Giá trị của biểu thức A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\) là....
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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
S
3 tháng 9 2017
C1: dễ nên tự làm nhé
C2: \(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{2}+\frac{5}{2}\right)\)
\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(=6-5-3+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)\)
\(=-2-\frac{1}{2}=\frac{-4}{2}-\frac{1}{2}=\frac{-5}{2}\)
HT
18 tháng 8 2015
Cách 1:
A = \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
A = \(\frac{35}{6}-\frac{31}{6}-\frac{19}{6}\)
A = \(\frac{-15}{6}=\frac{-5}{2}\)
Cách 2:
A = \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
A = \(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
A = \(6-5-3-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}+\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\)
A = \(\left(6-5-3\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)+\left(\frac{1}{2}-\frac{3}{2}-\frac{5}{2}\right)\)
A = \(-2-0+\left(2-\frac{5}{2}\right)\)
A = \(-2+\left(2-\frac{5}{2}\right)\)
A = \(-2+2-\frac{5}{2}\)
A = \(0-\frac{5}{2}\)
A = \(\frac{-5}{2}\)
23 tháng 2 2017
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(3A=1+\frac{1}{3}+...+\frac{1}{3^5}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)\)
\(2A=1-\frac{1}{3^6}=\frac{3^6-1}{3^6}=\frac{728}{729}\)
\(\Rightarrow A=\frac{728}{729}:2=\frac{364}{729}\)
23 tháng 2 2017
Mong các bạn giúp tớ, tớ sẽ k cho, cảm ơn các bạn.......ek
22 tháng 8 2016
Cách 1:A=(6−2/3+1/2)−(5+5/3−3/2)−(3−7/3+5/2)
= ( 36/6 - 4/6 + 3/6)-(30/6 + 10/6 - 9/6) -(18/6 -14/6 +15/6)
= 35/6 - 31/6 - 19/6
= -15/2
Cách 2: A=(6−2/3+1/2)−(5+5/3−3/2)−(3−7/3+5/2)
= 6- 2/3 +1/2 - 5 - 5/3 + 3/2 - 3 + 7/3 -5/2
= (6-5-3) + ( -2/3-5/3+7/3) + (1/2+3/2-5/2)
= -2 + 0 - 1/2
= -2 - 1/2 = -4/2 - 1/2 = -5/2
NT
21 tháng 12 2018
mình rút gọn đc \(\frac{9x-18}{\left(x-3\right)\left(x+3\right)}\)
DA
7 tháng 9 2021
\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
\(A=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(A=\left(6-5-3\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
\(A=-2+\frac{-1}{2}\)
\(A=-\frac{5}{2}\)
Vậy A= -5/2
8 tháng 12 2019
a)Với x \(\ne\)-1
Ta có: x2 + x = 0
=> x(x + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-1\left(ktm\right)\end{cases}}\)
Với x = 0 => A = \(\frac{0-3}{0+1}=-3\)
b) Ta có: B = \(\frac{3}{x-3}+\frac{6x}{9-x^3}+\frac{x}{x+3}\)
B = \(\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
B = \(\frac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
B = \(\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
B = \(\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\)
B = \(\frac{x+3}{x-3}\)
c) Với x \(\ne\)\(\pm\)3; x \(\ne\)-1
Ta có: P = AB = \(\frac{x-3}{x+1}\cdot\frac{x+3}{x-3}=\frac{x+3}{x+1}=\frac{\left(x+1\right)+2}{x+1}=1+\frac{2}{x+1}\)
Để P \(\in\)Z <=> 2 \(⋮\)x + 1
<=> x + 1 \(\in\)Ư(2) = {1; -1; 2; -2}
<=> x \(\in\){0; -2; 1; -3}
24 tháng 6 2016
Cách 1: A= \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)\)\(-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
= \(\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=\frac{-15}{6}\)=\(\frac{-5}{2}\)
Cách 2: A= \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)\)\(-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
= \(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
= \(\left(6-5-3\right)\)\(+\left(-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}\right)\)\(+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
= \(-2+0+\left(\frac{-1}{2}\right)\)=\(\frac{-5}{2}\)
3A=\(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
3A-A=\(1-\frac{1}{3^6}\)
2A=\(\frac{3^6-1}{3^6}\)
A=\(\frac{\frac{3^6-1}{3^6}}{2}\)
A=\(\frac{364}{729}\)
3A= 3.(1/2+1/3^2+1/3^3+...+1/3^6)
3A= 1+1/3+1/3^2+1/3^3+...+1/3^5
3A-A=(1+1/3+1/3^2+...+1/3^5)-(1/3+1/3^2+..+1/3^6)
2A=1-1/3^6
2A=1-1/729
2A=728/729
A=364/729
k nhé