https://i.imgur.com/PfmYiMQ.jpg
Bài 2: Hệ hai phương trình bậc nhất hai ẩn. Luyện tập
Các câu hỏi tương tự
\(\left\{{}\begin{matrix}\frac{\sqrt{5}x}{\sqrt{5}+1}=\frac{y}{\sqrt{5}-1}\\x+y=\sqrt{5}\end{matrix}\right.\)
Giải hệ phương trình
\(\left\{{}\begin{matrix}2\sqrt{\left(x+2\right)\left(3x-y\right)}=y+6\\\sqrt{x+2}+\sqrt{y+2}=\sqrt{xy+5x}-12\end{matrix}\right.\)
Giải hpt:
\(\left\{{}\begin{matrix}2\sqrt{\left(x+2\right)\left(3x-y\right)}=y+6\\\sqrt{x+2}+\sqrt{y-2}=\sqrt{xy+5x-12}\end{matrix}\right.\)\(\)
giải hpt:
a)\(\left\{{}\begin{matrix}\dfrac{10}{\sqrt{12x-3}}+\dfrac{5}{\sqrt{4y+1}}=1\\\dfrac{7}{\sqrt{12x-3}}+\dfrac{8}{\sqrt{4y+1}}=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=4\\x\left(1+4y\right)+y=2\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2+x+1=3y\\y^2+y+1=3x\end{matrix}\right.\)
Giải hệ \(\left\{{}\begin{matrix}9+\left(y-5\right)^2=x+y\\\sqrt{y}-4+x=\frac{\sqrt{x+y}+\sqrt{x+y-9}+2}{\sqrt{y}-x+4}\end{matrix}\right.\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}x+y+\sqrt{2y-1}+\sqrt{x-y}=5\\y^2+2=xy+y\end{matrix}\right.\)
giải hệ: a, left{{}begin{matrix}x^2+frac{1}{y^2}+frac{x}{y}3x+frac{1}{y}+frac{x}{y}3end{matrix}right.
b, left{{}begin{matrix}sqrt[]{x-1}+sqrt[]{y-1}2frac{1}{x}+frac{1}{y}1end{matrix}right.
c,left{{}begin{matrix}xsqrt[]{x}+ysqrt[]{y}35xsqrt[]{y}+ysqrt[]{x}30end{matrix}right.
d,left{{}begin{matrix}x^2+xy+y^23x+xy+y-1end{matrix}right.
e,left{{}begin{matrix}left(frac{x}{y}right)^3+left(frac{x}{y}right)^212left(xyright)^2+xy6end{matrix}right.
Đọc tiếp
giải hệ: a, \(\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}\sqrt[]{x-1}+\sqrt[]{y-1}=2\\\frac{1}{x}+\frac{1}{y}=1\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}x\sqrt[]{x}+y\sqrt[]{y}=35\\x\sqrt[]{y}+y\sqrt[]{x}=30\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}x^2+xy+y^2=3\\x+xy+y=-1\end{matrix}\right.\)
e,\(\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\)
giải hệ pt \(\left\{{}\begin{matrix}x-y=1\left(1\right)\\\sqrt{x^2+2y+3}=3x-5\left(2\right)\end{matrix}\right.\left(x,y\in R\right)}\)
\(\left\{{}\begin{matrix}\dfrac{5x-1}{x+3}-\dfrac{y+5}{y-2}\\\dfrac{x+4}{x+3}+\dfrac{2y-1}{y-2}\end{matrix}\right.\)