tính tổng
S= 1.2 + 2.3+ 3.4+............ + 49 . 50
(1/1.2+1/2.3+1/3.4+…+1/49.50)x=49/50
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{49.50}\right)x=\frac{49}{50}\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)x=\frac{49}{50}\)
\(\left(1-\frac{1}{50}\right)x=\frac{49}{50}\)
\(\frac{49}{50}x=\frac{49}{50}\)
\(x=\frac{\frac{49}{50}}{\frac{49}{50}}\)
\(x=1\)
Vậy \(x=1\)
Gọi A=1/1.2+1/2.3+1/3.4+...+1/49.50
A=1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50
A=1-1/50
A=49/50
Viết lại ta có: (1/1.2+1/2.3+1/3.4+...+1/49.50)x=49/50
49/50x=49/50
=> x=1
Chứng tỏ rằng : 1/1.2+1/2.3+1/3.4+...+1/49+50<1
1/1.2 +1/2.3 +1/3.4 +...+ 1/49.50
=1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50
=1-1/50<1
1/1.2 + 1/2.3 +1/3.4 + ... + 1/49.50 ( chỗ này 49.50 chứ ko phải 49+50 đâu nha)
= 1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50 (-1/2+1/2 là hết cứ như z thì chỉ còn lại 1-1/50)
=1-1/50 <1
Sửa lại đề: Chứng tỏ rằng \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{49.50}< 1\)
Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...........+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}< 1\left(đpcm\right)\)
Chứng minh rằng 1/1.2 + 1/2.3 + 1/3.4 +........+1/49+50 = 1/26 + 1/27 +1/28 +.....+ 1/50
1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+...+1/50
=1/1-1/2+1/3-1/4+...+1/49-1/50
=(1/1+1/3+...+1/49)-(1/2+1/4+...+1/50)
=(1/1+1/2+1/3+...+1/49+1/50)-2(1/2+1/4+...+1/50)
=1/1+1/2+1/3+...+1/50-1-1/2-1/3-...-1/25
=1/26+1/27+...+1/50 (đpcm)
Tính hợp lý :
1, 25 . 12
2, 37 . 102
3, 125.24
4, A= 1+2+3+...+49+50
5, B = 1.2 + 2.3 +3.4 +...+20.21
Ai nhanh mk sẽ tích cho
1, 25x12= 25x10+25x2
= 250+50
= 300
2, 37x102=37x100+37x2
= 3700 + 74
= 3774
3, 125x24=125x20+125x4
= 2500+500
= 3000
4, A= 1+2+3+...+50
= [(50-1)/1+1]x(50+1)/2
= 1275
Tính a
A=1.2+2.3+3.4+......+49+50=?????
Giúp mình nha mình tick cho
Ta có : A = 1.2 + 2.3 + 3.4 + ..... + 49.50
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .... + 49.50.51
=> 3A = 49.50.51
= >A = 49.50.51/3 = 41650
3A= 1.2.3 + 2.3.(4-1) +3.4.(5-2) +...+ 99.100.(101-98)
3A= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +...+-99.100.98
3A=99.100.101
A= 33.100.101
CMR: 1/1.2+1/2.3+1/3.4+....+1/49.50=1/26+1/27+.....+1/49+1/50
1/1.2 + 1/2.3 + ...... + 1/49.50
= 1/1 - 1/2 + 1/2 - - .... - 1/50 = 1 - 1/50 = 49/50
Cho A= 1/1.2 +1/2.3 +1/3.4 +......+1/49+50. Chứng minh rằng 7/ 12 < A < 5/6
\(A=\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{49.50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\frac{1}{1}-\frac{1}{50}\)
\(A=\frac{49}{50}\)
Vì \(\frac{245}{420}< \frac{245}{294}< \frac{245}{250}\)
Vậy \(\frac{7}{12}< \frac{49}{50}< \frac{5}{6}\)
Cho :A=1/1.2 +1/2.3 + 1/3.4 + .......+1/49+50. Chứng minh rằng 7/12 < A < 5/6.
Hình như phân số cuối sai đề bn nhỉ?
Tính tổng
A=1/1.2+1/2.3+1/3.4+..........+1/49+1/50
AI LÀM NHANH NHẤT MÌNH SẼ TICK