giải hệ phương trình \(\left\{{}\begin{matrix}\left(x-y\right)\sqrt{y}+y\sqrt{x-y}+1=x+\sqrt{xy-y^2}\\x^2y-3x+2+\left(2x^2-3x\right)\sqrt{y-1}=0\end{matrix}\right.\)
Những câu hỏi liên quan
Giải các hệ phương trình sau:
a) left{{}begin{matrix}4x^2-4xy-14x-3y^2+y+1005sqrt{xy}+2x+2y6sqrt{y}-8end{matrix}right.
b) left{{}begin{matrix}2x^4+3x^2y+4x^2-2y^2+3y+20sqrt{xleft(y-1right)}+2y+2sqrt{y-1}3x+2sqrt{x}+2end{matrix}right.
c) left{{}begin{matrix}x^6+3x^2-y^3-6y^2-15y-140sqrt{xy+2x-y-2}+6x-2y10end{matrix}right.
d) left{{}begin{matrix}xy+x+yx^2-2y^2xsqrt{2y}-ysqrt{x-1}2x-2yend{matrix}right.
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Giải các hệ phương trình sau:
a) \(\left\{{}\begin{matrix}4x^2-4xy-14x-3y^2+y+10=0\\5\sqrt{xy}+2x+2y=6\sqrt{y}-8\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2x^4+3x^2y+4x^2-2y^2+3y+2=0\\\sqrt{x\left(y-1\right)}+2y+2\sqrt{y-1}=3x+2\sqrt{x}+2\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}x^6+3x^2-y^3-6y^2-15y-14=0\\\sqrt{xy+2x-y-2}+6x-2y=10\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
giải hệ phương trình
a) left{{}begin{matrix}sqrt{2x^2+2y^2}+sqrt{frac{4}{3}left(x^2+xy+y^2right)}2left(x+yright)sqrt{3x+1}+sqrt{5x+4}3xy-y+3end{matrix}right.
b) left{{}begin{matrix}sqrt{5x^2+2xy+2y^2}+sqrt{2x^2+2xy+5y^2}3left(x+yright)sqrt{x+2y+1}+2sqrt[3]{12x+7y+8}2xy+x+5end{matrix}right.
c)left{{}begin{matrix}x^2+xy+x+30left(x+1right)^2+3left(y+1right)+2left(xy-sqrt{x^2y+2y}right)0end{matrix}right.
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giải hệ phương trình
a) \(\left\{{}\begin{matrix}\sqrt{2x^2+2y^2}+\sqrt{\frac{4}{3}\left(x^2+xy+y^2\right)}=2\left(x+y\right)\\\sqrt{3x+1}+\sqrt{5x+4}=3xy-y+3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\\\sqrt{x+2y+1}+2\sqrt[3]{12x+7y+8}=2xy+x+5\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2+xy+x+3=0\\\left(x+1\right)^2+3\left(y+1\right)+2\left(xy-\sqrt{x^2y+2y}\right)=0\end{matrix}\right.\)
b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
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caau a) binh phuong len ra no x=y tuong tu
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c)
ĐK $y \geqslant 0$
Hệ đã cho tương đương với
$\left\{\begin{matrix} 2x^2+2xy+2x+6=0\\ (x+1)^2+3(y+1)+2xy=2\sqrt{y(x^2+2)} \end{matrix}\right.$
Trừ từng vế $2$ phương trình ta được
$x^2+2+2\sqrt{y(x^2+2)}-3y=0$
$\Leftrightarrow (\sqrt{x^2+2}-\sqrt{y})(\sqrt{x^2+2}+3\sqrt{y})=0$
$\Leftrightarrow x^2+2=y$
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Giải các hệ phương trình sau:a) \(\left\{{}\begin{matrix}\left(2x-y\right)^2-6x+3y=0\\x+2y=0\end{matrix}\right.\);b) \(\left\{{}\begin{matrix}\sqrt{\dfrac{2x-y}{x+y}}+\sqrt{\dfrac{x+y}{2x-y}}=2\\3x+y=14\end{matrix}\right.\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2-3\left(2x-y\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(2x-y-3\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-y=0\\x+2y=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y-3=0\\x+2y=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
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b.
ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)
Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:
\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)
\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)
\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)
Thay xuống pt dưới:
\(6y+y=14\Rightarrow y=2\)
\(\Rightarrow x=4\)
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Giải hệ phương trình:
1, left{{}begin{matrix}left(17-3xright)sqrt{5-x}+left(3y-14right)sqrt{4-y}02sqrt{2x+y+5}+3sqrt{3x+2y+11}x^2+6x+13end{matrix}right.
2, left{{}begin{matrix}xleft(x+yright)+sqrt{x+y}sqrt{2y}left(sqrt{2y^3}+1right)x^2y-5x^2+7left(x+yright)-46sqrt[3]{xy-x+1}end{matrix}right.
3, left{{}begin{matrix}sqrt{x}+sqrt[4]{32-x}-y^2+30sqrt[4]{x}+sqrt{32-x}+6y-240end{matrix}right.
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Giải hệ phương trình:
1, \(\left\{{}\begin{matrix}\left(17-3x\right)\sqrt{5-x}+\left(3y-14\right)\sqrt{4-y}=0\\2\sqrt{2x+y+5}+3\sqrt{3x+2y+11}=x^2+6x+13\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x\left(x+y\right)+\sqrt{x+y}=\sqrt{2y}\left(\sqrt{2y^3}+1\right)\\x^2y-5x^2+7\left(x+y\right)-4=6\sqrt[3]{xy-x+1}\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt[4]{32-x}-y^2+3=0\\\sqrt[4]{x}+\sqrt{32-x}+6y-24=0\end{matrix}\right.\)
1/PT (1) cho ta nhân tử x - y - 1:)
\(\left\{{}\begin{matrix}\left(17-3x\right)\sqrt{5-x}+\left(3y-14\right)\sqrt{4-y}=0\left(1\right)\\2\sqrt{2x+y+5}+3\sqrt{3x+2y+11}=x^2+6x+13\left(2\right)\end{matrix}\right.\)
ĐK: \(x\le5;y\le4\); \(2x+y+5\ge0;3x+2y+11\ge0\)
PT (1) \(\Leftrightarrow\left(17-3x\right)\left(\sqrt{5-x}-\sqrt{4-y}\right)-3\left(x-y-1\right)\sqrt{4-y}=0\)
\(\Leftrightarrow\left(3x-17\right)\left(\frac{x-y-1}{\sqrt{5-x}+\sqrt{4-y}}\right)-3\left(x-y-1\right)\sqrt{4-y}=0\)
\(\Leftrightarrow\left(x-y-1\right)\left(\frac{3x-17}{\sqrt{5-x}+\sqrt{4-y}}-3\sqrt{4-y}\right)=0\)
Dễ thấy cái ngoặc to < 0
Do đó x= y + 1
Thay xuống PT (2):\(y^2+8y+20=2\sqrt{3y+7}+3\sqrt{5y+14}\)\(\left(y+1\right)\left(y+2\right)=y^2+3y+2\)
ĐK: \(y\ge-\frac{7}{3}\) (để các căn thức được thỏa mãn)
PT (2) \(\Leftrightarrow y^2+3y+2+2\left(y+3-\sqrt{3y+7}\right)+3\left(y+4-\sqrt{5y+14}\right)=0\)
\(\Leftrightarrow\left(y^2+3y+2\right)\left(1+\frac{2}{y+3+\sqrt{3y+7}}+\frac{3}{y+4+\sqrt{5y+14}}\right)=0\)
Cái ngoặc to > 0 =>...
P/s: Is that true? Ko đúng thì chịu thua-_- Mất nửa tiếng đồng hồ để gõ bài này đấy:(
2/ĐK: \(x\ge-y;y\ge0\)
PT (1) \(\Leftrightarrow x\left(x+y\right)+\sqrt{x+y}=2y^2+\sqrt{2y}\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)+y\left(x-y\right)+\sqrt{x+y}-\sqrt{2y}=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+2y+\frac{1}{\sqrt{x+y}+\sqrt{2y}}\right)=0\)
Cái ngoặc to \(\ge y+\frac{1}{\sqrt{x+y}+\sqrt{2y}}>0\).
Do đó x = y \(\ge0\)
Thay xuống pt dưới: \(x^3-5x^2+14x-4=6\sqrt[3]{x^2-x+1}\)
Lập phương hai vế lên ra pt bậc 6, tuy nhiên cứ yên tâm, nghiệm rất đẹp: x = 1:)
Em đưa kết quả luôn: \(\left(x-1\right)\left(x^2-4x+7\right)\left(x^6-10x^5+56x^4-160x^3+272x^2-64x+40\right)=0\)
P/s: khúc cuối em ko còn cách nào khác nên đành lập phương:((
Giải hệ phương trình:
\(a,\left\{{}\begin{matrix}2x^3+x^2y+2x^2+xy+6=0\\x^2+3x+y=1\end{matrix}\right.\)
\(b,\left\{{}\begin{matrix}x^2=\left(2-y\right)\left(2+y\right)\\2x^3=\left(x+y\right)\left(4-xy\right)\end{matrix}\right.\)
\(c,\left\{{}\begin{matrix}\sqrt[3]{x+2y}=4-x-y\\\sqrt[3]{x+6}+\sqrt{2y}=2\end{matrix}\right.\)
a/
\(\Leftrightarrow\left\{{}\begin{matrix}x^2\left(2x+y\right)+x\left(2x+y\right)=-6\\x^2+x+2x+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+x\right)\left(2x+y\right)=-6\\x^2+x+2x+y=1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2+x=a\\2x+y=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}ab=-6\\a+b=1\end{matrix}\right.\) với
Theo Viet đảo, a và b là nghiệm của:
\(t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x=3\\2x+y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x=-2\left(vn\right)\\2x+y=3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x-3=0\\y=-2x-2\end{matrix}\right.\) (bấm casio)
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b/
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=4-y^2\\2x^3=\left(x+y\right)\left(4-xy\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=4\\2x^3=\left(x+y\right)\left(4-xy\right)\end{matrix}\right.\)
\(\Rightarrow2x^3=\left(x+y\right)\left(x^2+y^2-xy\right)\)
\(\Leftrightarrow2x^3=x^3+y^3\)
\(\Leftrightarrow x^3=y^3\Rightarrow x=y\)
Thay vào pt đầu:
\(2x^2=4\Rightarrow x^2=2\Rightarrow x=y=\pm\sqrt{2}\)
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Giải hệ phương trình:
1, left{{}begin{matrix}x^2+1+y^2+xyyx+y-2frac{y}{1+x^2}end{matrix}right.
2, left{{}begin{matrix}x^3+8y^3-4xy^212x^4+8y^4-2x-y0end{matrix}right.
3, left{{}begin{matrix}x^2+y^2frac{1}{5}4x^2+3x-frac{57}{25}-yleft(3x+1right)end{matrix}right.
4, left{{}begin{matrix}sqrt{12-y}+sqrt{yleft(12-xright)}12x^3-8x-12sqrt{y-2}end{matrix}right.
5, left{{}begin{matrix}left(1-yright)sqrt{x-y}+x2+left(x-y-1right)sqrt{y}2y^2-3x+6y+12sqrt{x-2y}-sqrt{4x-5y-3}end{matrix}right.
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Giải hệ phương trình:
1, \(\left\{{}\begin{matrix}x^2+1+y^2+xy=y\\x+y-2=\frac{y}{1+x^2}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^3+8y^3-4xy^2=1\\2x^4+8y^4-2x-y=0\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}x^2+y^2=\frac{1}{5}\\4x^2+3x-\frac{57}{25}=-y\left(3x+1\right)\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\sqrt{12-y}+\sqrt{y\left(12-x\right)}=12\\x^3-8x-1=2\sqrt{y-2}\end{matrix}\right.\)
5, \(\left\{{}\begin{matrix}\left(1-y\right)\sqrt{x-y}+x=2+\left(x-y-1\right)\sqrt{y}\\2y^2-3x+6y+1=2\sqrt{x-2y}-\sqrt{4x-5y-3}\end{matrix}\right.\)
1) \(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}2x^2+y^2-3xy+3x-2y+1=0\\4x^2-y^2+x+4=\sqrt{2x+y}+\sqrt{x+4y}\end{matrix}\right.\)
Giải hệ phương trình:
1. left{{}begin{matrix}x+32sqrt{left(3y-xright)left(y+1right)}sqrt{3y-2}-sqrt{dfrac{x+5}{2}}xy-2y-2end{matrix}right.
2. left{{}begin{matrix}sqrt{2y^2-7y+10-xleft(y+3right)}+sqrt{y+1}x+1sqrt{y+1}+dfrac{3}{x+1}x+2yend{matrix}right.
3. left{{}begin{matrix}sqrt{4x-y}-sqrt{3y-4x}12sqrt{3y-4x}+yleft(5x-yright)xleft(4x+yright)-1end{matrix}right.
4. left{{}begin{matrix}9sqrt{dfrac{41}{2}left(x^2+dfrac{1}{2x+y}right)}3+40xx^2+5xy+6y4y^2+9x+9end{matrix}right.
5. left{{}begin{mat...
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Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}x+3=2\sqrt{\left(3y-x\right)\left(y+1\right)}\\\sqrt{3y-2}-\sqrt{\dfrac{x+5}{2}}=xy-2y-2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{2y^2-7y+10-x\left(y+3\right)}+\sqrt{y+1}=x+1\\\sqrt{y+1}+\dfrac{3}{x+1}=x+2y\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\sqrt{4x-y}-\sqrt{3y-4x}=1\\2\sqrt{3y-4x}+y\left(5x-y\right)=x\left(4x+y\right)-1\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}9\sqrt{\dfrac{41}{2}\left(x^2+\dfrac{1}{2x+y}\right)}=3+40x\\x^2+5xy+6y=4y^2+9x+9\end{matrix}\right.\)
5. \(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left[y+\sqrt{xy}+x\left(1-x\right)\right]=4\end{matrix}\right.\)
6. \(\left\{{}\begin{matrix}x^4-x^3+3x^2-4y-1=0\\\sqrt{\dfrac{x^2+4y^2}{2}}+\sqrt{\dfrac{x^2+2xy+4y^2}{3}}=x+2y\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)
Giải hệ phương trình:
\(\left\{{}\begin{matrix}x^3+2y^2+xy^2=2+x-2x^2\\4y^2=\left(\sqrt{y^2+1}+1\right)\left(y^2-x^3+3x-2\right)\end{matrix}\right.\)
phân tích pt1 thành (x+2)(x2+y2-1)=0
\(\Rightarrow\)x= -2 hoặc y2=1-x2
Nếu x=-2 thay vào pt2
Nếu y2=1-x2.Thay vào pt2 để đưa về biến x
Nhân liên hợp 2 vế vs \(\sqrt{2-x^2}-1\)
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