-\left(1-2m\right)^2=10\\ \Leftrightarrow4m^2-4m^2+4m-1=10\\ \Leftrightarrow m=\dfrac{11}{4}\)
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\(\left\{{}\begin{matrix}3x-y=2m-1\\x+2y=3m+2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2y=4m-2\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y+x+2y=4m-2+3m+2\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\x+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\m+2y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\2y=2m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=m+1\end{matrix}\right.\)
\(x^2+y^2+3\\ =m^2+\left(m+1\right)^2+3\\ =m^2+m^2+2m+1+3\\ =2m^2+2m+4\\ =2\left(m^2+m+2\right)\)
\(=2\left(m^2+m+\dfrac{1}{4}+\dfrac{7}{4}\right)\)
\(=2\left[\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]\)
\(=2\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{2}\ge\dfrac{7}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow m=-\dfrac{1}{2}\)
Vậy ...
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Ta có
3 x − y = 2 m + 1 x + 2 y = − m + 2 ⇔ 6 x − 2 y = 4 m + 2 x + 2 y = − m + 2 ⇔ 7 x = 3 m + 4 x + 2 y = − m + 2 ⇔ x = 3 m + 4 7 3 m + 4 7 + 2 y = − m + 2 ⇔ x = 3 m + 4 7 2 y = − 7 m + 14 7 − 3 m + 4 7 ⇔ x = 3 m + 4 7 y = − 5 m + 5 7
hệ phương trình có nghiệm duy nhất ( x ; y ) = 3 m + 4 7 ; − 5 m + 5 7
Để x – y = 1 thì 3 m + 4 7 − − 5 m + 5 7 = 1 ⇔ 8m – 1 = 7 ⇔ 8m = 8 m = 1
Vậy với m = 1 thì hệ phương trình có nghiệm duy nhất (x; y) thỏa mãn x − y = 1
Đáp án: C
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\(HPT\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4m+6\\x+2y=3m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x=7m+7\\x+2y=3m+1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=m+1\\m+1+2y=3m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=m+1\\y=m\end{matrix}\right.\)
\(x^2+y^2=5\Leftrightarrow m^2+2m+1+m^2=5\\ \Leftrightarrow2m^2+2m-4=0\\ \Leftrightarrow m^2+m-2=0\\ \Leftrightarrow\left[{}\begin{matrix}m=1\\m=-2\end{matrix}\right.\)
3x+2y = 2
=> y= \(\frac{2-3x}{2}\) thay vào x^3 + y^2 = 12 ta có :
\(x^3+\left(\frac{2-3x}{2}\right)^2=12\)
<=> \(4x^3+4-12x+9x^2=48\)
<=> \(4x^3+9x^2-12x-44=0\)
<=> \(\left(x-2\right)\left(4x^2+17x+22\right)=0\)
<=> x = 2
=> y = -2
Vậy ( 2 ; -2 ) là nghiệm của hpt