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chứng minh rằng A= 12/1*4*7+12/4*7*10+12/7*10*13+...+12/54*57*60<1/2
giải giup minh nha minh tich cho
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Đặt: \(\hept{\begin{cases}A=\frac{10^{10}+1}{10^{11}+1}\\B=\frac{10^{11}-1}{10^{12}-1}\end{cases}}\)
Ta có:
\(\hept{\begin{cases}10A=\frac{10^{11}+10}{10^{11}+1}=1+\frac{9}{10^{11}+1}\\10B=\frac{10^{12}-10}{10^{12}-1}=1-\frac{9}{10^{12}-1}\end{cases}}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
\(1)\frac{11}{12}-\frac{5}{12}\left(0,8-\frac{1}{10}\right):\left(-\frac{7}{12}\right)\)
\(=\frac{11}{12}-\frac{5}{12}\times\frac{7}{10}:\left(-\frac{7}{12}\right)\)
\(=\frac{11}{12}-\frac{7}{24}:\left(-\frac{7}{12}\right)\)
\(=\frac{11}{12}-\left(-\frac{1}{2}\right)\)
\(=\frac{17}{12}\)
a, \(B=\dfrac{10^{12}+1}{10^{12}+1}=1\)
+) Xét \(n>12\Rightarrow A>1=B\)
+) Xét \(n< 12\Rightarrow A< B=1\)
Vậy...
b, \(\overline{abc}-\overline{deg}⋮7\)
\(\Rightarrow\left\{{}\begin{matrix}\overline{abc}⋮7\\\overline{deg}⋮7\end{matrix}\right.\)
Ta có: \(\overline{abcdeg}=1000\overline{abc}+\overline{deg}⋮7\) ( do \(\left(1000;7\right)=1\) )
\(\Rightarrowđpcm\)
Bài1:
a)Ta có:
\(-203< 0;\dfrac{1}{2017}>0\)
Nên \(-203< \dfrac{1}{2017}\)
b)\(\dfrac{7}{29}và\dfrac{12}{47}\)
c)Đặt \(A=\dfrac{10^{11}+1}{10^{12}+1}\);\(B=\dfrac{10^{12}+1}{10^{13}+1}\)
Ta có:\(10A=\dfrac{10^{12}+1+9}{10^{12}+1}=1+\dfrac{9}{10^{12}+1}\)
\(10B=\dfrac{10^{13}+1+9}{10^{13}+1}=1+\dfrac{9}{10^{13}+1}\)
Do đó:\(10A>10B\Rightarrow A>B\)
Bài2:
a)\(500>2^x>100\)
Ta có:\(100< 2^7< 2^8< 500\)
\(\Rightarrow x\in\left\{7;8\right\}\)
Vậy...
Câu sau tương tự
a) Ta có: \(-203< 0;\dfrac{1}{2017}>0\)
\(\Rightarrow\dfrac{1}{2017}>-203\)
\(A=\frac{10^{11}-1}{10^{12}-1}\)
\(\Leftrightarrow10A=\frac{10\left(10^{11}-1\right)}{\left(10^{12}-1\right)}=\frac{10^{12}-10}{10^{12}-1}=1-\frac{9}{10^{12}-1}\left(1\right)\)
\(B=\frac{10^{10}+1}{10^{11}+1}\)
\(\Leftrightarrow10B=\frac{10\left(10^{10}+1\right)}{10^{11}+1}=\frac{10^{11}+10}{10^{11}+1}=\frac{9}{10^{11}+1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A< B\)
Nếu có 1 phân số a/b < 1 thì a/b < a+n/b+n.
Tương tự ta có: A < (10^11 -1)+11/(10^12 -1)+10
A < 10^11+10/10^12+10
A < 10(10^10+1)/10(10^11+1)
A < 10(10^10+1)/10(10^11+1)
A < 10^10+1/10^11+1
Vậy A < B
\(x+\frac{1}{10}+x+\frac{1}{10}+x+\frac{1}{12}=x+\frac{1}{13}+x+\frac{1}{14}\)
\(\Rightarrow3x+\left(\frac{1}{10}+\frac{1}{10}+\frac{1}{12}\right)=2x+\left(\frac{1}{13}+\frac{1}{14}\right)\)
\(\Rightarrow3x+\left(\frac{6}{60}+\frac{6}{60}+\frac{5}{60}\right)=2x+\left(\frac{14}{182}+\frac{13}{182}\right)\)
\(\Rightarrow3x+\frac{17}{60}=2x+\frac{27}{182}\)
\(\Rightarrow3x-2x=\frac{27}{182}-\frac{17}{60}\)(dùng quy tắc chuyển vế)
\(\Rightarrow x=\frac{810}{5460}-\frac{1547}{5460}\)
\(\Rightarrow x=\frac{-737}{5460}\)
bạn nhớ thử lại nhé :)
Ta luôn có nếu a>0; b>0 thì \(\frac{a}{b}< \frac{a+m}{b+m}\left(m\in N\right)\)
Áp dụng vào bài toán ta thấy 1011-1 > 0 và 1012-1 > 0 nên
\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}=\frac{10^{11}+10}{10^{12}+10}=\frac{10.\left(10^{10}+1\right)}{10.\left(10^{11}+1\right)}=\frac{10^{10}+1}{10^{11}+1}=B\)
Vậy A < B
Xin lỗi bn nhé bài toán phụ phía trên đang còn 1 đk nữa là a<b
A<B
cách làm thì tự tìm hiểu nha
mak bài này thầy chữa rùi mak
\(-\dfrac{1}{12}-\left(-\dfrac{1}{10}\right)\\ =-\dfrac{1}{12}+\dfrac{1}{10}\\ =\dfrac{-5}{60}+\dfrac{6}{60}\\ =\dfrac{-5+6}{60}\\ =\dfrac{1}{60}\)
-1/12--1/10
=-5/60--6/60
=-11/60