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HPT : \(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=\frac{5}{36}\\\frac{4}{x}+\frac{3}{y}=\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{3}{x}+\frac{3}{y}=\frac{5}{12}\left(1\right)\\\frac{4}{x}+\frac{3}{y}=\frac{1}{2}\left(2\right)\end{cases}}\)
Từ (1) và (2), lấy vế trừ vế ta được :
\(\Leftrightarrow\left(\frac{4}{x}+\frac{3}{y}\right)-\left(\frac{3}{x}+\frac{3}{y}\right)=\frac{1}{2}-\frac{5}{12}\)
\(\Leftrightarrow\frac{1}{x}=\frac{1}{12}\)
\(\Leftrightarrow\frac{1}{y}=\frac{5}{36}-\frac{1}{x}=\frac{5}{36}-\frac{1}{12}=\frac{1}{18}\)
\(\Leftrightarrow\hept{\begin{cases}x=12\\y=18\end{cases}}\)
sin 650=cos 350
\(cos70^0=sin30^0\)
\(tan80^0=cot20^0\)
\(cot68^0=tan32^0\)
abc=100a+10b+c=n2-1(*)
cba=100c+10b+a=n2-4n+4(**)
(*)-(**)=99(a-c)=4n+5
=> 4n-5 chia hết cho 99
Mà \(100\le abc\le999\)
=> \(100\le n^2-1\le999\)
<=> \(101\le n^2\le1000\)=\(11< 31\)=\(39\le4n-5\le199\)
Vì 4n+5 chia hết cho 99
Nên 4n-5=99
4n=99+5
4n=104
n=104:4
n=26
Vậy abc=675
\(\Rightarrow\left\{{}\begin{matrix}x+y=71-xy\\xy\left(x+y\right)=880\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y=71-xy\\xy\left(71-xy\right)=880\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=71-xy\\x^2y^2-71xy+880=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y=71-xy\\\left(xy-16\right)\left(xy-55\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=71-xy\\\left[{}\begin{matrix}xy=16\\xy=55\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}xy=55\\x+y=16\end{matrix}\right.\\\left\{{}\begin{matrix}xy=16\\x+y=55\end{matrix}\right.\end{matrix}\right.\)
+ TH1: \(\left\{{}\begin{matrix}xy=55\\x+y=16\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=\frac{55}{x}\\x+\frac{55}{x}=16\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=\frac{55}{x}\\x^2-16x+55=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=\frac{55}{x}\\\left(x-5\right)\left(x-11\right)=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=5\\y=11\end{matrix}\right.\\\left\{{}\begin{matrix}x=11\\y=5\end{matrix}\right.\end{matrix}\right.\)
TH còn lại lm tương tự
cảm ơn bn nhìu