\(\frac{145.146-15}{145.145+130}=\frac{145.145+145-15}{145.145+130}=\frac{145.145+130}{145.145+130}=1\)

\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+.....+\frac{3^2}{97.100}\)

\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)

Ta thấy :

 \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)

\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)

\(.........\)

\(\frac{3}{97.100}=\frac{100-97}{97.100}=\frac{1}{97}-\frac{1}{100}\)

\(\Rightarrow A=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\right)\)

\(=3\left(1-\frac{1}{100}\right)=3\cdot\frac{99}{100}=\frac{297}{100}\)

đáp án = \(\frac{297}{100}\)

đúng không?

kết bạn với mh nha

Ta có\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)

= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

= \(1-\frac{1}{46}\)

Vì \(1-\frac{1}{46}< 1\)nên S<1

\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+.......+\frac{3}{43\cdot46}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+......+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}\)

Ta có \(1-\frac{1}{46}< 1\)=> S < 1

\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{40.43}+\frac{3}{43.46}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}\)

vì \(1-\frac{1}{46}< 1\Rightarrow S< 1\)

A:3=\(\frac{3}{1.4}+\frac{3}{4.7}\)\(+.....+\frac{3}{97.100}\)

A:3=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\)

A:3=\(\frac{1}{1}-\frac{1}{100}\)

A:3=\(\frac{99}{100}\)

A=\(\frac{99}{100}.3\)

A=\(\frac{297}{100}\)

\(A:3=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\)

\(A:3=\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A:3=\frac{1}{1}-\frac{1}{100}\)

\(A:3=\frac{99}{100}\)

\(A=\frac{99}{100}.3\)

\(A=\frac{297}{100}\)

\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{43}-\frac{1}{46}\)

\(S=1-\frac{1}{46}\)

Có \(1-\frac{1}{46}< 1\)

\(\Rightarrow S< 1\)

nhan xet:3/1.4=1/1-1/4

3/4.7=1/4-1/7

3/7.10=1/7-1/10

.....................

3/40.43=1/40-1/43

3/43.46=1/43-1/46

S=1/1-1/3+1/3-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

S=1/1-1/46

S=46/46-1/46

S=45/46<1

vay s<1

tớ ko biết

S = 3 - \(\frac{3}{100}\)= \(\frac{300}{100}-\frac{3}{100}=\frac{297}{100}\)

S=3/1.4+3/4.7+3/7.10+.....+3/97.100

S=1/1-1/4+1/4-1/7+1/7-1/10+.....+1/97-1/100

S=1-1/100

S=99/100

Trả lời

3/1.4+3/4.7+3/7.10

=3/1.3/10

=3/10

Chúc bạn học tốt #

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}\)

\(=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}\)

\(=\frac{1}{1}-\frac{1}{10}=\frac{10}{10}-\frac{1}{10}\)

\(=\frac{9}{10}\)