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ai làm cho mình bài này với:
tính:A=7/4.(3333/1212+3333/2020+3333/3030+3333/4242)
mình tích cho 3 luôn
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(A=\frac{7}{4}.\left(\frac{33.101}{12.101}+\frac{33.101}{20.101}+\frac{33.101}{30.101}+\frac{33.101}{42.101}\right)\)
\(A=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(A=\frac{7.11}{4}.\left(\frac{1}{4}+\frac{3}{20}+\frac{1}{10}+\frac{1}{14}\right)\)
\(A=\frac{77}{4}.\left(\frac{35}{140}+\frac{21}{140}+\frac{14}{140}+\frac{10}{140}\right)\)
\(A=\frac{77}{4}.\frac{80}{140}\)\(=\frac{77}{8}.\frac{20}{35}\)
\(A=11\)
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(A=\frac{7}{4}\cdot\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(A=\frac{7}{4}\cdot\left(33\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right)\)
\(A=\frac{7}{4}\cdot\left(33\cdot\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\right)\)
\(A=\frac{7}{4}\cdot\left(33\cdot\left(\frac{1}{3}-\frac{1}{7}\right)\right)\)
\(A=\frac{7}{4}\cdot\left(33\cdot\frac{4}{21}\right)\)
\(A=\frac{7}{4}\cdot\frac{132}{21}=11\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(A=\frac{3333}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{3333}{101}\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{3333}{101}.\frac{4}{21}=\frac{1111.4}{101.7}=\frac{4444}{707}\)
Bài 2
\(A=\frac{2^{10}+1}{2^{10}-1}=\frac{2^{10}-1+2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(B=\frac{2^{10}-1}{2^{10}-3}=\frac{2^{10}-3+4}{2^{10}-3}=1+\frac{4}{2^{10}-3}\)
Ta thấy \(2^{10}-1>2^{10}-3\Rightarrow\frac{2}{2^{10}-1}< \frac{2}{2^{10}-3}< \frac{4}{2^{10}-3}\)
Từ đó \(\Rightarrow1+\frac{2}{2^{10}-1}< 1+\frac{4}{2^{10}-3}\Rightarrow A< B\)
Bài 3\(P=\frac{\left(\frac{2}{3}-\frac{1}{4}\right)+\frac{5}{11}}{\frac{5}{12}+\left(1-\frac{7}{11}\right)}=\frac{\frac{5}{12}+\frac{5}{11}}{\frac{5}{12}+\frac{4}{11}}=\frac{\frac{55+60}{11.12}}{\frac{55+48}{12.11}}=\frac{115}{103}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=7/4.(3333/1212+3333/2020+3333/3030+3333/4242)
A=7/4 X 44/7
A=11
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(=\frac{7}{4}.\left[3333.\left(\frac{1}{1212}+\frac{1}{2020}+\frac{1}{3030}+\frac{1}{4242}\right)\right]\)
\(=\frac{7}{4}.\left[3333.\left(\frac{1}{12.101}+\frac{1}{20.101}+\frac{1}{30.101}+\frac{1}{42.101}\right)\right]\)
\(\frac{7}{4}.\left[3333.\frac{1}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]\)
\(=\frac{7}{4}.33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=33.\left[\frac{7}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\right]\)
\(=33.\left[\frac{7}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)
\(=33.\left(\frac{7}{4}.\frac{4}{7}\right)\)
\(=33.1\)
\(=33\)
Vậy \(A=33\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{7}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)=\frac{7}{4}\left(\frac{33\times101}{12\times101}+\frac{33\times101}{20\times101}+\frac{33\times101}{30\times101}+\frac{33\times101}{42\times101}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=\frac{7}{4}\times\frac{44}{7}=11\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 7/4 * ( 3333/1212 + 3333/2020 + 3333/3030 + 3333/4242) => A = 7/4* (33/12 + 33/20 + 33/30 + 33/42) => A = 7/4* ( 33/3*4 + 33/4*5 + 33/5*6 + 33/6*7) => A = 7/4* { 33/(4-3) * ( 1/4 - 1/5 + 1/5 - 1/6 + 1/6-1/7)} => A = 7/4*33 * ( 1/4 - 1/7) A = 231/4 * 3/28 =693/112.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=7/4.(333/1212+3333/2020+3333/3030+3333/4242)
=7/4.3333.(1/1212+1/2020+1/4242)
=7/4.3333.(1/12.101+1/20.101+1/42.101)
=7/4.3333.1/101.(1/12+1/20+1/42)
=7/4.33.101.1/101.(1/3.4+1/4.5+1/5.6+1/6.7)
=7/4.33.(1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7)
=7/4.33.(1/3-1/7)
=7/4.33.4/21
=7.33.4/4.21
=11
ko hieu tai sao co the dat ra mot cau hoi vo li nhu vay?
deo hieuluon