Ghpt \(\left\{{}\begin{matrix}x^2-5y+3+6\sqrt{y^2-7x+4}=0\\y\left(y-x+2\right)=3x+3\end{matrix}\right.\)
GHPT: \(\left\{{}\begin{matrix}x+\sqrt{x^2+2x+2}=\sqrt{y^2+1}-y-1\\x^3-\left(3x^2+2y-6\right)\sqrt{2x^2-y-2}=0\end{matrix}\right.\)
Từ pt thứ nhất: \(\Leftrightarrow x+1+\sqrt{\left(x+1\right)^2+1}=\left(-y\right)+\sqrt{\left(-y\right)^2+1}\)
Xét hàm \(f\left(t\right)=t+\sqrt{t^2+1}\Rightarrow f'\left(t\right)=1+\dfrac{t}{\sqrt{t^2+1}}=\dfrac{t+\sqrt{t^2+1}}{\sqrt{t^2+1}}\)
\(f'\left(t\right)>\dfrac{t+\sqrt{t^2}}{\sqrt{t^2+1}}=\dfrac{t+\left|t\right|}{\sqrt{t^2+1}}\ge0\Rightarrow f'\left(t\right)>0\) ; \(\forall t\)
\(\Rightarrow f\left(t\right)\) đồng biến trên R
\(\Rightarrow x+1=-y\Rightarrow y=-x-1\)
Thế xuống pt dưới:
\(x^3-\left(3x^2-2x-8\right)\sqrt{2x^2+x-1}=0\)
Bạn coi lại đề, pt vô tỉ này ko giải được
GHPT :
\(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\\3x\left(y-7\right)+10=\sqrt{10x-2}+2\sqrt{8y-3}\end{matrix}\right.\)
\(ĐK:x\ge\dfrac{1}{5};y\ge\dfrac{3}{8}\)
\(PT\left(1\right)\Leftrightarrow\dfrac{3x^2-3y^2}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}=3\left(x+y\right)\\ \Leftrightarrow3\left(x+y\right)\left(\dfrac{x-y}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+y=0\\\dfrac{x-y}{\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x-y=\sqrt{5x^2+2xy+2y^2}-\sqrt{2x^2+2xy+5y^2}\\ \Leftrightarrow\left(x-y\right)=\dfrac{3\left(x^2-y^2\right)}{\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}}\\ \Leftrightarrow\left(x-y\right)\left[\dfrac{3\left(x+y\right)}{\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}}-1\right]=0\)
\(\Leftrightarrow x=y\)
Với \(x+y=0\Leftrightarrow x=-y\), thay vào PT 2
\(\Leftrightarrow3\left(-y\right)\left(y-7\right)+10=\sqrt{10\left(-y\right)-2}+2\sqrt{8y-3}\\ \Leftrightarrow3y\left(7-y\right)+10=\sqrt{-10y-2}+2\sqrt{8y-3}\)
ĐK: \(\left\{{}\begin{matrix}-10y-2\ge0\\8y-3\ge0\end{matrix}\right.\Leftrightarrow y\in\varnothing\)
Với \(x-y=0\Leftrightarrow x=y\), thay vào PT 2
\(\Leftrightarrow3x^2-21x+10=\sqrt{10x-2}+2\sqrt{8x-3}\left(x\ge\dfrac{3}{8}\right)\\ \Leftrightarrow3x^2-24x+9=\sqrt{10x-2}-\left(x+1\right)+2\sqrt{8x-3}-2x\)
\(\Leftrightarrow3\left(x^2-8x+3\right)=\dfrac{-x^2+8x-3}{\sqrt{10x-2}+\left(x+1\right)}+\dfrac{2\left(-x^2+8x-3\right)}{\sqrt{8x-3}+x}\\ \Leftrightarrow\left(x^2-8x+3\right)\left(3+\dfrac{1}{\sqrt{10x-2}+x+1}+\dfrac{2}{\sqrt{8x-3}+x}\right)=0\)
Dễ thấy ngoặc lớn vô nghiệm với \(x\ge\dfrac{3}{8}>0\)
\(\Leftrightarrow x^2-8x+3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{13}\left(n\right)\\x=4-\sqrt{13}\left(n\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=4+\sqrt{13}\\y=4-\sqrt{13}\end{matrix}\right.\)
Vậy HPT có nghiệm \(\left(x;y\right)\in\left\{\left(4+\sqrt{13};4+\sqrt{13}\right);\left(4-\sqrt{13};4-\sqrt{13}\right)\right\}\)
Ghpt \(\left\{{}\begin{matrix}x^2+2y=xy+4\\x^2-x-3-x\sqrt{6-x}=\left(y-3\right)\sqrt{y-3}\end{matrix}\right.\)
\(ĐK:x\le6;y\ge3\\ \left\{{}\begin{matrix}x^2+2y=xy+4\left(1\right)\\x^2-x-3-x\sqrt{6-x}=\left(y-3\right)\sqrt{y-3}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2-4+2y-xy=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)-y\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-y+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=y-2\end{matrix}\right.\)
Từ đó thế vào PT(2)
Với \(x=y-2\Leftrightarrow x+2=y\)
\(\left(2\right)\Leftrightarrow x^2-x+3-x\sqrt{6-x}=\left(x-1\right)\sqrt{x-1}\left(1\le x\le6\right)\\ \Leftrightarrow2x^2-2x+6-2x\sqrt{6-x}=2\left(x-1\right)\sqrt{x-1}\\ \Leftrightarrow\left(x-\sqrt{6-x}\right)^2+x\left(x-1\right)=2\left(x-1\right)\sqrt{x-1}\\ \Leftrightarrow\left(x-\sqrt{6-x}\right)^2+\left(x-1\right)\left(x-2\sqrt{x-1}\right)=0\\ \Leftrightarrow\left(\dfrac{x^2-6+x}{x+\sqrt{6-x}}\right)^2+\dfrac{\left(x-1\right)\left(x^2-4x+4\right)}{x^2+2\sqrt{x-1}}=0\\ \Leftrightarrow\left[\dfrac{\left(x-2\right)\left(x+3\right)}{x+\sqrt{6-x}}\right]^2+\dfrac{\left(x-1\right)\left(x-2\right)^2}{x^2+2\sqrt{x-1}}=0\\ \Leftrightarrow\left(x-2\right)^2\left[\left(\dfrac{x+3}{x+\sqrt{6-x}}\right)^2+\dfrac{x-1}{x^2+2\sqrt{x-1}}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\left(\dfrac{x+3}{x+\sqrt{6-x}}\right)^2+\dfrac{x-1}{x^2+2\sqrt{x-1}}=0\left(1\right)\end{matrix}\right.\)
Dễ thấy \(\left(1\right)>0\) với \(x\ge1\)
Do đó \(x=2\Leftrightarrow y=4\)
Vậy HPT có nghiệm \(\left(x;y\right)=\left(2;4\right)\)
ai giúp t với
1:\(\left\{\begin{matrix}x\sqrt{12-y}+\sqrt{y\left(12-x^2\right)}=12\\x^3-8x-1=2\sqrt{y-2}\end{matrix}\right.\)
2:\(\left\{\begin{matrix}\left(1-y\right)\sqrt{x-y}+x=2+\left(x-y-1\right)\sqrt{y}\\2y^2-3x+6y+1=2\sqrt{x-2y}-\sqrt{4x-5y-3}\end{matrix}\right.\)
3:\(\left\{\begin{matrix}y\left(x^2+2x+2\right)=x\left(y^2+6\right)\\\left(y-1\right)\left(x^2+2x+7\right)=\left(x+1\right)\left(y^2+1\right)\end{matrix}\right.\)
4:\(\left\{\begin{matrix}x-2\sqrt{y+1}=3\\x^3-4x^2\sqrt{y+1}-9x-8y=-52-4xy\end{matrix}\right.\)
5:\(\left\{\begin{matrix}\frac{y-2x+\sqrt{y}-x}{\sqrt{xy}}+1=0\\\sqrt{1-xy}+x^2-y^2=0\end{matrix}\right.\)
Giải hệ phương trình
\(\left\{{}\begin{matrix}4\left(2x-y+3\right)-3\left(x-2y+3\right)=48\\3\left(3x-4y+3\right)+4\left(4x-2y-9\right)=48\end{matrix}\right.\)
\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}-2\left(2x+1\right)+1,5=3\left(y-2\right)-6x\\11,5-4\left(3-x\right)=2y-\left(5-x\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{8x-5y-3}{7}+\dfrac{11y-4x-7}{5}=12\\\dfrac{9x+4y-13}{5}-\dfrac{3\left(x-2\right)}{4}=15\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2\sqrt{3}x-\sqrt{5}y=2\sqrt{6}-\sqrt{15}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)
=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75
=>x=7; y=5
b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)
=>4x+9y=8 và -8x+3y=5
=>x=-1/4; y=1
c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)
=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5
=>2x-3y=-5,5 và 3x-2y=-4,5
=>x=-1/2; y=3/2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
=>\(x=\sqrt{2};y=\sqrt{3}\)
1.Giải hệ phương trình:
a.\(\left\{{}\begin{matrix}2\sqrt{2}x+y=2\sqrt{2}\\7x-3y=7\end{matrix}\right.\)
b.\(\left\{{}\begin{matrix}7x+y=-\frac{1}{7}\\-\frac{4}{3}x-2y=1\frac{1}{3}\end{matrix}\right.\)
c.\(\left\{{}\begin{matrix}2\sqrt{5}x+3y=\sqrt{2}\\\sqrt{5}x-y=3\sqrt{2}\end{matrix}\right.\)
d.\(\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y}=-5\\\frac{3}{x}-\frac{4}{y}=1\end{matrix}\right.\)
e.\(\left\{{}\begin{matrix}-\frac{5}{3x+1}+\frac{7}{2x+1}=\frac{5}{7}\\\frac{1}{3x+1}-\frac{1}{2y-3}=\frac{2}{7}\\\end{matrix}\right.\)
g.\(\left\{{}\begin{matrix}2x^2+5y^2=129\\-3x^2+y^2=13\end{matrix}\right.\)
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, \(\left\{{}\begin{matrix}\left(1-y\right)\sqrt{x-y}+x=2+\left(x-y-1\right)\sqrt{y}\\2y^2-3x+6y+1=2\sqrt{x-2y}-\sqrt{4x-5y-3}\end{matrix}\right.\)
GHPT sau: \(\left\{{}\begin{matrix}\dfrac{25}{9}+\sqrt{9x^2-4}=\dfrac{1}{9}\left(\dfrac{2}{x}+\dfrac{18x}{y^2-2y+2}+25y\right)\\7x^3+y^3+3xy\left(x-y\right)-12x^2+6x=1\end{matrix}\right.\)