tính giá trị biểu thức:A=[(11,81+8,19).0,02]:[(-9):11,25]:23^10-23^11/23^11+21.23^10
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Answer:
\(\left(x-\dfrac{1}{2}\right)^2=4\)
\(\left(x-\dfrac{1}{2}\right)^2=\left(\pm2\right)^2\)
TH1: \(\left(x-\dfrac{1}{2}\right)^2=2^2\)
\(x-\dfrac{1}{2}=2\)
\(x=2+\dfrac{1}{2}\)
\(x=\dfrac{5}{2}\)
TH2: \(\left(x-\dfrac{1}{2}\right)^2=\left(-2\right)^2\)
\(x-\dfrac{1}{2}=-2\)
\(x=-2+\dfrac{1}{2}\)
\(x=\dfrac{-3}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
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Answer:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{100}{100}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
\(\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.....\dfrac{99}{100 }\) Giải: \(=\dfrac{3}{1.4}.\dfrac{4}{1.5}.\dfrac{5}{1.6}...\dfrac{99}{1.100}\)
\(=\dfrac{4-1}{1.4}.\dfrac{5-1}{1.5}.\dfrac{6-1}{1.6}...\dfrac{100-1}{1.100}\)
\(=1-\dfrac{1}{4}.1-\dfrac{1}{5}.1-\dfrac{1}{6}...1-\dfrac{1}{100}\)
\(=1\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{100}\right)\)