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\(75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\frac{-1^2}{2}\)
\(=\frac{75}{100}-\frac{3}{2}+\frac{5}{10}.\frac{12}{5}-\frac{1}{2}\)
\(=\frac{3}{4}-\frac{3}{2}+\frac{6}{5}-\frac{1}{2}\)
\(=\frac{15}{20}-\frac{30}{20}+\frac{24}{20}-\frac{10}{20}\)
\(=\frac{15-30+24-10}{20}\)
\(=\frac{-1}{20}\).
Đặt A=\(\frac{1}{3}.5+\frac{1}{5}.7+...+\frac{1}{97}.99\)
=>A=\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
=>2A=\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
=>2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
=>2A=\(\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
=>A=\(\frac{32}{99}:2=\frac{32}{99}.\frac{1}{2}=\frac{32}{198}=\frac{16}{99}\)
\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-9999}{100^2}\)
\(=-\frac{3.8...9999}{2^2.3^2...100^2}=-\frac{1.3.2.4...99.101}{2.2.3.3...100.100}=-\frac{\left(1.2....99\right).\left(3.4...101\right)}{\left(2.3...100\right).\left(2.3...100\right)}=-\frac{1.101}{100.2}=-\frac{101}{200}\)
\(< -\frac{100}{200}=\frac{1}{2}=B\)
=> A < B
(1-1/2) x ( 1-1/3) x (1-1/4) x ..... x (1-1/2011) x (1-1/2012)
= 1/2 x 2/3 x 3/4 x ...... x 2010/2011 x 2011/2012
suy ra loại các nhân tử chung ( ko cần viết phần này )
= 1/2012
chúc học tốt
em mới học lớp 5 có j sai cho em xl
a) \(\frac{2}{3}x\times\frac{1}{2}=\frac{1}{10}\Rightarrow\frac{2}{3}x=\frac{1}{5}\Rightarrow x=\frac{3}{10}\)
\(\frac{1}{2.x}-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{45.46}=-2\)
\(\frac{1}{2.x}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{45.46}\right)=-2\)
\(\frac{1}{2.x}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{45}-\frac{1}{46}\right)=-2\)
\(\frac{1}{2.x}-\left(1-\frac{1}{46}\right)\)
\(\frac{1}{2.x}-\frac{45}{46}=-2\)
\(\frac{1}{2.x}=-2+\frac{45}{46}\)
\(\frac{1}{2.x}=\frac{-47}{46}\)
\(2x=\frac{46}{-47}\)
\(x=\frac{46}{-47}:2=\frac{-23}{47}\)