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Ta có :
\(A=\dfrac{10^{11}-1}{10^{12}-1}< 1\)
\(\Leftrightarrow A< \dfrac{10^{11}-1+11}{10^{12}-1+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}=\dfrac{10^{10}+1}{10^{11}+1}=B\)
Vậy \(\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{10}+1}{10^{11}+1}\)
Vậy...
\(a)\dfrac{-11}{12}và\dfrac{17}{-18}\) \(\Leftrightarrow\dfrac{-11}{12}và\dfrac{-17}{18}\) \(\Leftrightarrow\dfrac{-33}{36}và\dfrac{-34}{36}\)
Ta thấy rằng : \(-33>-34\Rightarrow\dfrac{-33}{36}>\dfrac{-34}{36}\)
Hay : \(\dfrac{-11}{12}>\dfrac{17}{-18}\)
\(b)\dfrac{-14}{-21}và\dfrac{-60}{-72}\)
Ta có : \(\dfrac{-14}{-21}\text{=}\dfrac{-14:-7}{-21:-7}\text{=}\dfrac{2}{3}\text{=}\dfrac{4}{6}\)
\(\dfrac{-60}{-72}\text{=}\dfrac{-60:-12}{-72:-12}=\dfrac{5}{6}\)
Do đó : \(\dfrac{-14}{-21}< \dfrac{-60}{-72}\)
\(c)\dfrac{2135}{13790}và\dfrac{4}{3}\)
Xét phân số : \(\dfrac{2135}{13790}\) ta thấy rằng : \(tử< mẫu\left(2135< 13790\right)\)
\(\Rightarrow\dfrac{2135}{13790}< 1\)
Xét phân số : \(\dfrac{4}{3}có\) : \(tử>mẫu\left(4>3\right)\)
\(\Rightarrow\dfrac{4}{3}>1\)
Do đó : \(\dfrac{2135}{13790}< \dfrac{4}{3}\)
\(d)\dfrac{2022}{2021}và\dfrac{10}{9}\)
Ta thấy rằng : \(\dfrac{2022}{2021}-\dfrac{1}{2021}\text{=}1\)
\(\dfrac{10}{9}-\dfrac{1}{9}\text{=}1\)
Mà : \(\dfrac{1}{9}>\dfrac{1}{2021}\)
\(\Rightarrow\dfrac{2022}{2021}< \dfrac{10}{9}\)
\(e)\dfrac{35}{36}và\dfrac{16}{17}\)
Ta có : \(\dfrac{35}{36}+\dfrac{1}{36}\text{=}1\)
\(\dfrac{16}{17}+\dfrac{1}{17}\text{=}1\)
Mà : \(\dfrac{1}{36}< \dfrac{1}{17}\)
\(\Rightarrow\dfrac{35}{36}>\dfrac{16}{17}\)
\(f)-1,3< -1,2\)
a) Ta có:
\(-\dfrac{11}{12}=\dfrac{1}{12}-1\)
\(-\dfrac{17}{18}=\dfrac{1}{18}-1\)
Mà: \(\dfrac{1}{12}>\dfrac{1}{18}\)
Hay: \(\dfrac{1}{12}-1>\dfrac{1}{18}-1\Rightarrow-\dfrac{11}{12}>-\dfrac{17}{18}\)
b) Ta có:
\(\dfrac{-14}{-21}=\dfrac{2}{3}=\dfrac{4}{6}\)
\(\dfrac{-60}{-72}=\dfrac{5}{6}\)
Mà: \(5>4\Rightarrow\dfrac{-60}{-72}>\dfrac{-14}{-21}\)
c) Ta có:
\(\dfrac{2135}{13790}=\dfrac{61}{394}< 1\) (tử nhỏ hơn mẫu)
\(\dfrac{4}{3}>1\) (tử lớn hơn mẫu)
Ta có: \(\dfrac{61}{394}< \dfrac{4}{3}\Rightarrow\dfrac{2135}{13790}< \dfrac{4}{3}\)
d) Ta có:
\(\dfrac{2022}{2021}=\dfrac{1}{2021}+1\)
\(\dfrac{10}{9}=\dfrac{1}{9}+1\)
Ta thấy: \(\dfrac{1}{2021}< \dfrac{1}{9}\Rightarrow\dfrac{1}{2021}+1< \dfrac{1}{9}+1\)
Hay \(\dfrac{2022}{2021}< \dfrac{10}{9}\)
e) Ta có:
\(\dfrac{35}{36}=1-\dfrac{1}{36}\)
\(\dfrac{16}{17}=1-\dfrac{1}{17}\)
Ta có: \(\dfrac{1}{36}< \dfrac{1}{17}\Rightarrow1-\dfrac{1}{36}>1-\dfrac{1}{17}\)
Hay \(\dfrac{35}{36}>\dfrac{16}{17}\)
f) Ta có: \(1,3>1,2\)
\(\Rightarrow-1,3< -1,2\)
\(B=\dfrac{1}{11}+\dfrac{1}{11^2}+\dfrac{1}{11^3}+...+\dfrac{1}{11^{99}}+\dfrac{1}{11^{100}}\\ 11B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{11^{98}}+\dfrac{1}{11^{99}}\\ 11B-B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{1^{99}0}-\dfrac{1}{11}-\dfrac{1}{11^2}-\dfrac{1}{11^3}-...-\dfrac{1}{11^{100}}\\ 10B=1-\dfrac{1}{11^{99}}\\ B=\dfrac{1-\dfrac{1}{11^{99}}}{10}\)
có : `1-1/(11^99)<1`
\(\Rightarrow\dfrac{1-\dfrac{1}{11^{99}}}{10}< \dfrac{1}{10}\)
hay `B<1/10`
Câu 1:
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\left(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}\right)\) - \(\left(\dfrac{x+1}{13}+\dfrac{x+1}{14}\right)=0\)
\(\Rightarrow\left(x+1\right).\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)\)= 0
Vì \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)
\(\Rightarrow x+1=0\)
=> x = 0 - 1
=> x = -1
Câu 2:
Ta có: \(A=\dfrac{3n+9}{n-4}=\dfrac{3n-3.4+9+12}{n-4}\)
\(=\dfrac{3.\left(n-4\right)+21}{n-4}=3+\dfrac{21}{n-4}\)
Để A có giá trị nguyên thì:
n - 4 \(\in\) Ư(21)
=> n - 4 \(\in\)
| n4 | 3 | -3 | 7 | -7 | -1 | 1 | -21 | 21 |
| n | 7 | 1 | 11 | -3 | 3 | 5 | -17 | 25 |
a) \(\dfrac{12}{47}\) và \(\dfrac{11}{53}\)
Ta có: \(\dfrac{11}{47}>\dfrac{11}{53}\) mà \(\dfrac{12}{47}>\dfrac{11}{47}\)
\(\Rightarrow\dfrac{12}{47}>\dfrac{11}{53}\)
a) Ta có :\(\dfrac{12}{47}>\dfrac{12}{48}=\dfrac{1}{4}=\dfrac{11}{44}>\dfrac{11}{53}\)
\(\Rightarrow\dfrac{12}{47}>\dfrac{11}{53}\)
b) Ta có : \(\dfrac{456}{461}=1-\dfrac{5}{461}\)
\(\dfrac{123}{128}=1-\dfrac{5}{128}\)
Vì \(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow1-\dfrac{5}{461}>1-\dfrac{5}{128}\)
\(\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)
c) Ta có :\(\dfrac{12}{47}>\dfrac{12}{48}=\dfrac{1}{4}=\dfrac{19}{76}>\dfrac{19}{77}\)
=> \(\dfrac{12}{47}>\dfrac{19}{77}\)
d) Ta có : \(13A=13.\dfrac{13^{15}+1}{13^{16}+1}=\dfrac{13^{16}+13}{13^{16}+1}=\dfrac{13^{16}+1+12}{13^{16}+1}=1+\dfrac{12}{13^{16}+1}\)
\(13B=13.\dfrac{13^{16}+1}{13^{17}+1}=\dfrac{13^{17}+13}{13^{17}+1}=\dfrac{13^{17}+1+12}{13^{17}+1}=1+\dfrac{12}{13^{17}+1}\)
Ta thấy : \(\dfrac{12}{13^{16}+1}>\dfrac{12}{13^{17}+1}\Rightarrow1+\dfrac{12}{13^{16}+1}>1+\dfrac{12}{13^{17}+1}\Rightarrow\dfrac{13^{15}+1}{13^{16}+1}>\dfrac{13^{16}+1}{13^{17}+1}\)
`a)` Xét tử số phân số M :
\(2012-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{2012}{2020}\\ =\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{2012}{2020}\right)\\ =\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{2020}\\ =24\left(\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}\right)\)
Ta được : \(M=\dfrac{24\left(\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}\right)}{\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}}=24\)
`b)` Xét tử số phân số N :
\(\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+...+\dfrac{1}{101.400}\\ =\dfrac{1}{299}.\left(\dfrac{299}{1.300}+\dfrac{299}{2.301}+\dfrac{299}{3.302}+...+\dfrac{299}{101.400}\right)\\ =\dfrac{1}{299}.\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)
Xét mẫu số phân số N :
\(\dfrac{1}{1.102}+\dfrac{1}{2.103}+\dfrac{1}{3.104}+...+\dfrac{1}{299.400}\\ =\dfrac{1}{101}.\left(\dfrac{101}{1.102}+\dfrac{101}{2.103}+\dfrac{101}{3.104}+...+\dfrac{101}{299.400}\right)\\ =\dfrac{1}{101}.\left(1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+\dfrac{1}{3}-\dfrac{1}{104}+...+\dfrac{1}{299}-\dfrac{1}{400}\right)\)
\(=\dfrac{1}{101}.\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)
Ta được: \(N=\dfrac{\dfrac{1}{299}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)}{\dfrac{1}{101}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)}\\ =\dfrac{\dfrac{1}{299}}{\dfrac{1}{101}}=\dfrac{101}{299}\)
tỉ số của a / b là (92 - 1/9 - 2/ 10 - 3/11 - ... - 92/100) trên 1/45 + 1/50 + ... + 1/500 :)) hay ngắn tắc hơn là A/B cho nhanh :)))))))))))))))
\(A=\left(1+1+...+1\right)-\left(\dfrac{1}{9}+\dfrac{2}{10}+...+\dfrac{92}{100}\right)\)𝓒𝓸́ 92 𝓼𝓸̂́ 1
\(A=\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\)
\(A=\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\)
\(A=8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)\)
\(B=\dfrac{1}{45}+\dfrac{1}{50}+...+\dfrac{1}{500}\)
\(B=\dfrac{1}{5}.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)}{\dfrac{1}{5}.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)}\\ \Rightarrow\dfrac{A}{B}=\dfrac{8}{\dfrac{1}{5}}=40\)
𝓥𝓪̣̂𝔂 𝓽𝓲̉ 𝓼𝓸̂́ 𝓬𝓾̉𝓪 𝓐 𝓿𝓪̀ 𝓑 𝓵𝓪̀ 40
a. Xét phân số trung gian là \(\dfrac{72}{78}\) , ta thấy:
\(\dfrac{72}{73}>\dfrac{72}{78}\)
\(\dfrac{58}{78}< \dfrac{72}{78}\)
\(\Rightarrow\dfrac{72}{73}>\dfrac{58}{78}\)
b. Xét phân số trung gian là \(\dfrac{n}{n+2}\) , ta thấy:
\(\dfrac{n}{n+3}< \dfrac{n}{n+2}\)
\(\dfrac{n}{n+2}< \dfrac{n+1}{n+2}\)
\(\Rightarrow\dfrac{n}{n+3}< \dfrac{n+1}{n+2}\)
c. Ta có: \(\dfrac{10^{11}-1}{10^{12}-1}< 1\) (vì tử < mẫu)
\(\Rightarrow\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{\left(10^{11}-1\right)+11}{\left(10^{12}-1\right)+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10^{10}+1}{10^{11}+1}\)
Vậy \(\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{10}+1}{10^{11}+1}\)
d. Xét phân số trung gian là \(\dfrac{1}{4}\) , ta thấy:
\(\dfrac{12}{47}>\dfrac{12}{48}=\dfrac{1}{4}\)
\(\dfrac{19}{77}< \dfrac{19}{76}=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{12}{47}>\dfrac{19}{77}\)