tính:
a,A= (1/2-1)*(1/3-1)*...*(1/10-1)
b,B= (1/4-1)*(1/9-1)*....*(1/100-1)
c,B= (1-1/2)*(1-1/3)*........*(1-1/n+1) với n thuộc N
các bn ơi giúp mk vs
mk cần gấp !!!!!!!
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Ta có :\(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\)(1)
\(\Rightarrow3N=1+\frac{1}{3}+...+\frac{1}{3^8}\)(2)
Lấy (2) - (1) ta có :
\(\Rightarrow2N=1-\frac{1}{3^9}\)
\(\Rightarrow N=\frac{1-\frac{1}{3^9}}{2}\)
ta có: \(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\)
\(\Rightarrow\frac{1}{3}N=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{10}}\)
\(\Rightarrow\frac{1}{3}N-N=\frac{1}{3^{10}}-\frac{1}{3}\)
\(\frac{2}{3}N=\frac{1}{3^{10}}-\frac{1}{3}\)
\(N=\frac{\frac{1}{3^{10}}-\frac{1}{3}}{\frac{2}{3}}\)
CHÚC BN HỌC TỐT!!!
\(B=1+5+5^2+5^3+...+5^{2008}+5^{2009}\)
\(\Rightarrow 5B=5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\)
Trừ theo vế:
\(5B-B=(5+5^2+5^3+5^4+...+5^{2009}+5^{2010})-(1+5+5^2+...+5^{2009})\)
\(4B=5^{2010}-1\)
\(B=\frac{5^{2010}-1}{4}\)
\(S=\frac{3^0+1}{2}+\frac{3^1+1}{2}+\frac{3^2+1}{2}+..+\frac{3^{n-1}+1}{2}\)
\(=\frac{3^0+3^1+3^2+...+3^{n-1}}{2}+\frac{\underbrace{1+1+...+1}_{n}}{2}\)
\(=\frac{3^0+3^1+3^2+..+3^{n-1}}{2}+\frac{n}{2}\)
Đặt \(X=3^0+3^1+3^2+..+3^{n-1}\)
\(\Rightarrow 3X=3^1+3^2+3^3+...+3^{n}\)
Trừ theo vế:
\(3X-X=3^n-3^0=3^n-1\)
\(\Rightarrow X=\frac{3^n-1}{2}\). Do đó \(S=\frac{3^n-1}{4}+\frac{n}{2}\)
\(a,\frac{62}{7}:x=\frac{29}{9}:\frac{3}{56}\)
\(\frac{62}{7}:x=\frac{1624}{27}\)
\(x=\frac{62}{7}:\frac{1624}{27}=\frac{837}{5684}\)
\(b,\frac{1}{5}:x=\frac{1}{5}-\frac{1}{7}\)
\(\frac{1}{5}:x=\frac{2}{35}\)
\(x=\frac{1}{5}:\frac{2}{35}=\frac{7}{2}\)
\(c,\frac{2}{3}.x-\frac{4}{7}=\frac{1}{7}\)
\(\frac{2}{3}.x=\frac{1}{7}+\frac{4}{7}=\frac{5}{7}\)
\(x=\frac{5}{7}:\frac{2}{3}=\frac{15}{14}\)
\(d,\frac{2}{7}-\frac{8}{9}.x=\frac{2}{3}\)
\(\frac{8}{9}.x=\frac{2}{7}-\frac{2}{3}=-\frac{8}{21}\)
\(x=-\frac{8}{21}:\frac{8}{9}=-\frac{3}{7}\)
\(e,\frac{4}{7}+\frac{5}{9}:x=\frac{1}{5}\)
\(\frac{5}{9}:x=\frac{1}{5}-\frac{4}{7}=-\frac{13}{35}\)
\(x=\frac{5}{9}:-\frac{13}{35}=\frac{175}{117}\)
\(i,\frac{2}{5}-\frac{2}{5}.x=\frac{2}{5}\)
\(\frac{2}{5}.\left(1-x\right)=\frac{2}{5}\)
\(1-x=\frac{2}{5}:\frac{2}{5}=1\)
\(x=1-1=0\)
\(g,\frac{2}{3}+\frac{1}{3}:x=-1\)
\(\frac{1}{3}:x=-1-\frac{2}{3}=-\frac{5}{3}\)
\(x=\frac{1}{3}:-\frac{5}{3}=-\frac{1}{5}\)
học tốt nha
a) A = 1002 - 992 + 982 - 972 + ... + 22 - 12
A = (1002 - 992) + (982 - 972) + ... + (22 - 12)
A = (100 - 99)(100 + 99) + (98 - 97)(98 + 97) + ... + (2 - 1)(2 + 1)
A = 1. 199 + 1. 195 + ... + 1.3
A = 199 + 195 + ... + 3
A = (199 + 3)[(199 - 3) : 4 + 1] : 2
A = 202 . 50 : 2
A = 5050
b) B = (202 + 182 + 162 + ... + 22) - (192 + 172 + 152 + ... + 12)
B = 202 + 182 + 162 + ... + 22 - 192 - 173 - 152 - ... - 12)
B = (202 - 192) + (182 - 172) + (162 - 152) + ... + (22 - 12)
B = (20 - 19)(20 + 19) + (18 - 17)(18 + 17) + ... + (2 - 1)(2 + 1)
B = 1. 39 + 1.35 + ... + 1.3
B = 39 + 35 + ... + 3
B = (39 + 3)[(39 - 3) : 4 + 1] : 2
B = 42 . 10 : 2
B = 210
#)Giải :
a)\(A=100^2-99^2+98^2-97^2+...+2^2-1\)
\(A=\left(100-99\right)+\left(98-97\right)+...+\left(2-1\right)\)
\(A=100+99+98+...+2+1\)
\(A=\frac{\left(1+100\right)100}{2}=5050\)
b)\(B=\left(20^2+18^2+16^2+...+2^2\right)-\left(19^2+17^2+15^2+...+1^2\right)\)
\(B=20^2-19^2+18^2-17^2+...+2^2-1\)
Giờ trở thành dạng của ý a) rùi nhé, tương tự mak làm theo
c)\(C=\left(-1\right)^n.\left(-1\right)^{2n+1}.\left(-1\right)^{n+1}\)
\(C=\left(-1\right)^n.\left(-1\right)^2.\left(-1\right)^n.\left(-1\right).\left(-1\right)^n.\left(-1\right)\)
\(C=\left[\left(-1\right)^n.\left(-1\right)^n.\left(-1\right)^n\right].1.\left(-1\right).\left(-1\right)\)
\(C=\left(-1\right)^n.1.1\)
\(C=\left(-1\right)^n\)
a) Ta có: \(A=\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{10}-1\right)=\frac{-1}{2}.\frac{-2}{3}...\frac{-9}{10}=\frac{-\left(1.2.3...9\right)}{2.3.4...10}=-\frac{1}{10}\)
b) Ta có : \(B=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)....\left(\frac{1}{100}-1\right)=\frac{-3}{4}.\frac{-8}{9}....\frac{-99}{100}=-\frac{3.8....99}{\left(2.3...10\right)\left(2.3...10\right)}\)
\(=-\frac{1.3.2.4...9.11}{\left(2.3....10\right)\left(2.3...10\right)}=\frac{\left(1.2.3...10\right).\left(3.4..10.11\right)}{\left(2.3...10\right).\left(2.3.4...10\right)}=\frac{11}{2}=5,5\)
c) Ta có : \(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{n+1}\right)=\frac{1}{2}.\frac{2}{3}...\frac{n}{n+1}=\frac{1.2...n}{2.3...\left(n+1\right)}=\frac{1}{n+1}\)