Giải hpt\(\left\{{}\begin{matrix}x^3+y^3=1+y-x+xy\\7xy+y-x=7\end{matrix}\right.\)
giải hpt:
\(\left\{{}\begin{matrix}x^3+y^3=1+y-x+xy\\7xy+x-y=7\end{matrix}\right.\)
giải hpt: a,\(\left\{{}\begin{matrix}x+y-\sqrt{xy}=3\\\sqrt{x+1}+\sqrt{y+1}=4\end{matrix}\right.\) b,\(\left\{{}\begin{matrix}x+y=5+\sqrt{\left(x-1\right)\left(y-1\right)}\\\sqrt{x-1}+\sqrt{y-1}=3\end{matrix}\right.\)
a.
ĐKXĐ: \(x;y\ge-1;xy\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y-3=\sqrt{xy}\\x+y+2\sqrt{xy+x+y+1}=14\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\ge0\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}u-3=\sqrt{v}\\u+2\sqrt{u+v+1}=14\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=u^2-6u+9\left(u\ge3\right)\\4\left(u+v+1\right)=\left(14-u\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\4u+4\left(u^2-6u+9\right)+4=\left(14-u\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\3u^2+8u-156=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\\left[{}\begin{matrix}u=6\\u=-\dfrac{26}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=6\\v=9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=6\\xy=9\end{matrix}\right.\) \(\Rightarrow x=y=3\)
b.
ĐKXĐ: \(x;y\ge1\)
Xét \(\sqrt{x-1}+\sqrt{y-1}=3\)
\(\Leftrightarrow x+y-2+2\sqrt{\left(x-1\right)\left(y-1\right)}=9\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(y-1\right)}=\dfrac{11-x-y}{2}\)
Thế vào pt đầu:
\(x+y=5+\dfrac{11-x-y}{2}\)
\(\Leftrightarrow x+y=7\Rightarrow y=7-x\)
Thế xuống pt dưới:
\(\sqrt{x-1}+\sqrt{6-x}=3\)
\(\Leftrightarrow5+2\sqrt{\left(x-1\right)\left(6-x\right)}=9\)
\(\Leftrightarrow\left(x-1\right)\left(6-x\right)=4\)
\(\Leftrightarrow...\)
giải hệ
\(\left\{{}\begin{matrix}x^3+y^3=1+y-x+xy\\7xy+y-x=7\end{matrix}\right.\)
Cộng vế vs vế của 2 phương trình ta được :
\(x^3+y^3+6xy=8\Leftrightarrow\left(x+y-2\right)\left(\frac{3\left(x-y\right)^2}{4}\right)+\left(\frac{\left(x+y\right)^2}{4}\right)+2\left(x+y\right)+4=0\)
Tới đây ta xét 2 TH : +) \(x+y=2\) bạn chắc tự giải được
\(\frac{3\left(x-y\right)^2}{4}+\frac{\left(x+y\right)^2}{4}+2\left(x+y\right)+4=0\)
Ta thấy : \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\frac{\left(x+y\right)^2}{4}+4\ge2|x+y|\ge2\left(x+y\right)\end{matrix}\right.\)
Dấu "=" xảy ra khi :
\(\left\{{}\begin{matrix}x-y=0\\\left(x+y\right)^2=4^2\\x+y< 0\end{matrix}\right.\)
Hay x = y = −2x = y = −2 không thoả mãn hệ phương trình.
Giải hệ pt
1/\(\left\{{}\begin{matrix}4x\sqrt{y+1}+8x=\left(4x^2-4x-3\right)\sqrt{x+1}\\\dfrac{x}{x+1}+x^2=\left(y+2\right)\sqrt{\left(x+1\right)\left(y+1\right)}\end{matrix}\right.\)
2/\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)
3/\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)
4/\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)
m.n giúp e mấy bài này vs ạ!!
giải hpt: a,\(\left\{{}\begin{matrix}x^2+y^2+xy=7\\x^4+y^4+x^2y^2=21\end{matrix}\right.\) b,\(\left\{{}\begin{matrix}x+y+\dfrac{1}{x}+\dfrac{1}{y}=7\\x^2-y^2+\dfrac{1}{x^2}-\dfrac{1}{y^2}=21\end{matrix}\right.\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\\left(x^2+y^2+xy\right)\left(x^2+y^2-xy\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\x^2+y^2-xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\xy=2\end{matrix}\right.\)
\(\Rightarrow x^2+\left(\dfrac{2}{x}\right)^2=5\)
\(\Leftrightarrow x^4-5x^2=4=0\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\\left(x+\dfrac{1}{x}\right)^2-\left(y+\dfrac{1}{y}\right)^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)\left(x+\dfrac{1}{x}-y-\dfrac{1}{y}\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\x+\dfrac{1}{x}-y-\dfrac{1}{y}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}=5\\y+\dfrac{1}{y}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+1=0\\y^2-2y+1=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
giải hpt:
1,\(\left\{{}\begin{matrix}x^2y^2-2x+y^2=0\\2x^2-4x+3+y^3=0\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\left(x^2-xy\right)\left(xy-y^2\right)=25\\\sqrt{x^2-xy}+\sqrt{xy-y^2}=3\left(x-y\right)\end{matrix}\right.\)
giải hpt: \(\left\{{}\begin{matrix}x+y+xy=5\\x^3+y^3=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=5\\\left(x+y\right)^3-3xy\left(x+y\right)=9\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}u+v=5\\u^3-3uv=9\end{matrix}\right.\)
\(\Rightarrow u^3-3u\left(5-u\right)=9\)
\(\Leftrightarrow u^3+3u^2-15u-9=0\)
\(\Leftrightarrow\left(u-3\right)\left(u^2+6u+3\right)=0\)
\(\Leftrightarrow...\)
giải hpt:
\(\left\{{}\begin{matrix}2x+y=1\\x^2+y^2-xy=3\end{matrix}\right.\)
2x + y = 1 <=> y = 1 - 2x
Thế vào pt còn lại thì:
x^2 + (1 - 2x)^2 - x(1 - 2x) = 3
<=> x^2 + 4x^2 - 4x + 1 - x + 2x^2 - 3 = 0
<=> 7x^2 - 5x - 2 = 0
<=> (x - 1)(7x + 2) = 0
<=> x = 1 hoặc x = -2/7
Với x = 1 <=> y = 1 - 2.1 = -1
Với x = -2/7 <=> y = 1 - 2.(-2/7) = 11/7
giải hệ:
\(\left\{{}\begin{matrix}x+2y=7\\x^2+y^2-2xy=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-y=2\\x^2+y^2+164\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-y+xy=-13\\x^2+y^2-x-y=32\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-y=3\\x^3-y^3=7\end{matrix}\right.\)
Câu 1:
Từ PT(1) suy ra $x=7-2y$. Thay vào PT(2):
$(7-2y)^2+y^2-2(7-2y)y=1$
$\Leftrightarrow 4y^2-28y+49+y^2-14y+4y^2=1$
$\Leftrightarrow 9y^2-42y+48=0$
$\Leftrightarrow (y-2)(9y-24)=0$
$\Leftrightarrow y=2$ hoặc $y=\frac{8}{3}$
Nếu $y=2$ thì $x=7-2y=3$
Nếu $y=\frac{8}{3}$ thì $x=7-2y=\frac{5}{3}$
Câu 3: Bạn xem lại PT(2) là -x+y đúng không?
Câu 4:
$x^3-y^3=7$
$\Leftrightarrow (x-y)^3-3xy(x-y)=7$
$\Leftrightarrow 3^3-9xy=7$
$\Leftrightarrow xy=\frac{20}{9}$
Áp dụng định lý Viet đảo, với $x+(-y)=3$ và $x(-y)=\frac{-20}{9}$ thì $x,-y$ là nghiệm của pt:
$X^2-3X-\frac{20}{9}=0$
$\Rightarrow (x,-y)=(\frac{\sqrt{161}+9}{6}, \frac{-\sqrt{161}+9}{6})$ và hoán vị
$\Rightarrow (x,y)=(\frac{\sqrt{161}+9}{6}, \frac{\sqrt{161}-9}{6})$ và hoán vị.