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\(\left\{{}\begin{matrix}\dfrac{x+2}{y-1}=\dfrac{x-4}{y+2}\\\dfrac{2x+3}{y-1}=\dfrac{4x+1}{2y+1}\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\left(x+2\right)\left(y+2\right)=\left(y-1\right)\left(x-\text{4}\right)\\\left(2x+3\right)\left(2y+1\right)=\left(y-1\right)\left(4x+1\right)\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}xy+2x+2y+4=xy-4y-x+4\\4xy+2x+6y+3=4xy-4x+y-1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)
\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)
Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))
a: ĐKXĐ: y<=1/2
\(\left\{{}\begin{matrix}3\left(x-1\right)-\sqrt{1-2y}=1\\\left(x-1\right)+2\sqrt{1-2y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6\left(x-1\right)-2\sqrt{1-2y}=2\\\left(x-1\right)+2\sqrt{1-2y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7\left(x-1\right)=7\\\left(x-1\right)+2\sqrt{1-2y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-1=1\\2\sqrt{1-2y}=5-1=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\\sqrt{1-2y}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\1-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
b:
ĐKXĐ: \(x\in R\)
\(\left\{{}\begin{matrix}\sqrt{x^2-2x+1}-3y=7\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{\left(x-1\right)^2}-3y=7\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left|x-1\right|-3y=7\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left|x-1\right|-6y=14\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2y=13\\\left|x-1\right|-3y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{13}{2}\\\left|x-1\right|=3y+7=3\cdot\dfrac{13}{2}+7=\dfrac{39}{2}+7=\dfrac{53}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{13}{2}\\x-1\in\left\{\dfrac{53}{2};-\dfrac{53}{2}\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{13}{2}\\x\in\left\{\dfrac{55}{2};-\dfrac{51}{2}\right\}\end{matrix}\right.\)
c: ĐKXĐ: y>=4
\(\left\{{}\begin{matrix}2\left(x^2-x\right)+\sqrt{y-4}=0\\3\left(x^2-x\right)-2\sqrt{y-4}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4\left(x^2-x\right)+2\sqrt{y-4}=0\\3\left(x^2-x\right)-2\sqrt{y-4}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7\left(x^2-x\right)=-7\\2\left(x^2-x\right)+\sqrt{y-4}=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-x=-1\\\sqrt{y-4}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-x+1=0\\y-4=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vôlý\right)\\y=8\end{matrix}\right.\)
=>\(\left(x,y\right)\in\varnothing\)
a, b và c có thể dùng phương pháp thế hoặc cộng trừ đại số
\(a,\left\{{}\begin{matrix}x=1-y\\1-y-y=-5\end{matrix}\right.=>\left\{{}\begin{matrix}x=1-y\\1-2y=-5\end{matrix}\right.=>\left\{{}\begin{matrix}x=1-y\\2y=6\end{matrix}\right.=>\left\{{}\begin{matrix}x=1-y\\y=3\end{matrix}\right.=>\left\{{}\begin{matrix}x=-2\\y=3\end{matrix}\right.\)
Kết luận hpt có 1 nghiệm duy nhất (x;y)=(-2;3)
b và c làm tương tự
a.\(\Leftrightarrow\left\{{}\begin{matrix}2x=-4\\x-y=-5\end{matrix}\right.\) ( cộng đại số bạn nhé )
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\-2-y=-5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=3\end{matrix}\right.\)
b.\(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\x-2y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2-2y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)
c.\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\9x-6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}13x=13\\9x-6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\9.1-6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)
1)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-27\\8x+6y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2y=5x+9\\23x=-23\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(-1;2\right)\)
2)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=4\\2x+4y=10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3y=-6\\x=5-2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;2\right)\)
3)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=14\\3x+6y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=4-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(2;1\right)\)
4)
HPT \(\Leftrightarrow\left\{{}\begin{matrix}5x+6y=17\\54x-6y=42\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}59x=59\\y=9x-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(1;2\right)\)
Câu a)
Có: \(\left\{\begin{matrix} (x+y)^2+3y^2=7\\ x+2y(x+1)=5\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x^2+4y^2+2xy=7\\ x+2y=5-2xy\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} x^2+4y^2+2xy=7\\ x^2+4y^2+4xy=(5-2xy)^2\end{matrix}\right.\)
Lấy PT(2) trừ PT(1) thu được:
\(2xy=(5-2xy)^2-7\)
\(\Leftrightarrow 2(xy)^2-11xy+9=0\)
\(\Rightarrow xy=\frac{9}{2}\) hoặc \(xy=1\) hay \(\left[\begin{matrix} 2xy=9\\ 2xy=2\end{matrix}\right.\)
Nếu \(2xy=9\Rightarrow x+2y=5-2xy=-4\)
Theo định lý Viete đảo thì $x,2y$ là nghiệm của PT:
\(X^2+4X+9=0\)\(\Leftrightarrow (X+2)^2+5=0\) (vl)
Nếu \(2xy=2\Rightarrow x+2y=5-2xy=3\)
Theo định lý Viete đảo thì $x,2y$ là nghiệm của PT:
\(X^2-3X+2=0\Rightarrow (x,2y)=(2,1); (1,2)\)
\(\Rightarrow (x,y)=(2,\frac{1}{2}); (1; 1)\)
Câu b:
\(\left\{\begin{matrix} x(y-1)+2y=x(x+1)(1)\\ \sqrt{2x-1}+xy-3y+1=0(2)\end{matrix}\right.\)
Từ \((1)\Leftrightarrow y(x+2)=x(x+1)+x\)
\(\Leftrightarrow y(x+2)=x(x+2)\Leftrightarrow (x+2)(y-x)=0\)
\(\Rightarrow \left[\begin{matrix} x=-2\\ x=y\end{matrix}\right.\)
Nếu \(x=-2\) thay vào (2) thấy ngay vô lý vì ĐKXĐ là \(x\geq \frac{1}{2}\)
Nếu \(x=y\), thay vào (2): \(\sqrt{2x-1}+x^2-3x+1=0\)
\(\Leftrightarrow (\sqrt{2x-1}-x)+(x^2-2x+1)=0\)
\(\Leftrightarrow \frac{2x-1-x^2}{\sqrt{2x-1}+x}+(x-1)^2=0\)
\(\Leftrightarrow (x-1)^2\left[1-\frac{1}{\sqrt{2x-1}+x}\right]=0\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \sqrt{2x-1}+x=1\end{matrix}\right.\)
Với trường hợp \(\sqrt{2x-1}+x=1(x\leq 1)\Rightarrow \sqrt{2x-1}=1-x\)
\(\Rightarrow 2x-1=(1-x)^2=x^2-2x+1\)
\(\Leftrightarrow x^2-4x+2=0\Rightarrow x=2\pm \sqrt{2}\). Vì \(\frac{1}{2}\leq x\leq 1\Rightarrow x=2-\sqrt{2}\)
Vậy \((x,y)=(1,1); (2-\sqrt{2}; 2-\sqrt{2})\)