Giải hệ pt: \(\left\{{}\begin{matrix}\frac{15}{x}-\frac{7}{y}=9\\\frac{4}{x}+\frac{9}{y}=35\end{matrix}\right.\)
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NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2020
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge16\\y\ge9\end{matrix}\right.\)
Từ pt thứ nhất của hệ:
\(\frac{8xy}{x^2+y^2+6xy}+\frac{17}{8}\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{21}{4}\)
\(\Leftrightarrow\frac{8}{\frac{x}{y}+\frac{y}{x}+6}+\frac{17}{8}\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{21}{4}\)
Đặt \(\frac{x}{y}+\frac{y}{x}=t\ge2\)
\(\Rightarrow\frac{8}{6+t}+\frac{17}{8}t=\frac{21}{4}\)
\(\Leftrightarrow\frac{17}{8}t^2+\frac{15}{2}t-\frac{47}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-\frac{94}{17}< 0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\frac{x}{y}+\frac{y}{x}=2\Leftrightarrow x^2+y^2=2xy\)
\(\Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x=y\)
Thay xuống pt dưới:
\(\sqrt{x-16}+\sqrt{x-9}=7\)
\(\Leftrightarrow\sqrt{x-16}-3+\sqrt{x-9}-4=0\)
\(\Leftrightarrow\frac{x-25}{\sqrt{x-16}+3}+\frac{x-25}{\sqrt{x-9}+4}=0\)
\(\Leftrightarrow...\)
9 tháng 2 2020
a) \(\left\{{}\begin{matrix}x+2y=-1\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
Vậy..............................................................................
b) \(\left\{{}\begin{matrix}\frac{5}{x}-\frac{6}{y}=3\\\frac{4}{x}+\frac{9}{y}=7\end{matrix}\right.\)ĐKXĐ: x,y≠0
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{20}{x}-\frac{24}{y}=12\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\frac{69}{y}=23\\\frac{20}{x}+\frac{45}{y}=35\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=10\end{matrix}\right.\)
Vậy...................................................................................
c) \(\left\{{}\begin{matrix}3\sqrt{x+1}+\sqrt{y-1}=1\\\sqrt{x+1}-\sqrt{y-1}=-2\end{matrix}\right.\)ĐKXĐ:\(\left\{{}\begin{matrix}x\ge-1\\y\ge1\end{matrix}\right.\)
\(\Rightarrow4\sqrt{x+1}\)\(=-1\)(vô nghiệm)
Vậy hệ pt vô nghiệm
d) Nhân 3 pt đầu rồi thu gọn
23 tháng 2 2020
Ta có hệ pt :
\(\left\{{}\begin{matrix}\frac{y}{3}=\frac{x}{4}\\\frac{z}{9}=\frac{x}{4}\\7x-3y+2z=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3x}{4}\\z=\frac{9x}{4}\\7x-\frac{3.3x}{4}+\frac{2.9x}{4}=90\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3x}{4}\\z=\frac{9x}{4}\\28x-9x+18x=360\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3x}{4}\\z=\frac{9x}{4}\\37x=360\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{3}{4}.\frac{360}{37}\\z=\frac{9}{4}.\frac{360}{37}\\x=\frac{360}{37}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\frac{270}{37}\\z=\frac{810}{37}\\x=\frac{360}{37}\end{matrix}\right.\)
Vậy . . . . . . . . .
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2019
1/ ĐKXĐ:...
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{12}{x}+\frac{3}{y-2}=3\end{matrix}\right.\) \(\Rightarrow\frac{10}{x}=-1\Rightarrow x=-10\)
\(\frac{4}{-10}+\frac{1}{y-2}=1\Rightarrow\frac{1}{y-2}=\frac{7}{5}\Rightarrow y-2=\frac{5}{7}\Rightarrow y=\frac{19}{7}\)
2/ ĐKXĐ:...
Đặt \(\left\{{}\begin{matrix}\frac{1}{2x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a-b=0\\3a-6b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{9}\\b=\frac{2}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{2x-y}=\frac{1}{9}\\\frac{1}{x+y}=\frac{2}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-y=9\\x+y=\frac{9}{2}\end{matrix}\right.\) \(\Rightarrow...\)
3/ \(\Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-6y-15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-x+6y=-19\end{matrix}\right.\) \(\Rightarrow...\)
4/ Bạn tự giải
Đặt \(\frac{1}{x}=a,\frac{1}{y}=b\)
Ta có hệ phương trình:
\(\left\{{}\begin{matrix}15a-7b=9\\4a+9b=35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}60a-28b=36\\60a+135b=525\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-163b=-489\\4a+9b=35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=3\\4a+9.3=35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=3\\4a=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=3\\a=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}=2\\\frac{1}{y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất là (x;y) = (\(\frac{1}{2};\frac{1}{3}\))