Bài 1:tính
a, A = 25/6+25/6.11+25/11.16+...+25/41.46
b, B= 1/10+1/15+1/21+...+1/120
Bài 2: Chứng minh rằng:
1/2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50
bài 3: tìm y
(y - 1/2).(1/2+1/6+1/12+1/20+.....+1/90) = 1/3
Những câu hỏi liên quan
Bài 1 Tính giá trị biểu thức :
A = 3/1.4 + 5/4.9 + 7/9.16 + 9/16.25 + 11/25.36
B = 3/1.4 + 3/4.7 + ... + 3/100.103
C = 3/1.4 + 6/4.10 + 9/10.19 + 12/19.31 + 15/31.46 + 18/46.64
Bài 2 Chứng minh rằng :
1/1.2 + 1/3.4 + 1/5.6 + ... + 1/49.50 = 1/26 + 1/27 + 1/28 + ... + 1/50
Bài 1:
\(A=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+\dfrac{9}{16.25}+\dfrac{11}{25.36}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{36}\)
\(=1-\dfrac{1}{36}=\dfrac{35}{36}\)
\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)
\(C=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}+\dfrac{15}{31.46}+\dfrac{18}{46.64}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{64}\)
\(=1-\dfrac{1}{64}=\dfrac{63}{64}\)
Bài 2:
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\left(đpcm\right)\)
Đúng 1
Bình luận (1)
Chứng minh rằng:
a) 1.2 - 1 phần 2! + 2.3 -1 phần 3! + 3.4 -1/4! + ... + 99.100 -1 /100! < 2
b) 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/49.50 = 1/26 + 1/27 + 1/28 + ... + 1/50
chứng tở rằng 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 + 1/17 + 1/18 + 1/ 19 + 1/20 + 1/21 + 1/22 + 1/ 23 + 1/24 + 1/25 + 1/26 + 1/27 + 1/28 + 1/29 + 1/30 + 1/31 + 1/32 < 3
ta có
1/2<1/1.2
1/3<1/2.3
...
1/32<1/31.32
=>1/2+1/3+...+1/32<1/1.2+1/2.3+...+1/31.32
=>1/2+1/3+...+1/32<1/1-1/2+1/2-1/3+...+1/31-1/32
=>1/2+1/3+...+1/32<1/1-1/32=31/32
vì 31/32<1
=>tổng đó <1
ta lại có 1+1=2 mà 2 <3
=>tổng đó <3
vậy:-------(bn tự lm nha)
k cho mik vs nha
Đúng 0
Bình luận (0)
Chứng minh rằng : \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
Chứng minh rằng : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Mk làm câu a thôi nhé :)
Vì \(\frac{1}{5^2}< \frac{1}{4.5}\)
\(\frac{1}{6^2}< \frac{1}{5.6}\)
...
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(=>\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(< \)\(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)(1)
Vì \(\frac{1}{5^2}>\frac{1}{5.6}\)
\(\frac{1}{6^2}>\frac{1}{6.7}\)
...
\(\frac{1}{100^2}>\frac{1}{100.101}\)
\(=>\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}\)(2)
Từ (1) và (2) => ĐPCM
a) 21/47 + 9/45 +26/47 + 4/5 (21/47 + 26/47) + (9/45 + 4/5)1+1 2b) 15/12 + 5/13 - 3/12 - 18/13 (15/12 - 3/12) + (18/13 - 5/13)1+1 2c) 13/25 + 6/41 - 38/25 + 35/41 - 1/2 (38/25 - 13/25) + (35/41 + 6/41) + 1/2 1+1 + 1/2 5/2d) 27/23 + 5/21 - 4/23 + 6/21 + 1/2 (27/23 - 4/23) + (6/21 - 5/21) + 1/2 ..........................Đúng không? Phần sau mình tính sao ra kết quả lạ lắm, tính giúp mình.
Đọc tiếp
a) 21/47 + 9/45 +26/47 + 4/5
= (21/47 + 26/47) + (9/45 + 4/5)
=1+1 = 2
b) 15/12 + 5/13 - 3/12 - 18/13
= (15/12 - 3/12) + (18/13 - 5/13)
=1+1 = 2
c) 13/25 + 6/41 - 38/25 + 35/41 - 1/2
= (38/25 - 13/25) + (35/41 + 6/41) + 1/2
= 1+1 + 1/2 = 5/2
d) 27/23 + 5/21 - 4/23 + 6/21 + 1/2
= (27/23 - 4/23) + (6/21 - 5/21) + 1/2
= ..........................
Đúng không? Phần sau mình tính sao ra kết quả lạ lắm, tính giúp mình.
(-3/4+2/3):5/11+(-1/4+1/3):5/11
a)|x-13|-35-45=-47
b)(4+x).(x+3)=0
c)25-(5-x)=120-(28-5)
chứng tở rằng 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 + 1/17 + 1/18 + 1/ 19 + 1/20 + 1/21 + 1/22 + 1/ 23 + 1/24 + 1/25 + 1/26 + 1/27 + 1/28 + 1/29 + 1/30 + 1/31 + 1/32 < 3
giai day du nha!
bn tham khảo câu trả lời của mik ở bn hoàng thu trang nhabây h mik ghi lại dài dòng lắm
Đúng 0
Bình luận (0)
Chứng Minh Rằng: 1/1.2+1/3.4/1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50
a) A = 1/1.2+ 1/3.4+ 1/5.6+...+ 1/99.100
CMR: 7/12<A< 5/6
b) CMR: 1/1.2+ 1/3.4+ 1/5.6+...+1/49.50 = 1/26+ 1/27+ 1/28+...+1/50
a)A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) > 1 / (1*2) + 1 / (3*4) = 1 / 2 + 1 / 12 = 7 / 12 ♦
A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100) =
(1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 <
1 - 1 / 2 + 1 / 3 = 5 / 6 ♥
♦, ♥ => 7 / 12 < A < 5 / 6
b)ta có:
1/1.2+1/3.4+1/5.6+...+1/49.50
=>1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50
=>(1+1/3+1/5+1/7+...+1/49)-(1/2+1/4+1/6+...+1/50)
=>(1+1/2+1/3+...+1/49+1/50)-(1/2+1/4+1/6+...+1/50).2
=>(1+1/2+1/3+...+1/49+1/50) -( 1+1/2+1/3+...+1/25)
=>1/26+1/27+1/28+...+1/50=1/26+1/27+1/28+...+1/50
hay 1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50
Đúng 0
Bình luận (0)
1) 3x-5y/2=7y-3z/3=5z-7x/4 và x+y+z=17
Tìm x,y,z
2) CM: 1/1.2+1/3.4+1/5.6+.............+1/49.50=1/26+1/27+1/28+.........+1/50
\(2\)
CMR
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)\)
Đúng 0
Bình luận (0)
