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\(C=\frac{2}{4.7}-\frac{3}{5.9}+\frac{2}{7.10}-\frac{3}{9.13}+...+\frac{2}{301.304}-\frac{3}{401.405}\)
\(C=\left(\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{301.304}\right)-\left(\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{401.405}\right)\)
\(C=\frac{2}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{301.304}\right)-\frac{3}{4}\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{401.405}\right)\)
\(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+..+\frac{1}{401}-\frac{1}{405}\right)\) \(C=\frac{2}{3}\left(\frac{1}{4}-\frac{1}{304}\right)-\frac{3}{4}\left(\frac{1}{5}-\frac{1}{405}\right)\)
\(C=\frac{25}{152}-\frac{4}{27}\)
\(C=\frac{67}{4104}\)
Bài 1:
\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)
\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)
Bài 2:
\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)
<=> \(\frac{7}{8}-x=\frac{27}{40}\)
<=> \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)
Vậy...
\(A=\frac{3n+9}{n-4}=\frac{3n-12+21}{n-4}=\frac{3\left(n-4\right)+21}{n-4}=3+\frac{21}{n-4}\)
\(\Rightarrow n-4\inƯ\left(21\right)\Rightarrow n-4\in\left\{-21;-7;-3;-1;1;3;7;21\right\}\)
\(\Rightarrow n\in\left\{-17;3;1;3;5;7;11;25\right\}\)
( giá trị là chỗ n-4 \(\in\){ -21;-7;...;21 } rồi + 3 nha bạn )
\(B=\frac{6n+5}{2n-1}=\frac{6n-3+8}{2n-1}=\frac{3\left(2n-1\right)+8}{2n-1}=3+\frac{8}{2n-1}\)
\(\Rightarrow2n-1\inƯ\left(8\right)\Rightarrow2n-1\in\left\{-1;1\right\}\)( vì 2n - 1 là số lẻ )
\(\Rightarrow n\in\left\{0;1\right\}\)
( giá trị là chỗ 2n-1 \(\in\){ -1;1 } rồi + 3 nha bạn )
- \(A=\frac{3n+9}{n-4}=\frac{3n-12+21}{n-4}=\frac{3\left(n-4\right)+21}{n-4}=\frac{3\left(n-4\right)}{n-4}+\frac{21}{n-4}=3+\frac{21}{n-4}\)
Để A nguyên thì \(\frac{21}{n-4}\) nguyên
=>21 chia hết cho n-4
=>n-4\(\in\)Ư(21)
=>n-4\(\in\left\{-21;-7;-3;-1;1;3;7;21\right\}\)
=>n\(\in\left\{-17;-3;1;3;5;7;11;25\right\}\)(1)
- \(B=\frac{6n+5}{2n-1}=\frac{6n-3+8}{2n-1}=\frac{3\left(2n-1\right)+8}{2n-1}=\frac{3\left(2n-1\right)}{2n-1}+\frac{8}{2n-1}=3+\frac{8}{2n-1}\)
Để B nguyên thì \(\frac{8}{2n-1}\) nguyên
=>8 chia hết cho 2n-1
=>2n-1\(\in\)Ư(8)
=>2n-1\(\in\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
=>2n\(\in\left\{-7;-3;-1;0;2;3;5;9\right\}\)
=>n\(\in\left\{\frac{-7}{2};\frac{-3}{2};\frac{-1}{2};0;1;\frac{3}{2};\frac{5}{2};\frac{9}{2}\right\}\)
Vì n là số nguyên nên n\(\in\left\{0;1\right\}\)(2)
Từ (1) và (2) => n=1 thì A và B nguyên
n=1 => \(A=3+\frac{21}{n-4}=3+\frac{21}{1-4}=3+\frac{21}{-3}=3+\left(-7\right)=-4\)
\(B=3+\frac{8}{2n-1}=3+\frac{8}{2.1-1}=3+\frac{8}{1}=3+8=11\)
Kết luận:n=1 thì A=-4 và B=11
1/h=1/2(1/a+1/b)=1/2a+1/2b=(a+b)/2ab
=>(a+b/)2ab-1/h=0
quy dong len ta co
(a+b)h/2abh-2ab/2abh=0=> (ah+bh-2ab)/2abh=0 =>ah+bh-2ab=0
=>ah+bh-ab-ab=0
=>a(h-b)-b(a-h)=0
=>a(h-b)=b(a-h)
=>a/b=(a-h)(h-b)
a) \(A=4+4^2+4^3+...+4^{200}\)
\(4A=4^2+4^3+...+4^{201}\)
\(4A-A=3A=4^{201}-4\)
\(A=\frac{4^{201}-4}{3}\)
b) \(B=1+5+5^2+...+5^{2017}\)
\(5B=5+5^2+5^3+...+5^{2018}\)
\(5B-B=4B=5^{2018}-1\)
\(B=\frac{5^{2018}-1}{4}\)
c) \(C=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{500}}\)
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{499}}\)
\(3C-C=2C=1-\frac{1}{3^{500}}=\frac{3^{500}-1}{3^{500}}\)
\(C=\frac{\left(\frac{3^{500}-1}{3^{500}}\right)}{2}\)
T_i_c_k cho mình nha,có j ko hiểu cứ hỏi mình nhé ^^
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
Ta có :
\(C=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n-1}{n!}\)
\(=\left(\frac{2}{2!}+\frac{3}{3!}+\frac{4}{4!}+...+\frac{n}{n!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{n!}\right)\)
\(=\left(1+\frac{1}{2!}+\frac{1}{3!}+....+\frac{1}{\left(n-1\right)!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+....+\frac{1}{n!}\right)\)
\(=1+\frac{1}{2!}+\frac{1}{3!}+....+\frac{1}{\left(n-1\right)!}-\frac{1}{2!}-\frac{1}{3!}-\frac{1}{4!}-....-\frac{1}{n!}\)
\(=1-\frac{1}{n!}=\frac{n!-1}{n!}\)