Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(S=\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{30}}\)
\(\Rightarrow4S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{29}}\)
\(\Rightarrow3S=4S-S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{29}}-\dfrac{1}{4}-\dfrac{1}{4^2}-...-\dfrac{1}{4^{30}}=1-\dfrac{1}{4^{30}}\)
\(\Rightarrow S=\dfrac{1-\dfrac{1}{4^{30}}}{3}\)
a
= { 1*( 1+1/2+1/3+1/4) } / { 1 * ( 1-1/2 +1/3-1/4)} : { 3*(1+1/2+1/3+1/4)} / { 2*( 1-1/2 +1/3-1/4)}
Sau đó bn tự tính ra nhé cứ tính nhu bình thường sẽ ra.
Mà mình thấy máy câu này yêu cầu tính chứ có bảo tính theo cách hợp lí đâu? Vì thế bn cứ lấy máy tính tính như bình thường là được .
Giải:
a)A=1/56+1/72+1/90+1/110+1/132+1/156
A=1/7.8+1/8.9+1/9.10+1/10.11+1/11.12+1/12.13
A=1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12+1/12-1/13
A=1/7-1/13
A=6/91
b)B=4/21+4/77+4/165+4/285+4/437+4/621
B=4/3.7+4/7.11+4/11.15+4/15.19+4/19.23+4/23.27
B=1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23-1/27
B=1/3-1/27
B=8/27
c) C=1/21+1/77+1/165+1/285+1/437+1/621
C=1/3.7+1/7.11+1/11.15+1/15.19+1/19.23+1/23.27
C=1/4.(4/3.7+4/7.11+4/11.15+4/15.19+4/19.23+4/23.27)
C=1/4.(1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23-1/27)
C=1/4.(1/3-1/27)
C=1/4.8/27
C=2/27
d) D=1/1.6+1/6.11+1/11.16+1/16.21+1/21.26+1/26.31
D=1/5.(5/1.6+5/6.11+5/11.16+5/16.21+5/21.26+5/26.31)
D=1/5.(1/1-1/6+1/6-1/11+1/11-1/16+1/16-1/21+1/21-1/26+1/26-1/31)
D=1/5.(1/1-1/31)
D=1/5.30/31
D=6/31
Nếu câu d cậu viết thiếu thì làm như vầy nhé!
Chúc bạn học tốt!
Nếu như câu d ko chép sai thì làm thế này nha:
d) D=1/1.6+1/6.11+1/11.16+1/16.21+1/26.31
D=1/5.(5/1.6+5/6.11+5/11.16+5/16.21)+1/806
D=1/5.(1/1-1/6+1/6-1/11+1/11-1/16+1/16-1/21)+1/806
D=1/5.(1/1-1/21)+1/806
D=1/5.20/21+1/806
D=4/21+1/806
D=3245/16926
Chúc bạn học tốt!
Ta có một số phân tích sau : \(a^4\)\(+\)\(4\)\(=\)\(\left(a^2-2a+2\right)\)\(\left(a^2+2a+2\right)\)
Nhân mỗi biểu thức trong ngoặc ở cả tử thức với \(16\)\(=\)\(2^4\), ta được :
\(A\)\(=\)\(\frac{\left(1+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(29^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(30^4+\frac{1}{4}\right)}\)
\(A\)\(=\)\(\frac{\left(2^4+4\right)\left(6^4+4\right)\left(10^4+4\right)...\left(58^4+4\right)}{\left(4^4+4\right)\left(8^4+4\right)\left(12^4+4\right)...\left(60^4+4\right)}\)
Kết hợp với phân tích nêu trên, khi đó :
\(A\)\(=\)\(\frac{\left(2^2-2.2+2\right)\left(2^2+2.2+2\right)\left(6^2-2.6+2\right)\left(6^2+2.6+2\right)....\left(58^2-2.58+2\right)\left(58^2+2.58+2\right)}{\left(4^2-2.4+2\right)\left(4^2+2.4+2\right)\left(8^2-2.8+2\right)\left(8^2+2.8+2\right)....\left(60^2-2.60+2\right)\left(60^2+2.60+2\right)}\)
\(\Rightarrow\)\(A\)\(=\)\(\frac{2.10.26.50.82.122....3250.3482}{10.26.50.82.122....3482.3722}\)\(=\)\(\frac{2}{3722}\)\(=\)\(\frac{1}{1861}\)
Đặt :
\(PHUC=\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)\left(5^4+\dfrac{1}{4}\right)..........\left(11^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right).........\left(12^4+\dfrac{1}{4}\right)}\)
\(\Leftrightarrow PHUC=\dfrac{\left(1^2+1+\dfrac{1}{2}\right)\left(1^2-1+\dfrac{1}{2}\right)......\left(11^2-11+\dfrac{1}{2}\right)}{\left(2^2+2+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right)........\left(12^2-12+\dfrac{1}{2}\right)}\)
\(\Leftrightarrow PHUC=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right).........\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(1.2+\dfrac{1}{2}\right).........\left(12.13+\dfrac{1}{2}\right)}\)
\(\Leftrightarrow PHUC=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(\Leftrightarrow PHUC=\dfrac{1}{313}\)
\(A=\dfrac{\left(1+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(51^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right).....\left(52^4+\dfrac{1}{4}\right)}\)
\(=\dfrac{\left(1+1+\dfrac{1}{2}\right)\left(1-1+\dfrac{1}{2}\right)......\left(11^2-11+\dfrac{1}{2}\right)}{\left(2+2^2+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right)........\left(12^2-12+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right).......\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right).......\left(12.13+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(=\dfrac{1}{313}\)
a, Ta có: \(\dfrac{1}{n}.\dfrac{1}{n+4}=\dfrac{1}{n.\left(n+4\right)}=\dfrac{1}{4}.\dfrac{4}{n.\left(n+1\right)}=\dfrac{1}{4}.\left(\dfrac{1}{n}-\dfrac{1}{n+4}\right)\)
Vậy \(\dfrac{1}{n}.\dfrac{1}{n+1}=\dfrac{1}{4}.\left(\dfrac{1}{n}-\dfrac{1}{n+4}\right)\)
b, \(A=\dfrac{4}{3}.\dfrac{4}{7}+\dfrac{4}{7}.\dfrac{4}{11}+...+\dfrac{4}{95}.\dfrac{4}{99}=4.\left(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{95.99}\right)\)
\(=4.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{99}\right)\)
\(=4.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)=4.\dfrac{32}{99}=\dfrac{128}{99}\)
Vậy \(A=\dfrac{128}{99}\)
1/4 , 3/2 = 6/4 , 1 , 1/2 = 2/4
nên 1/4 , 2/4 , 1, 6/4
\(\dfrac{1}{4};\dfrac{6}{4};\dfrac{4}{4};\dfrac{2}{4}\)
\(=>\dfrac{1}{4}< \dfrac{2}{4}< \dfrac{4}{4}< \dfrac{6}{4}\)