giúp dùm nghen cần gấp
tính
A=\(\frac{2^3+1}{2^3-1}.\frac{3^3+1}{3^3-1}.\frac{4^3+1}{4^3-1}......\frac{10^3+1}{10^3-1}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
JA
26 tháng 7 2016
Tong quat: a^3+1=(a+1)[a^2-a+1]=(a+1)[(a-0,5)^2+0,75]
a^3-1=(a-1)[a^2+a+1]=(a-1)[(a+0,5)^2+0,75]
Tu so cua A=(2+1).[(2-0,5)^2+0,75].(3+1).[(3-0,5)^2+0,75].(4+1).[(4-0,75)^2+0,75]....(10+1).[(10-0,5)^2+0,75]
=3.[1,5^2+0,75].4.[2,5^2+0,75].5.[3,5^2+0,75]....11.[9,5^2+0,75]
Mau so cua A= (2-1).[(2+0,5)^2+0,75].(3-1).[(3+0,5)^2+0,75].(4-1).[(4+0,75)^2+0,75]....(10-1).[(10+0,5)^2+0,75]
=[2,5^2+0,75].2.[3,5^2+0,75].3.[4,5^2+0,75]....9.[10,5^2+0,75]
Vay A=3.[1,5^2+0,75].4.[2,5^2+0,75].5.[3,5^2+0,75]....11.[9,5^2+0,75]/[2,5^2+0,75].2.[3,5^2+0,75].3.[4,5^2+0,75]....9.[10,5^2+0,75]
=(3.4.5...11/1.2.3...9).[(1,5^2+0,75)(2,5^2+0,75)(3,5^2+0,75)...(9,5^2+0,75)/(2,5^2+0,75)(3,5^2+0,75)(4,5^2+0,75)...(10,5^2+0,75)]
=11.10.(1,5^2+0,75)/2.(10,5^2+0,75)
Con bao nhieu ban tu tinh tiep nha
Tai vi may minh bi lag nen khong danh phan so duoc vi vay minh phai tach mau, tu ra. sorry
21 tháng 7 2019
a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)
=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
=> x + 1 = 0
=> x = -1
21 tháng 7 2019
b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)
=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)
=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)
=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
=> x - 2021 = 0
=> x = 2021
c) \(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)
=> \(-\frac{1}{12}x+6=7\)
=> \(-\frac{1}{12}x=1\)
=> x = -12
Tui làm đại nghen không biết đúng không nữa.
Dùng hằng đẳng thức:
\(a^3+1=\left(a+1\right)\left(a^2-a+1\right)=\left(a+1\right)\left[\left(a-0,5\right)^2+0,75\right]\)
\(a^3-1=\left(a-1\right)\left(a^2+a+1\right)=\left(a-1\right)\left[\left(a+0,5\right)^2+0,75\right]\)
Ta có: \(A=\frac{2^3+1}{2^3-1}.\frac{3^3+1}{3^3-1}.\frac{4^3+1}{4^3-1}...\frac{10^3+1}{10^3-1}\)
\(=\frac{\left(2^3+1\right)\left(3^3+1\right)\left(4^3+1\right)...\left(10^3+1\right)}{\left(2^3-1\right)\left(3^3-1\right)\left(4^3-1\right)...\left(10^3-1\right)}\)
Đặt \(P=\left(2^3+1\right)\left(3^3+1\right)\left(4^3+1\right)...\left(10^3+1\right)\)
<=> P = (2+1)[(2-0,5)2 + 0,75] . (3+1)[(3-0,5)2 + 0,75] . (4+1)[(4-0,5)2 + 0,75] ... (10+1)[(10-0,5)2 + 0,75]
= 3.(1,52 + 0,75) . 4(2,52 + 0,75) . 5(3,52 + 0,75)... 11(9,52 + 0,75)
Đặt \(Q=\left(2^3-1\right)\left(3^3-1\right)\left(4^3-1\right)...\left(10^3-1\right)\)
<=> Q = (2-1)[(2+0,5)2 + 0,75] . (3-1)[(3+0,5)2 + 0,75] . (4-1)[(4+0,5)2 + 0,75] ... (10-1)[(10+0,5)2 + 0,75]
= (2,52 + 0,75) . 2(3,52 + 0,75) . 3(4,52 + 0,75)... 9(10,52 + 0,75)
=> \(A=\frac{P}{Q}\)\(=\frac{\text{3.(1,5^2 + 0,75) . 4(2,5^2 + 0,75) . 5(3,5^2 + 0,75)... 11(9,5^2 + 0,75)}}{\left(2,5^2+0,75\right).2\left(3,5^2+0,75\right).3\left(4,5^2+0,75\right)...9\left(10,5^2+0,75\right)}\)
\(=\frac{3.4.5...11}{1.2.3...9}.\frac{\left(1,5^2+0,75\right)\left(2,5^2+0,75\right)\left(3,5^2+0,75\right)...\left(9,5^2+0,75\right)}{\left(2,5^2+0,75\right)\left(3,5^2+0,75\right)\left(4,5^2+0,75\right)...\left(10,5^2+0,75\right)}\)
\(=\frac{10.11.\left(1,5^2+0,75\right)}{2.\left(10,5^2+0,75\right)}=\frac{55}{37}\)
Vậy: \(A=\frac{55}{37}\)
K NHA!!
0,75 ở đâu ra z chỉ dới