Giải hệ phương trình :
\(\left\{{}\begin{matrix}2\sqrt{x+y}=y^2+y-x\\x\left(y^2+y\right)=\left(y^4-y^2\right)^2-2\end{matrix}\right.\)
Giải các hệ phương trình sau
a,\(\left\{{}\begin{matrix}\sqrt{3}x-y=\sqrt{2}\\x-\sqrt{2}y=\sqrt{3}\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}\dfrac{x+2}{y-1}=\dfrac{x-4}{y+2}\\\dfrac{2x+3}{y-1}=\dfrac{4x+1}{2y+1}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{x+2}{y-1}=\dfrac{x-4}{y+2}\\\dfrac{2x+3}{y-1}=\dfrac{4x+1}{2y+1}\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\left(x+2\right)\left(y+2\right)=\left(y-1\right)\left(x-\text{4}\right)\\\left(2x+3\right)\left(2y+1\right)=\left(y-1\right)\left(4x+1\right)\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}xy+2x+2y+4=xy-4y-x+4\\4xy+2x+6y+3=4xy-4x+y-1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(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) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))
Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=6\\\sqrt{x}-\sqrt{y}=4,5\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}=1\\\sqrt{y}+\sqrt{x+1}=1\end{matrix}\right.\)
1) Ta có: \(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}-3\sqrt{y}=15\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11\sqrt{x}=33\\3\sqrt{x}-\sqrt{y}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)
2) Ta có: \(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2\sqrt{x+3}+4\sqrt{y+1}=-4\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y+1}=0\\\sqrt{x+3}-2\sqrt{y+1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y+1}=0\\\sqrt{x+3}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+1=0\\x+3=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
4. Đk: \(x,y\ge0\)
\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}=1\\\sqrt{y}+\sqrt{x+1}=1\end{matrix}\right.\left(1\right)\)
Ta có: \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}\ge0+1=1\\\sqrt{y}+\sqrt{x+1}\ge0+1=1\end{matrix}\right.\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=0,\sqrt{x+1}=1\\\sqrt{y}=0,\sqrt{y+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)<tmđk>
Vậy hệ pt có nghiệm \(\left(x,y\right)=\left(0;0\right)\)
Giải hệ phương trình bằng phương pháp lượng giác hóa:
\(\left\{{}\begin{matrix}\left(2-x\right).\left(2+y\right)=8\\x\sqrt{4-y^2}+y\sqrt{4-x^2}=4\end{matrix}\right.\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}\sqrt{xy+\dfrac{x-y}{x^2+y^2+1}}+\sqrt{x}=y+\sqrt{y}\\\left|x-1\right|+\left|y-2\right|=1+x^2-y^2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{xy+\dfrac{x-y}{x^2+y^2+1}}+\sqrt{x}=y+\sqrt{y}\left(1\right)\\\left|x-1\right|+\left|y-2\right|=1+x^2-y^2\left(2\right)\end{matrix}\right.\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\y\ge0\\xy+\dfrac{x-y}{x^2+y^2+1}\ge0\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{xy+\dfrac{x-y}{x^2+y^2+1}}-y=\sqrt{y}-\sqrt{x}\)
\(\Leftrightarrow\dfrac{y\left(x-y\right)+\dfrac{x-y}{x^2+y^2+1}}{\sqrt{xy+\dfrac{x-y}{x^2+y^2+1}}+y}=\dfrac{x-y}{-xy}\Leftrightarrow\left(x-y\right)\left[\dfrac{y+\dfrac{1}{x^2+y^2+1}}{\sqrt{xy+\dfrac{x-y}{x^2+y^2+1}}+y}+xy\right]=0\Leftrightarrow x=y\).
Thay x = y vào (2) ta có \(\left|y-1\right|+\left|y-2\right|=1\). (*)
Ta có \(\left|y-1\right|+\left|y-2\right|=\left|y-1\right|+\left|2-y\right|\ge y-1+2-y=1\).
Mà đẳng thức xảy ra ở (1) nên ta phải có \(1\le y\le2\). (TMĐK)
Vậy pt đã cho có vô số nghiệm \(x=y=k\) với \(1\le k\le2\)
Giải hệ phương trình sau : \(\left\{{}\begin{matrix}\sqrt{x-2}+\left(y-1\right)^2=4\\2\sqrt{x-2}-\left(2y-y^2\right)=2\end{matrix}\right.\)
Lời giải:
Lấy PT (2) trừ đi 2* PT(1) ta có:
$-(2y-y^2)-2(y-1)^2=-6$
$\Leftrightarrow y^2-2y-4=0$
Vì $y^2-2y=4$ nên $2\sqrt{x-2}=2+(2y-y^2)=2-(y^2-2y)=2-4=-2<0$ (vô lý)
Do đó hpt vô nghiệm.
Giải hệ phương trình \(\left\{{}\begin{matrix}\left(x^2+y^2\right)\left(x+y\right)=x+y-2xy\\x^2+11x+6=2\sqrt{9y-5}+\sqrt{x+y}\end{matrix}\right.\)
ĐKXĐ: \(\left\{{}\begin{matrix}9y-5\ge0\\x+y\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\ge\dfrac{5}{9}\\x+y\ge0\end{matrix}\right.\).
Phương trình (1) tương đương với:
\(\left(x^2+y^2\right)\left(x+y\right)-\left(x+y\right)+2xy=0\)
\(\Leftrightarrow\left(x^2+y^2\right)\left(x+y\right)-\left(x^2+y^2\right)+x^2+y^2-\left(x+y\right)+2xy=0\)
\(\Leftrightarrow\left(x^2+y^2\right)\left(x+y-1\right)+\left(x+y\right)^2-\left(x+y\right)=0\)
\(\Leftrightarrow\left(x^2+y^2\right)\left(x+y-1\right)+\left(x+y\right)\left(x+y-1\right)=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x^2+y^2+x+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y-1=0\\x^2+y^2+x+y=0\end{matrix}\right.\)
- Với \(x^2+y^2+x+y=0\) có \(x+y=0\) (theo điều kiện)
suy ra \(x=y=0\) (không thỏa mãn).
- Với \(x+y-1=0\Leftrightarrow y=1-x\) thế vào phương trình (2) ta được:
\(x^2+11x+6=2\sqrt{9\left(1-x\right)-5}+\sqrt{1}\)
\(\Leftrightarrow x^2+11x+5-2\sqrt{14-9x}=0\)
\(\Rightarrow\left(x^2+11x+5\right)^2=4\left(14-9x\right)\)
\(\Leftrightarrow x^4+22x^3+131x^2+146x-31=0\)
Bạn giải phương trình trên, thử lại ta được nghiệm của bài toán.
Đáp án ra số khá xấu nên thầy không ghi ra đây.
Em có thể tham khảo cách làm nhé.
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.
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2-3\left(2x-y\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(2x-y-3\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-y=0\\x+2y=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y-3=0\\x+2y=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
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:
\(6y+y=14\Rightarrow y=2\)
\(\Rightarrow x=4\)
giải hệ phương trình
\(\left\{{}\begin{matrix}\sqrt{x-2}+\sqrt{y-3}=3\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}+\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=4\end{matrix}\right.\)
a.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\5\sqrt{x-2}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\\sqrt{x-2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-3}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)
b.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}\sqrt{x+2}.\left(x-y+3\right)=\sqrt{y}\\x^2+\left(x+3\right)\left(2x-y+5\right)=x+16\end{matrix}\right.\)
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\) thì pt đầu trở thành:
\(a\left(a^2-b^2+1\right)=b\)
\(\Leftrightarrow a\left(a-b\right)\left(a+b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+1\right)=0\)
\(\Leftrightarrow a=b\Rightarrow\sqrt{x+2}=\sqrt{y}\Rightarrow y=x+2\)
Thay xuống pt dưới:
\(x^2+\left(x+3\right)\left(x+3\right)=x+16\)
\(\Leftrightarrow2x^2+5x-7=0\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=3\\x=-\dfrac{2}{7}\left(loại\right)\end{matrix}\right.\)
giải hệ pt
c)\(\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\2\sqrt{x-1}-\sqrt{y}=4\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\left(x-1\right)+\left(y+2\right)=2\\4\left(x-1\right)+3\left(y+2\right)=7\end{matrix}\right.\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\4x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+4y=4\\4x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)