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Ta có \(\frac{1}{m}\)+\(\frac{n}{6}\)=\(\frac{1}{2}\)
\(\frac{1}{m}\)=\(\frac{1}{2}\)-\(\frac{n}{6}\)
\(\frac{1}{m}\)=\(\frac{3}{6}\)-\(\frac{n}{6}\)
\(\frac{1}{m}\)=\(\frac{3-n}{6}\)
=>m*(3-n)=6
=>3-nEƯ(6)
Ta có bảng giá trị
| 3-n | 1 | 2 | 3 | 6 | -1 | -2 | -3 | -6 |
| m | 6 | 3 | 2 | 1 | -6 | -3 | -2 | -1 |
| n | 2 | 1 | 0 | -3 | 4 | 5 | 6 | 9 |
Ta có :
\(B=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot2}+\frac{1}{2\cdot15}+\frac{13}{15\cdot4}\)
\(=>\frac{B}{7}=\frac{5}{2\cdot7}+\frac{4}{7\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
\(=>\frac{B}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(=>\frac{B}{7}=\frac{1}{2}-\frac{1}{28}=\frac{14}{28}-\frac{1}{28}=\frac{13}{28}\)
\(=>B=\frac{13}{28}\cdot7=\frac{13}{4}\)
Câu e) là hỗn số 5 và 8 phần mừi bảy CHIA x đấy nhá!!! 
a... h..a..ha ha < cười ngượng> mk vit hơi xấu các bn thông cảm cho
Mình k mag máy tính cầm tay nên chịu. Nhưng mấy bài này dễ mà : câu c bạn chỉ cần đổi vế theo thứ tự thôi, câu d và e thì áp dụng tính chất kết hợp
\(3^{x-1}=\frac{1}{243}\)
\(\Rightarrow3^x=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
a.
\(28\div2x=7\)
\(2x=\frac{28}{7}\)
\(2x=4\)
\(x=\frac{4}{2}\)
\(x=2\)
b.
\(\left(115+3x\right)-\left(17+x\right)=214\)
\(115+3x-17-x=214\)
\(3x-x=214-115+17\)
\(2x=116\)
\(x=\frac{116}{2}\)
\(x=58\)
Ta có: \(\frac{a}{b}< \frac{a+1}{b+1}\)
\(B=\frac{10^{2013}+1}{10^{2014}+1}< \frac{10^{2013}+1+9}{10^{2014}+1+9}=\frac{10^{2013}+10}{10^{2014}+10}=\frac{10\left(10^{2012}+1\right)}{10\left(10^{2013}+1\right)}=\frac{10^{2012}+1}{2^{2013}+1}=A\)
Vậy: \(A>B\)
Ta có:
\(10A=\frac{10\left(10^{2012}+1\right)}{10^{2013}+1}=\frac{10^{2013}+10}{10^{2013}+1}=\frac{10^{2013}+1+9}{10^{2013}+1}=\frac{10^{2013}+1}{10^{2013}+1}+\frac{9}{10^{2013}+1}=1+\frac{9}{10^{2013}+1}\)
\(10B=\frac{10\left(10^{2013}+1\right)}{10^{2014}+1}=\frac{10^{2014}+10}{10^{2014}+1}=\frac{10^{2014}+1+9}{10^{2014}+1}=\frac{10^{2014}+1}{10^{2014}+1}+\frac{9}{10^{2014}+1}=1+\frac{9}{10^{2014}+1}\)
Vì 102013+1<102014+1
\(\Rightarrow\frac{9}{10^{2013}+1}>\frac{9}{10^{2014}+1}\)
\(\Rightarrow1+\frac{9}{10^{2013}+1}>1+\frac{9}{10^{2014}+1}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
\(\frac{1}{7}\)B=\(\frac{5}{2.7.1}+\frac{4}{1.7.11}+\frac{3}{11.2.7}+\frac{1}{2.7.15}+\frac{13}{15.4.7}\)
\(\frac{1}{7}\)B=\(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{1}{7}B=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(\frac{1}{7}B=\frac{1}{2}-\frac{1}{28}\)
\(\frac{1}{7}B=\frac{13}{28}\)
B=\(\frac{13}{28}:\frac{1}{7}\)
B=\(\frac{13}{4}\)
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
\(\frac{2323}{2424}=\frac{23.101}{24.101}=\frac{23}{24}\)
\(\frac{20132013}{20142014}=\frac{2013.10001}{2014.10001}=\frac{2013}{2014}\)
Ta có:
\(1-\frac{23}{24}=\frac{24}{24}-\frac{23}{24}=\frac{1}{24}\)
\(1-\frac{2013}{2014}=\frac{2014}{2014}-\frac{2013}{2014}=\frac{1}{2014}\)
Vì \(\frac{1}{24}>\frac{1}{2014}\) nên \(\frac{23}{24}< \frac{2013}{2014}\)
Vậy \(\frac{2323}{2424}< \frac{20132013}{20142014}\)
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