giúp dùm nghen cần gấp
tính
A=\(\frac{2^3+1}{2^3-1}.\frac{3^3+1}{3^3-1}\frac{4^3+1}{4^3-1}.....\frac{10^3+1}{10^3-1}\)
Những câu hỏi liên quan
giúp dùm nghen cần gấp
tính
A=\(\frac{2^3+1}{2^3-1}.\frac{3^3+1}{3^3-1}.\frac{4^3+1}{4^3-1}......\frac{10^3+1}{10^3-1}\)
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!!
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Huân Nguyễn
ta có: \(a^3+1=\left(a+1\right)\left(a^2-a+1\right)=\left(a+1\right)\left(a^2-a+0,25+0,75\right)\)
\(=\left(a+1\right)+\left[\left(a-0,5\right)^2+0,75\right]\)
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\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
Giúp mình với, mình đang cần gấp lắm nhé các bạn!
Tính
a)\(\frac{\left(\frac{3}{4}\right)^2-\left(-\frac{4}{5}\right)^2.\left(-1\right)^7}{\left(\frac{2}{3}\right)^3.\left(-\frac{5}{2}\right)^3}\)
b) \(\frac{16^{13}.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
c) \(\frac{1}{3}-\frac{1}{18}-\frac{1}{54}-\frac{1}{108}-\frac{1}{180}-\frac{1}{270}-\frac{1}{378}\)
GIÚP MÌNH NHA!!!!! ĐANG CẦN GẤP
Làm ơn giúp mình với!!
Tính giá trị biểu thức:
\(A=\frac{2^3+1}{2^3+1}.\frac{3^3+1}{3^3-1}.\frac{4^3+1}{4^3-1}...\frac{10^3+1}{10^3-1}\)
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
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left(frac{-5}{12}+frac{7}{4}-frac{3}{8}right)-left[4frac{1}{2}-7frac{1}{3}right]-left(frac{1}{4}-frac{5}{2}right)left[2frac{1}{4}-5frac{3}{2}right]-left(frac{3}{10}-1right)-5frac{1}{2}+left(frac{1}{3}-frac{5}{6}right)frac{4}{7}-left(3frac{2}{5}-1frac{1}{2}right)-frac{5}{21}+left[3frac{1}{2}-4frac{2}{3}right]frac{1}{8}-1frac{3}{4}+left(frac{7}{8}-3frac{7}{2}+frac{3}{4}right)-left[frac{7}{4}-frac{5}{8}right]left(frac{3}{5}-2frac{1}{10}+frac{11}{20}right)-left[frac{-3}{4}+1frac{7}{2}right]left[-2fr...
Đọc tiếp
\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)
\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)
\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)
\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)
\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)
\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)
\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)
\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)
\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)
\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)
\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)
\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)
\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)
\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)
\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)
\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)
\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)
\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)
TRÌNH BÀY GIÚP MÌNH NHA
\(\frac{1}{2}-\frac{1}{3}-\frac{2}{3}+\frac{1}{4}+\frac{2}{4}+\frac{3}{4}-\frac{1}{5}-\frac{2}{5}-\frac{3}{5}-\frac{4}{5}+...+\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+...+\frac{9}{10}\)\(+\)\(\frac{9}{10}\)
Đề bài là gì bạn ơi có chỗ ...
chưa có đề bạn ơi! Y_Y
Hok tốt ^_^
Quy luật của dãy số này là gì bạn có thể nói rõ ra đc ko
Xem thêm câu trả lời
a) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right)\div\left(1+\frac{2}{5}+\frac{2}{3}=-\frac{5}{4}\right)\)
b) \(-\frac{3x}{4}\times\left(\frac{1}{x}+\frac{2}{7}\right)=0\)
c) \(3-\frac{1-\frac{1}{2}}{1+\frac{1}{x}}=2\frac{2}{3}\)
Help me!!! Cần gấp trong 10 phút.
tìm số nguyên x
a) (x-3)+(x-2)+(x-1)+....+10+11=11
b) \(\frac{2}{3}\).\(\left[\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right]\)<= x <=\(4\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{6}\right]\)
giúp mình nha mai mình đi học rồi. minh cần gấp lắm bạn nào giải được thì mau giải giúp mình nha
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}}b.frac{x-1}{2020}+frac{x-2}{2019}-frac{x-3}{2018}frac{x-4}{2017}c.left(frac{3}{4}x+3right)-left(frac{2}{3}x-4right)-left(frac{1}{6}x+1right)left(frac{1}{3}x+4right)-left(frac{1}{3}x-3right)giúp mk vs mk đg cần gấpThank you
Đọc tiếp
a.\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+_{\frac{x-11}{12}}\)
b.\(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)
c.\(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
giúp mk vs mk đg cần gấp
Thank you
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
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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
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