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\(3x^2-75=0\)
\(3\left(x^2-25\right)=0\)
\(x^2-25^2=0\)
\(\left(x-5\right)\left(x+5\right)=0\)
\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
\(\dfrac{1}{12}h=5p\)
\(\dfrac{1}{2}\) yến \(=5kg\)
\(m=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+......+\dfrac{1}{97.99}\)
\(m.2=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+.....+\dfrac{2}{97.99}\)
\(m.2=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+.....+\dfrac{1}{97}-\dfrac{1}{99}\)
(Đã triệt tiêu hết tất cả các phân số ở giữa)
\(m.2=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)
\(m=\dfrac{32}{99}:2=\dfrac{16}{99}\)
This is about gathering information from others, not introducing yourself
527 = (53)9 = 1259 < 1289 = (27)9 = 263
263 = (29)7 = 5127 < 6257 = (54)7 = 528