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Võ Thiện Tuấn viết tổng quát kết quả hay phép đề bài hả bạn ?
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7} +....+\frac{1}{100}-\frac{1}{103}\)
\(=1-\frac{1}{103}\)
\(=\frac{102}{103}\)
\(S=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{2002\cdot2005}\)
\(3S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{2002\cdot2005}\)
\(3S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)
\(3S=\frac{1}{1}-\frac{1}{2005}\)
\(3S=\frac{2004}{2005}\)
\(S=\frac{2004}{2005}\div3=\frac{668}{2005}\)
Ta có:
\(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2002.2005}\)
\(\Rightarrow S=\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2002.2005}\right)\)
\(\Rightarrow S=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2002}-\frac{1}{2005}\right)\)
\(\Rightarrow S=\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{2005}\right)=\frac{1}{3}.\frac{2004}{2005}=\frac{668}{2005}\)
\(\frac{x}{1.4}+\frac{x}{4.7}+\frac{x}{7.10}+...+\frac{x}{36.39}=1\)
\(\frac{x}{3}.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{36.39}\right)=1\)
\(\frac{x}{3}.[(\frac{1}{1}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{10})+...+(\frac{1}{36}-\frac{1}{39})]=1\)
\(\frac{x}{3}.(\frac{1}{1}-\frac{1}{39})=1\)
\(\frac{x}{3}.\frac{38}{39}=1\)
\(\frac{x}{3}=1:\frac{38}{39}\)
\(\frac{x}{3}=\frac{39}{38}\)
\(\Rightarrow x=.....\)
Mình tính vội nên không tính kết quả đúng chưa, cậu kiểm tra lại nha, còn cách làm thế là chuẩn rồi! Học tốt!
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
Đến đây ta suy được ra S<1
Ta có :
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
\(S=\frac{45}{46}< 1\)
Vậy \(S< 1\)
\(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\left(1-\frac{1}{100}\right)\)
\(A=\frac{2}{3}.\frac{99}{100}\)
\(A=\frac{33}{50}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{49.50}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
\(S=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
\(T=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(T=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+....+\frac{43-40}{40.43}+\frac{46-43}{43.46}\)
\(T=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(T=\frac{1}{1}-\frac{1}{46}=\frac{45}{46}\)
1) Ta có : \(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Vậy T = \(=\frac{99}{100}\)
2) Ta có : \(T=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}=\frac{45}{46}\)
Vậy T = \(\frac{45}{46}\)
3. ( 1/1.4 +1/4.7 +1/7.10 +...+ 1/x.(x+3)
3/1.4 +1/4.7+1/7.10 + ...+ 3/ x . (x+3)
1/1 - 1/4 + 1/4 - 1/6 + 1/7 - 1/10 + ...+ 1/x-1/x+3
1/1 - 1/x+3
x+3/x+3 - 1/x+3
x+2/x+3
\(B=3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+........+\frac{1}{27.30}\right)\)
\(B=3.\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-.......-\frac{1}{27}+\frac{1}{27}-\frac{1}{30}\right)\)
\(B=1.\left(\frac{1}{1}-\frac{1}{30}\right)\)
\(B=\frac{29}{30}\)
B =\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{27.30}\)
B = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{27}-\frac{1}{30}\)
B =\(\frac{1}{1}-\frac{1}{30}\)
B =\(\frac{29}{30}\)
cho mk một tk đi bà con ơi
ủng hộ mk đi làm ơn
1/1.4 + 1/4.7 + 1/7.10 + ... + 1/35.38
= 1/3 ( 1/1 - 1/4 + 1/4 - 1/7 + 1/7 - ... - 1/35 + 1/35 - 1/38)
= 1/3 ( 1- 1/38 )
= 1/3 . 37/38
= 37/114
k cho mình nha mí bạn