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\(\left\{{}\begin{matrix}\left(x+1\right)^2-y^2+6y-9=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y^2-6y+9\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y-3\right)^2=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+1+y-3\right)\left(x+1-y+3\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+y-2\right)\left(x-y+4\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
TH1: x+y-2=0
=>x=-y+2
\(2x^2+y^2-6y-1=0\)
=>\(2\left(-y+2\right)^2+y^2-6y-1=0\)
=>\(2\left(y^2-4y+4\right)+y^2-6y-1=0\)
=>\(3y^2-14y+7=0\)
\(\Delta=\left(-14\right)^2-2\cdot3\cdot7=196-42=154>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}y=\dfrac{14-\sqrt{154}}{6}\\y=\dfrac{14+\sqrt{154}}{6}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-y+2=\dfrac{-2+\sqrt{154}}{6}\\x=\dfrac{-2-\sqrt{154}}{6}\end{matrix}\right.\)
TH2: x-y+4=0
=>x=y-4
\(2x^2+y^2-6y-1=0\)
=>\(2\left(y-4\right)^2+y^2-6y-1=0\)
=>\(2\left(y^2-8y+16\right)+y^2-6y-1=0\)
=>\(3y^2-22y+31=0\)
\(\Delta=\left(-22\right)^2-4\cdot3\cdot31=112>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}y_1=\dfrac{22-\sqrt{112}}{2\cdot3}=\dfrac{11-\sqrt{28}}{3}\\y_2=\dfrac{22+\sqrt{112}}{2\cdot3}=\dfrac{11+\sqrt{28}}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=y-4=\dfrac{11-\sqrt{28}}{3}-4=\dfrac{-1-\sqrt{28}}{3}\\x=y-4=\dfrac{11+\sqrt{28}}{3}-4=\dfrac{-1+\sqrt{28}}{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x+1\right)^2-y^2+6y-9=0\\2x^2+y^2-6y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y^2-6y+9\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y-3\right)^2=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y-2\right)\left(x-y+4\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y-2=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\\\left\{{}\begin{matrix}x-y+4=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2-y\\2\cdot\left(2-y\right)^2+y^2-6y-1=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=y-4\\2\cdot\left(y-4\right)^2+y^2-6y-1=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2-y\\3y^2-14y+7=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=y-4\\3y^2-22y+31=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-\dfrac{1+2\sqrt{7}}{3}\\y=\dfrac{7+2\sqrt{7}}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{-1+2\sqrt{7}}{3}\\y=\dfrac{7-2\sqrt{7}}{3}\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{-1+2\sqrt{7}}{3}\\y=\dfrac{11+2\sqrt{7}}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{1+2\sqrt{7}}{3}\\y=\dfrac{11-2\sqrt{7}}{3}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
Vậy các cặp (x;y) thỏa mãn là: \(\left(-\dfrac{1+2\sqrt{7}}{3};\dfrac{7+2\sqrt{7}}{3}\right);\left(\dfrac{-1+2\sqrt{7}}{3};\dfrac{7-2\sqrt{7}}{3}\right);\left(\dfrac{-1+2\sqrt{7}}{3};\dfrac{11+2\sqrt{7}}{3}\right);\left(-\dfrac{1+2\sqrt{7}}{3};\dfrac{11-2\sqrt{7}}{3}\right)\)
Ta có: \(x^2+6x+9-y^2\)
\(=\left(x+3\right)^2-y^2\)
=(x+3-y)(x+3+y)
3 số đầu bạn gộp thành hàng đẳng thức nhé
\(x^2+6y+9-y^2\)
\(\Rightarrow\left(x+3\right)^2-y^2\)
\(\Rightarrow\left(x+3-y\right).\left(x+3+y\right)\)
theo tính chất dãy tỉ số bằng nhau ta có: \(\frac{\left(7x-5\right)+\left(6x-4\right)}{6+4}=7x+6y-9\Leftrightarrow\frac{7x+6y-9}{10}=7x+6y-9\Leftrightarrow63x+54y-81=0\)
lại có: \(\frac{7x-5}{6}=\frac{6y-4}{4}\Rightarrow28x-20=36y-24\Rightarrow7x=9y-1\)
nên \(63x+54y-81=0\Leftrightarrow7x\cdot9+54y-81=0\Leftrightarrow9\left(9y-1\right)+54y-81=0\Leftrightarrow81y-9+54y-81=0\Leftrightarrow135y-90=0\Leftrightarrow y=\frac{90}{135}=\frac{2}{3}\Rightarrow x=\frac{9y-1}{7}=\frac{5}{7}\)
\(7x+3y⋮9\)
Ta có
\(9x+9y⋮9\)
\(\Rightarrow\left(9x+9y\right)-\left(7x+3y\right)=2x+6y⋮9\)
Theo bài ra ta có:
\(\left\{{}\begin{matrix}9x+9y⋮9\\2x+6y⋮9\end{matrix}\right.\)
⇔ 9\(x\) + 9y - (2\(x\) + 6y) ⋮ 9
⇔ 9\(x\) + 9y - 2\(x\) - 6y ⋮ 9
⇔ (9\(x\) - 2\(x\)) + (9y - 6y)⋮ 9
⇔ 7\(x\) + 3y ⋮ 9(đpcm)
\(=\left(5xy+8xy-95xy+8xy\right)+\left(-3y-6y-3y-6y\right)+6+6-9\)
\(=-74xy-18y+3\)
Đề bài sai, ví dụ \(x=11;y=3\) thì \(3x-8y=9\) chia hết cho 9 nhưng \(4x-6y=26\) không chia hết cho 9
1) \(x^2+6y-9-y^2=x^2-\left(y^2-6y+9\right)=x^2-\left(y-3\right)^2=\left(x-y+3\right)\left(x+y-3\right)\)
2) \(9y^2-6y+1-25x^2=\left(3y\right)^2-2.3y+1-\left(5x\right)^2=\left(3y-1\right)^2-\left(5x\right)^2\)
\(=\left(3y-1-5x\right)\left(3y-1+5x\right)\)
3) \(a^2-9+6x-x^2=a^2-\left(x^2-6x+9\right)=a^2-\left(x-3\right)^2=\left(a-x+3\right)\left(a+x-3\right)\)