\(3\frac{7}{10}+9\frac{6}{28}=?\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{A}{2}=\frac{3^3}{2}-\frac{5^3}{6}+\frac{7^3}{12}-\frac{9^3}{20}+\frac{11^3}{30}-\frac{13^3}{42}+\frac{15^3}{56}-\frac{17^3}{72}+...+\frac{199^3}{9900}\)
\(=3^2.\left(1+\frac{1}{2}\right)-5^2.\left(\frac{1}{2}+\frac{1}{3}\right)+7^2.\left(\frac{1}{3}+\frac{1}{4}\right)-9^2.\left(\frac{1}{4}+\frac{1}{5}\right)+...+199^2.\left(\frac{1}{99}+\frac{1}{100}\right)\)
\(=3^2+\left(\frac{3^2}{2}-\frac{5^2}{2}\right)-\left(\frac{5^2}{3}-\frac{7^2}{3}\right)+\left(\frac{7^2}{4}-\frac{9^2}{4}\right)-\left(\frac{9^2}{5}-\frac{11^2}{5}\right)+...+\left(\frac{197^2}{99}-\frac{199^2}{99}\right)+\frac{199^2}{100}\)
\(=3^2-8+8-8+...+8+\frac{199^2}{100}=3^2+\frac{199^2}{100}< 3^2+\frac{199.200}{100}=9+398=407\)
\(\Rightarrow A< 407.2=814\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ \(\frac{9.5^{20}.27^9-3.9^{15}.25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)
\(=\frac{5^{20}.3^{29}-3^{31}.5^{18}}{7.3^{29}.5^{18}-3^{29}.5^{19}}=\frac{3^{29}.5^{18}.\left(25-9\right)}{3^{29}.5^{18}.\left(7-5\right)}=\frac{16}{2}=8\)
CÁC BÀI CÒN LẠI TƯƠNG TỰ HẾT NHÉ E
![](https://rs.olm.vn/images/avt/0.png?1311)
Sử dụng khá nhiều kiến thức hằng đẳng thức lớp 8, lớp 7 bó tay
\(\frac{A}{2}=\frac{3^3}{2}-\frac{5^3}{6}+\frac{7^3}{12}-\frac{9^3}{20}+...-\frac{197^3}{9702}+\frac{199^3}{9900}\)
\(\frac{A}{2}=\frac{3^3}{1.2}-\frac{5^3}{2.3}+\frac{7^3}{3.4}-\frac{9^3}{4.5}+...+\frac{199^3}{99.100}\)
\(\frac{A}{2}=3^3\left(1-\frac{1}{2}\right)-5^3\left(\frac{1}{2}-\frac{1}{3}\right)+7^3\left(\frac{1}{3}-\frac{1}{4}\right)-...+199^3\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{A}{2}=3^3-\frac{3^3+5^3}{2}+\frac{5^3+7^3}{3}-\frac{7^3+9^3}{4}+...+\frac{197^3+199^3}{99}-\frac{199^3}{100}\)
\(\frac{A}{2}=3^3-\frac{199^3}{100}-\left(16.2^2+12\right)+\left(16.3^2+12\right)-\left(16.4^2+12\right)+...+\left(16.99^2+12\right)\)
\(\frac{A}{2}=3^3-\frac{199^3}{100}+16\left(3^2-2^2+5^2-4^2+7^2-6^2+...+99^2-98^2\right)\)
\(\frac{A}{2}=3^3-\frac{199^3}{100}+16\left(2+3+4+5+...+98+99\right)\)
\(\frac{A}{2}=3^3-\frac{199^3}{100}+16\left(99.50-1\right)\)
\(\Rightarrow A=16.99.100-\frac{199^3}{50}+22\) (đến đây bấm máy ra kết quả so sánh cũng được)
\(\Rightarrow A=\frac{2^3.100^2\left(100-1\right)-199^3}{50}+22\)
\(A=\frac{200^3-199^3-2.200^2}{50}+22\)
\(A=\frac{200^2+200.199+199^2-2.200^2}{50}+22\)
\(A=\frac{199^2-200^2+200.199}{50}+22\)
\(A=\frac{-199-200+200.199}{50}+22=\frac{199^2}{50}+18\)
\(A< \frac{199.200}{50}+18=814\)
Vậy \(A< 814\)
![](https://rs.olm.vn/images/avt/0.png?1311)