1)\(\begin{cases}\sqrt{4x^2+\left(4x-9\right)\left(x-3y\right)}+\sqrt{3xy}=9y\\4\sqrt{\left(x+2\right)\left(3y+2x\right)}=3x+9\end{cases}\) 4)\(\begin{cases}\left(x^2+y\right)\sqrt{x-y+6}=2x^2-x+3y-2\\\sqrt{10x-xy-12}+1=\frac{y-x}{\sqrt{y-4}+\sqrt{6-x}}\end{cases}\)
2)\(\begin{cases}x^2+\left(y-6\right)^2=y+13x+27\\\sqrt{9x^2+\left(2x-3\right)\left(x-y\right)}+4\sqrt{xy}=7y\end{cases}\) 5)\(\begin{cases}\sqrt{4xy+\left(3\sqrt{xy}-7\right)\left(x-y\right)}+2\sqrt{xy}=4y\\\left(2x+1\right)\left[12y-1+9\sqrt{xy}-x^2-x\right]=27\left(x+1\right)\end{cases}\)
3)\(\begin{cases}\sqrt{\left(x+2\right)\left(y+1\right)+\left(x-y+1\right)\sqrt{y^2+1}}+\sqrt{x+2}=y+\sqrt{y+1}+1\\\sqrt{3x+1}-\sqrt{y+1}=2x^2+4x-y-1\end{cases}\)
1)\(\begin{cases}x^2-y\left(x+y\right)+1=0\\\left(x^2+1\right)\left(x+y-2\right)+y=0\end{cases}\)
2)\(\begin{cases}x^2-4x+y^4+4y^2=2\\xy^2+2y^2+6x=23\end{cases}\)
3)\(\begin{cases}2x+\frac{1}{x+y}=3\\4x^2+4y^2+4xy+\frac{3}{\left(x+y\right)^2}=7\end{cases}\)
4)\(\begin{cases}y^6+x^9+3y^4+3y^2=8\\4y^2-3x^3y^2+x^3=2\end{cases}\)
5)\(\begin{cases}\sqrt{x+y}-2\sqrt{x-y}=1\\x+\sqrt{x^2+y^2}=8\end{cases}\)
6) \(\begin{cases}x+y-2=\frac{y}{x^2+1}\\x^2+y^2+xy=y-1\end{cases}\)
7) \(\begin{cases}4x-1=\sqrt{\left(2x+y\right).\left(2y+1\right)}\\\sqrt{x+2y+1}-\sqrt{x+y-1}=\sqrt{x-1}\end{cases}\)
8) \(\begin{cases}\left(x+y\right).\left(x+4y^2+y\right)+3y^4=0\\\sqrt{x+2y^2+1}-y^2+y+1=0\end{cases}\)
ôi trờiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
CÂU 1 :\(\hept{\begin{cases}x^5+xy^4=x^{10}+y^6\\\sqrt{4x+5}+\sqrt{y^2+8}=6\end{cases}}\)
CÂU 2:\(\hept{\begin{cases}x^2\left(y^2+1\right)+2y\left(x^2+x+1\right)=3\\\left(x^2+x\right)\left(y^2+y\right)=1\end{cases}}\)
CÂU 3: \(\hept{\begin{cases}x^3-3x^2y+4y^3=\left(x-2y\right)^2\\\sqrt{x-2y}+\sqrt{3x+2y}=4x-4\end{cases}}\)
Giải hệ phương trình:
1) \(\hept{\begin{cases}\sqrt[3]{x-y}=\sqrt{x-y}\\x+y=\sqrt{x+y+2}\end{cases}}\)
2) \(\hept{\begin{cases}x-\frac{1}{x}=y-\frac{1}{y}\\2y=x^3+1\end{cases}}\)
3) \(\hept{\begin{cases}\left(x-y\right)\left(x^2+y^2\right)=13\\\left(x+y\right)\left(x^2-y^2\right)=25\end{cases}\left(x;y\in R\right)}\)
4) \(\hept{\begin{cases}3y=\frac{y^2+2}{x^2}\\3x=\frac{x^2+2}{y^2}\end{cases}}\)
5) \(\hept{\begin{cases}x+y-\sqrt{xy}=3\\\sqrt{x+1}+\sqrt{y+1}=4\end{cases}\left(x;y\in R\right)}\)
6) \(\hept{\begin{cases}x^3-8x=y^3+2y\\x^2-3=3\left(y^2+1\right)\end{cases}\left(x;y\in R\right)}\)
7) \(\hept{\begin{cases}\left(x^2+1\right)+y\left(y+x\right)=4y\\\left(x^2+1\right)\left(y+x-2\right)=y\end{cases}\left(x;y\in R\right)}\)
8) \(\hept{\begin{cases}y+xy^2=6x^2\\1+x^2y^2=5x^2\end{cases}}\)
\(\begin{cases}2\sqrt{x^2+3x+2}-\sqrt{x+1}=2y\sqrt{y^2+1}+9-y-6y^2\\\sqrt{x^2+3x+2}+3\sqrt{x+1}=y\sqrt{y^2+1}-6+3y+4y^2\end{cases}\)
\(\begin{cases}x^2-y-1=2\sqrt{2x-1}\\y^3-8x^3+3y^2+4y-2x+2=0\end{cases}\)
\(\begin{cases}\left(x+\sqrt{x^2+4}\right)\left(y+\sqrt{y^2+1}\right)=2\\27x^6=x^3+4x+2\end{cases}\)
\(\begin{cases}x-\sqrt{3y-2}=\sqrt{9y^2-6y}-x\sqrt{x^2+2}\\x+y+\sqrt{y+3}=4\end{cases}\)
2)ĐK:x\(\ge\frac{1}{2}\)
pt(2)\(\Leftrightarrow\left(y+1\right)^3\)+(y+1)=\(\left(2x\right)^3\)+2x
Xét hàm số: f(t)=\(t^3\)+t
f'(t)=3\(t^2\)+1>0,\(\forall\)t
\(\Rightarrow\)hàm số liên tục và đồng biến trên R
\(\Rightarrow\)y+1=2x
Thay y=2x-1 vào pt(1) ta đc:
\(x^2\)-2x=2\(\sqrt{2x-1}\)
\(\Leftrightarrow\left(x^2-4x+2\right)\left(1+\frac{4}{2x-2+2\sqrt{2x-1}}\right)=0\)
\(\Leftrightarrow x^2\)-4x+2=0(do(...)>0)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2+\sqrt{2}\Rightarrow y=3+2\sqrt{2}\\x=2-\sqrt{2}\Rightarrow y=3-2\sqrt{2}\end{array}\right.\)
4)ĐK:\(y\ge\frac{2}{3}\)
pt(1)\(\Leftrightarrow x-\sqrt{3y-2}=\sqrt{3y\left(3y-2\right)}-x\sqrt{x^2+2}\)
\(\Leftrightarrow x\left(\sqrt{x^2+2}+1\right)=\sqrt{3y-2}\left(\sqrt{3y}+1\right)\)
Xét hàm số:\(f\left(t\right)=t\left(\sqrt{t^2+2}+1\right)\)
\(\Rightarrow\)hàm số liên tục và đồng biến trên R
\(\Rightarrow x=\sqrt{3y-2}\)
Thay vào pt(2) ta đc:\(\sqrt{3y-2}+y+\sqrt{y+3}=4\)
\(\Leftrightarrow\sqrt{3y-2}-1+\sqrt{y+3}-2+y-1=0\)
\(\Leftrightarrow\left(y-1\right)\left(\frac{3}{\sqrt{3y-2}+1}+\frac{1}{\sqrt{y+3}+2}+1\right)=0\)
\(\Leftrightarrow y=1\Rightarrow x=1\)(do...)>0)
KL:...
1,\(\hept{\begin{cases}x^2-2y^2-xy=0\\\sqrt{2x}+\sqrt{y+1}=2\end{cases}}\)
2,\(\hept{\begin{cases}\left(x-y\right)\left(x+y+y^2\right)=x\left(y+1\right)\\\sqrt{x}+\sqrt{y+1}=2\end{cases}}\)
3,\(\hept{\begin{cases}2y^3-\left(x+4\right)y^2+8y+x^2-4x=0\\\sqrt{\frac{1-x}{2}}+\sqrt{x+2y+3}=\sqrt{5}\end{cases}}\)
1,\(x^2-2y^2-xy=0\)
<=> \(\left(x-2y\right)\left(x+y\right)=0\)
<=> \(\orbr{\begin{cases}x=2y\\x=-y\end{cases}}\)
Sau đó bạn thế vào PT dưới rồi tính
3. ĐKXĐ \(x\le1\); \(x+2y+3\ge0\)
.\(2y^3-\left(x+4\right)y^2+8y+x^2-4x=0\)
<=> \(\left(2y^3-xy^2\right)+\left(x^2-4y^2\right)-\left(4x-8y\right)=0\)
<=> \(\left(x-2y\right)\left(-y^2+x+2y-4\right)=0\)
Mà \(-y^2+2y-4=-\left(y-1\right)^2-3\le-3\); \(x\le1\)nên \(-y^2+x+2y-4< 0\)
=> \(x=2y\)
Thế vào Pt còn lại ta được
\(\sqrt{\frac{1-x}{2}}+\sqrt{2x+3}=\sqrt{5}\)ĐK \(-\frac{3}{2}\le x\le1\)
<=> \(\frac{1-x}{2}+2x+3+2\sqrt{\frac{\left(1-x\right)\left(2x+3\right)}{2}}=5\)
<=> \(\sqrt{2\left(1-x\right)\left(2x+3\right)}=-\frac{3}{2}x+\frac{3}{2}\)
<=> \(\sqrt{2\left(1-x\right)\left(2x+3\right)}=-\frac{3}{2}\left(x-1\right)\)
<=> \(\orbr{\begin{cases}x=1\\\sqrt{2\left(2x+3\right)}=\frac{3}{2}\sqrt{1-x}\end{cases}}\)=> \(\orbr{\begin{cases}x=1\\x=-\frac{3}{5}\end{cases}}\)(TMĐK )
Vậy \(\left(x;y\right)=\left(1;\frac{1}{2}\right),\left(-\frac{3}{5};-\frac{3}{10}\right)\)
2,ĐKXĐ \(x\ge0\); \(y\ge-1\)
\(\left(x-y\right)\left(x+y+y^2\right)=x\left(y+1\right)\)
<=> \(x^2-y^3+xy^2-y^2=xy+x\)
<=> \(\left(x^2+xy^2\right)-\left(xy+y^3\right)-\left(x+y^2\right)=0\)
<=> \(\left(x+y^2\right)\left(x-y-1\right)=0\)
<=> \(\orbr{\begin{cases}x+y^2=0\\x=y+1\end{cases}}\)
+ x+y^2=0
Mà \(x\ge0;y^2\ge0\)
=> \(x=y=0\)(loại vì không thỏa mãn PT 2)
+ \(x=y+1\)
Thế vào PT 2 ta có
\(2\sqrt{x}=2\)=> \(x=1\)=> \(y=0\)
Vậy x=1;y=0
1)\(\begin{cases}y^3\left(3x^2-4x-23\right)=8-8y\\y^2\left(x^3+10x+27\right)=8x+6y\end{cases}\)
2\(\begin{cases}2\sqrt{x^2+5x-y+2}-2=\sqrt{y^2+8x}+x\\2y-\sqrt{x+1}=x+5\end{cases}\)
2)ĐK:\(\begin{cases}x\ge-1\\...\\y^2+8x\ge0\end{cases}\)
pt(1)\(\Leftrightarrow2\left[\sqrt{x^2+5x-y+2}-\left(x+2\right)\right]+\left(x+2-\sqrt{y^2+8x}\right)=0\)
\(\Leftrightarrow\left(x-y-2\right)\left(\frac{2}{\sqrt{x^2+5x-y+2}+x+2}+\frac{x+y-2}{x+2+\sqrt{y^2+8x}}\right)=0\)
\(\Rightarrow\)y=x-2
Thay vào pt(2) ta được:x-9=\(\sqrt{x+1}\)
\(\Leftrightarrow\begin{cases}x\ge9\\x^2-19x+80=0\end{cases}\Leftrightarrow x=\frac{19+\sqrt{41}}{2}}\)
\(\Rightarrow\)(x;y)=(\(\frac{19+\sqrt{41}}{2};\frac{15+\sqrt{41}}{2}\))(t/m)
bạn tách từng câu ra mik suy nghĩ từng câu
giải hpt:1)\(\begin{cases}\text{x+y+xy(2x+y)=5xy }\\\text{x+y+xy(3x-y)=4xy}\end{cases}\)
2)\(\begin{cases}\left(2x+y+1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{cases}\)
3)\(\begin{cases}\sqrt{9x+\frac{y}{x}}+2.\sqrt{y+\frac{2x}{y}}=4\\\left(\frac{2x}{y^2}-1\right)\left(\frac{y}{x^2}-9\right)=18\end{cases}\)
1. \(\begin{cases}x+y+xy\left(2x+y\right)=5xy\\x+y+xy\left(3x-y\right)=4xy\end{cases}\) \(\Leftrightarrow\begin{cases}2y-x=1\\x+y+xy\left(2x+y\right)=5xy\end{cases}\) (trừ 2 vế cho nhau)
\(\Leftrightarrow\begin{cases}x=2y-1\\\left(2y-1\right)+y+\left(2y-1\right)y\left(4y-2+y\right)=5\left(2y-1\right)y\end{cases}\) \(\Leftrightarrow\begin{cases}x=2y-1\\10y^3-19y^2+10y-1=0\end{cases}\) \(\Leftrightarrow\begin{cases}x=1\\y=1\end{cases}\)
