giải hệ phương trình sau
\(\left\{{}\begin{matrix}2-\sqrt{x^2y^4+2xy^2-y^4+1}=2\left(3-\sqrt{2}-x\right)y^2\\\sqrt{x-y^2}+x=3\end{matrix}\right.\)
giải hệ phương trình
a) \(\left\{{}\begin{matrix}\sqrt{2x^2+2y^2}+\sqrt{\frac{4}{3}\left(x^2+xy+y^2\right)}=2\left(x+y\right)\\\sqrt{3x+1}+\sqrt{5x+4}=3xy-y+3\end{matrix}\right.\)
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)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
caau a) binh phuong len ra no x=y tuong tu
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
$x^2+2+2\sqrt{y(x^2+2)}-3y=0$
$\Leftrightarrow (\sqrt{x^2+2}-\sqrt{y})(\sqrt{x^2+2}+3\sqrt{y})=0$
$\Leftrightarrow x^2+2=y$
Giải hệ phương trình:
a,\(\left\{{}\begin{matrix}\sqrt{x+y}\left(\sqrt{y}+1\right)=\sqrt{x^2+y^2}+2\\x\sqrt{y-1}+y\sqrt{x-1}=\dfrac{x^2+4y-4}{2}\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}x^3+2y^2=x^2y+2xy\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)
a.Hệ thứ nhất kì quặc thật:
\(\Leftrightarrow\sqrt{y^2+xy}+\sqrt{x+y}=\sqrt{x^2+y^2}+2\)
\(\Leftrightarrow\sqrt{x^2+y^2}-\sqrt{y^2+xy}=\sqrt{x+y}-2\)
\(\Leftrightarrow\dfrac{x\left(x-y\right)}{\sqrt{x^2+y^2}+\sqrt{y^2+xy}}=\dfrac{x+y-4}{\sqrt{x+y}+2}\)
\(\Rightarrow\left(x-y\right)\left(x+y-4\right)=\left(\dfrac{\sqrt{x^2+y^2}+\sqrt{y^2+xy}}{x\sqrt{x+y}+2x}\right)\left(x+y-4\right)^2\ge0\) (1)
\(2.\dfrac{x}{2}\sqrt{y-1}+2.\dfrac{y}{2}\sqrt{x-1}\le\dfrac{x^2}{4}+y-1+\dfrac{y^2}{4}+x-1\)
\(\Rightarrow\dfrac{x^2+4y-4}{2}\le\dfrac{x^2+y^2+4x+4y-8}{4}\)
\(\Leftrightarrow x^2-y^2+4y-4x\le0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y-4\right)\le0\) (2)
(1);(2) \(\Rightarrow\left(x-y\right)\left(x+y-4\right)=0\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=2\)
b.
\(x^3-x^2y+2y^2-2xy=0\)
\(\Leftrightarrow x^2\left(x-y\right)-2y\left(x-y\right)=0\)
\(\Leftrightarrow\left(x^2-2y\right)\left(x-y\right)=0\)
\(\Leftrightarrow y=x\) (loại \(x^2-2y=0\) do ĐKXĐ \(x^2-2y-1\ge0\))
Thế vào pt dưới
\(2\sqrt{x^2-2x-1}+\sqrt[3]{x^3-14}=x-2\)
\(\Leftrightarrow2\sqrt{x^2-2x-1}+\dfrac{x^3-14-\left(x-2\right)^3}{\sqrt[3]{\left(x^3-14\right)^2}+\left(x-2\right)\sqrt[3]{x^3-14}+\left(x-2\right)^2}=0\)
\(\Leftrightarrow\sqrt[]{x^2-2x-1}\left(2+\dfrac{6\sqrt[]{x^2-2x-1}}{\sqrt[3]{\left(x^3-14\right)^2}+\left(x-2\right)\sqrt[3]{x^3-14}+\left(x-2\right)^2}\right)=0\)
\(\Leftrightarrow\sqrt{x^2-2x-1}=0\)
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 bằng phương pháp phân tích nhân tử
a) \(\left\{{}\begin{matrix}x^2+2y=xy+4\\x^2-x-3-x\sqrt{6-x}=\left(y-3\right)\sqrt{y-3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2-2xy+x+y=0\\x^4-4x^2y+3x^2+y^2=0\end{matrix}\right.\)
Giải hệ phương trình sau bằng phương pháp thế
a)
\(\left\{{}\begin{matrix}\sqrt{5}+2)x+y=3-\sqrt{5}\\-x+2y=6-2\sqrt{5}\end{matrix}\right.\)
b)
\(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-5y\right)-12\end{matrix}\right.\)
Giải các hệ phương trình sau bằng phương pháp cộng đại số:
a) \(\left\{{}\begin{matrix}\sqrt{2}x-y=3\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{x}{2}-2y=\dfrac{3}{4}\\2x+\dfrac{y}{3}=-\dfrac{1}{3}\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\dfrac{2x-3y}{4}-\dfrac{x+y-1}{5}=2x-y-1\\\dfrac{x+y-1}{3}+\dfrac{4x-y-2}{4}=\dfrac{2x-y-3}{6}\end{matrix}\right.\)
a) Ta có: \(\left\{{}\begin{matrix}\sqrt{2}x-y=3\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}x-y=3\\\sqrt{2}x+2y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-3y=1\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}\cdot\dfrac{-1}{3}=\dfrac{4\sqrt{2}}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{2}}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}\dfrac{x}{2}-2y=\dfrac{3}{4}\\2x+\dfrac{y}{3}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-8y=3\\2x+\dfrac{1}{3}y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{25}{3}y=\dfrac{10}{3}\\2x-8y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{2}{5}\\2x=3+8y=3+8\cdot\dfrac{-2}{5}=-\dfrac{1}{5}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{2x-3y}{4}-\dfrac{x+y-1}{5}=2x-y-1\\\dfrac{x+y-1}{3}+\dfrac{4x-y-2}{4}=\dfrac{2x-y-3}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5\left(2x-3y\right)}{20}-\dfrac{4\left(x+y-1\right)}{20}=\dfrac{20\left(2x-y-1\right)}{20}\\\dfrac{4\left(x+y-1\right)}{12}+\dfrac{3\left(4x-y-2\right)}{12}=\dfrac{2\left(2x-y-3\right)}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x-15y-4x-4y+4=40x-20y-20\\4x+4y-4+12x-3y-6=4x-2y-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-19y+4-40x+20y+20=0\\16x+y-10-4x+2y+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-34x+y=-24\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-102x+3y=-72\\12x+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-114x=-76\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\12\cdot\dfrac{2}{3}+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\3y=4-8=-4\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
giải hệ phương trình:
1, \(\left\{{}\begin{matrix}\sqrt{3+2x^2y-x^4y^2}+x^2\left(1-2x^2\right)=y^4\\1+\sqrt{1+\left(x-y\right)^2}=-x^2\left(x^4+1-2x^2-2xy^2\right)\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\sqrt{x-1}+\sqrt{x}\left(3\sqrt{x}-y\right)+x\sqrt{x}=3y+\sqrt{y-1}\\3xy^2+4=4x^2+2y+x\end{matrix}\right.\)
giải giúp mik bt này vs mn!
1)\(\left\{{}\begin{matrix}2x^2+y^2+x=3\left(xy+1\right)+2y\\\dfrac{2}{3+\sqrt{2x-y}}+\dfrac{2}{3+\sqrt{4-5x}}=\dfrac{9}{2x-y+9}\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}\left(x+3y+1\right)\sqrt{2xy+2y}=y\left(3x+4y+3\right)\\\left(\sqrt{x+3}-\sqrt{2y-2}\right)\left(x-3+\sqrt{x^2+x+2y-4}\right)=4\end{matrix}\right.\)
3)\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}\sqrt{2x-3}=\left(y^2+2011\right)\left(5-y\right)+\sqrt{y}\\y\left(y-x+2\right)=3x+3\end{matrix}\right.\)
5)\(\left\{{}\begin{matrix}x^3+2x^2=x^2y+2xy\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14=x-2}\end{matrix}\right.\)
5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)
Thay từng TH rồi làm nha bạn
3,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\)
thay nhá
Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)
PT(1) \(\Leftrightarrow\left(x-y-1\right)\left(2x-y+3\right)=0\)
+) Với y = x - 1 thay vào pt (2):
\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))
Anh quy đồng lên đê, chắc cần vài con trâu đó:))
+) Với y = 2x + 3...
Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}\sqrt{x}+2\sqrt{-1}=5\\4\sqrt{x}-\sqrt{y-1}=2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{3x-1}-\sqrt{2y+1}=1\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\sqrt{x-2}+\sqrt{y-3}=3\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}2\sqrt{x+1}-3\sqrt{-2}=5\\4\sqrt{x+1}+\sqrt{y-2}=17\end{matrix}\right.\)
2) Ta có: \(\left\{{}\begin{matrix}\sqrt{3x-1}-\sqrt{2y+1}=1\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{3x-1}-2\sqrt{2y+1}=2\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-5\sqrt{2y+1}=-10\\\sqrt{3x-1}-\sqrt{2y+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2y+1}=2\\\sqrt{3x-1}-2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y+1=4\\3x-1=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y=3\\3x=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{2}\\x=\dfrac{10}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{3}{2}\end{matrix}\right.\)
3) Ta có: \(\left\{{}\begin{matrix}\sqrt{x-2}+\sqrt{y-3}=3\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-2}+2\sqrt{y-3}=6\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y-3}=10\\\sqrt{x-2}+\sqrt{y-3}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y-3}=2\\\sqrt{x-2}+2=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-3=4\\x-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=7\\x=3\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)