Tìm giá trị của biểu thức \(P=\frac{2}{1.3}-\frac{4}{3.5}+\frac{6}{5.7}+\frac{8}{7.9}+...-\frac{96}{95.97}+\frac{98}{97.99}\)
Tính giá trị biêut hức;B=2/1.3-4/3.5+6/5.7-8/7.9+...-96/95.97+98/97.99
Tính nhanh:
S = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{95.97}+\frac{1}{97.99}\)
\(S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
S=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{95.97}+\frac{1}{97.99}\)
S=\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{99}\right)\)
S=\(\frac{1}{2}.\left(1-\frac{1}{99}\right)\)
S=\(\frac{1}{2}.\frac{98}{99}\)
S=\(\frac{49}{99}\)
S = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
= \(\frac{1}{2}\) . (\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\))
= \(\frac{1}{2}\). (\(1-\frac{1}{99}\))
= \(\frac{1}{2}\). \(\frac{98}{99}\) = \(\frac{49}{99}\)
tính giá trị của biểu thức
a) A=\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + \(\frac{1}{4.5}\) + ...+\(\frac{1}{99.100}\)
b) B= \(\frac{2}{1.3}\)+\(\frac{2}{3.5}\) + \(\frac{2}{5.7}\)+\(\frac{2}{7.9}\) +...+\(\frac{2}{97.99}\)
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2.\left(1-\frac{1}{99}\right)\)
\(=2.\frac{98}{99}\)
\(=\frac{196}{99}=1\frac{97}{99}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(=1-\frac{1}{99}\)
\(=\frac{98}{99}\)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
=>\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=>\(\frac{1}{1}-\frac{1}{100}\)
=>\(\frac{99}{100}\)
B=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{97.99}\)
=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\)
=>\(\frac{1}{1}-\frac{1}{99}\)
=>\(\frac{98}{99}\)
Tính nhanh :
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{95.97}+\frac{2}{97.99}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
Tự tính
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
32/99
k với nghe bạn
và chúc chueeuf nay thi tốt
Đây là bài 0,5đ đề 45' trường mình, các bn lm thử nhé
\(A=\frac{1}{1.3}-\frac{2}{3.5}+\frac{3}{5.7}-\frac{4}{7.9}+...-\frac{48}{95.97}+\frac{49}{97.99}\)
\(CMR:A>\frac{1}{4}\)
(Dấu "." là dấu "x" nhé)
A có tổng cộng 49 số hạng, nhóm 2 số hạng liên tiếp với nhau được:
\(A=\left(\frac{1}{1.3}-\frac{2}{3.5}\right)+\left(\frac{3}{5.7}-\frac{4}{7.9}\right)+...+\left(\frac{47}{93.95}-\frac{48}{95.97}\right)+\frac{49}{97.99}\)
\(A=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{93.97}+\frac{49}{97.99}\)=> \(4A=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{93.97}+\frac{196}{97.99}=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{93}-\frac{1}{97}+\frac{196}{97.99}\)
=> \(4A=1-\frac{1}{97}+\frac{196}{97.99}=\frac{96}{97}+\frac{196}{97.99}=\frac{9700}{97.99}=\frac{100}{99}>1\)
\(4A>1=>A>\frac{1}{4}\)
Bn trừ 2 PS kiểu gì hay zậy?
Giúp mình nhá
B1 : Tính giá trị biểu thức :
S = \(\frac{2}{1.3}\)- \(\frac{4}{3.5}\)+ \(\frac{6}{5.7}\)- \(\frac{8}{7.9}\)+ ... - \(\frac{96}{95.97}\)+ \(\frac{98}{97.99}\)
B2 : Chứng minh rằng A = \(\frac{1}{4}\)+ \(\frac{1}{16}\)+ \(\frac{1}{36}\)+ \(\frac{1}{64}\)+ \(\frac{1}{100}\)+ \(\frac{1}{144}\)+ \(\frac{1}{196}\)+ \(\frac{1}{256}\)+ \(\frac{1}{324}\)< \(\frac{1}{2}\)
B2 : \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{114}+\frac{1}{196}+\frac{1}{256}+\frac{1}{324}\)
\(=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{18^2}\)
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{4^2}< \frac{1}{2\cdot4}\)
\(\frac{1}{6^2}< \frac{1}{4\cdot6}\)
...
\(\frac{1}{18}< \frac{1}{16\cdot18}\)
\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{18^2}< \frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{16}-\frac{1}{18}\right)\)
\(\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{18^2}< \frac{1}{2}< \frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{18}\right)\)
Tính giá trị của biểu thức
A =\(\left(\frac{1}{2}+1\right)\times\left(\frac{1}{3}+1\right)\times\left(\frac{1}{4}+1\right)\times....\times\left(\frac{1}{99}+1\right)\)
Chứng tỏ
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{97.99}>32\%\)
A =(1/2 +1)×(1/3 +1)×(1/4 +1)×....×(1/99 +1)
=3/2x4/3x...............x100/99
=2-1/99
=197/99
A= \(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot.....\cdot\frac{100}{99}\)
A=\(\frac{\left(3\cdot4\cdot5\cdot....\cdot99\right)\cdot100}{2\cdot\left(3\cdot4\cdot5\cdot...\cdot99\right)}\)
A=\(\frac{100}{2}=50\)
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
=> \(\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)>\(\frac{32}{100}\)=32%
Câu đầu tiên:
\(A=\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot...\cdot\left(\frac{1}{99}+1\right)\)
\(A=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{100}{99}=\frac{3\cdot4\cdot5\cdot...\cdot99\cdot100}{3\cdot4\cdot5\cdot...\cdot99\cdot2}=\frac{100}{2}=50\)
Câu thứ 2:
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}>\frac{32}{100}\)
Gía trị biểu thức:
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+\frac{1}{5.7}+\frac{1}{6.8}+\frac{1}{7.9}+\frac{1}{8.10}\)
=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\right)\)
= \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right)\)
= \(\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)
=\(\frac{29}{45}\)
<br class="Apple-interchange-newline"><div id="inner-editor"></div>S=11.3 −12.4 +13.5 −14.6 +15.7 −16.8 +17.9 −18.10
(11.3 +13.5 +15.7 +17.9 )−12.4 −14.6 −16.8 −18.10
=12 (21.3 +23.5 +25.7 +27.9 )−(12.4 +14.6 +16.8 +18.10 )
=12 (21.3 +23.5 +25.7 +27.9 )−12 (22.4 +24.6 +26.8 +28.10 )
=12 (1−13 +13 −15 +15 −17 +17 −19 )−12 (12 −14 +14 −16 +16 −18 +18 −110 )
=12 (1−19 )−12 (12 −110 )
=12 (1−19 −12 +110 )
Tính giá trị biểu thức
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+......1\frac{2}{41.43}\)
2/3.5 + 2/5.7 + 2/7.9 + ... + 2/41.43
= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/41 - 1/43
= 1/3 - 1/43
= 40/129
ỦNG HỘ NHA
2/3.5 + 2/5.7 + 2/7.9 +......+ 2/41.43
= 1/3-1/5 + 1/5-1/7 + 1/7-1/9 +.....+ 1/41-1/43
= 1/3-1/43
= 40/129.