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Sửa đề : \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{42}{43}\)
=2/3(3/1*4+3/4*7+...+3/97*100)
=2/3(1-1/4+1/4-1/7+...+1/97-1/100)
=2/3*99/100
=198/300
=66/100
=33/50
`1 - 2 + 3 - 4 + 5 - 6 +...+ 2021 - 2022`
`= (1 - 2) + (3 - 4) + (5 - 6) +...+ (2021 - 2022)`
`= (-1) + (-1) + (-1) + ... + (-1) ` [có `2022 : 2 = 1011` nhóm]
`= (-1) xx 1011 = -1011`
Có A=\(\frac{4}{1.4}+\frac{4}{4.7}+\frac{4}{7.10}+.........+\frac{4}{67.70}\)
A=\(\frac{4}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+............+\frac{3}{67.70}\right)\)
A=\(\frac{4}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-..........-\frac{1}{70}\right)\)
A=\(\frac{4}{3}.\left(1-\frac{1}{70}\right)\)
A=\(\frac{4}{3}.\frac{69}{70}=\frac{46}{35}\)
Vì \(\frac{46}{35}>\frac{9}{7}\) nên A>\(\frac{9}{7}\)
\(A=\frac{4}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-....-\frac{1}{70}\right)\)
\(A=\frac{4}{3}.\left(1-\frac{1}{70}\right)=\frac{4}{3}\cdot\frac{69}{70}=\frac{46}{35}>\frac{9}{7}\)
Vậy A >9/7
\(4\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{79.82}\right)\)
\(=\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{79.8}\right)\)
\(=\frac{4}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{79}-\frac{1}{82}\right)\)
\(=\frac{4}{3}\left(\frac{1}{4}-\frac{1}{82}\right)\)
\(=\frac{4}{3}.\frac{39}{164}=\frac{13}{41}\)