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Cách tui lẹ hơn cách bạn Nguyễn Duy Khánh nè!
Ta có: \(C=\frac{7^{28}+7^{24}+....+7^4+7^0}{\left(7^{30}+7^{26}+...+7^6+7^2\right)+\left(7^{28}+7^{24}+...+7^4+7^0\right)}\)
\(=\frac{7^{28}+7^{24}+...+7^4+7^0}{7^2\left(7^{28}+7^{24}+...+7^4+7^0\right)+\left(7^{28}+7^{24}+...+7^4+7^0\right)}\)
\(=\frac{7^{28}+7^{24}+...+7^4+7^0}{\left(7^{28}+7^{24}+...+7^4+7^0\right)\left(7^2+1\right)}=\frac{1}{7^2+1}=\frac{1}{50}\)
P/s: Easy đúng không?
\(C=\frac{7^{28}+7^{2\text{4}}+...+7^{\text{4}}+7^0}{7^{30}+7^{28}+...+7^2+7^0}\)
Đặt A là tử số ,B là mẫu số.Ta có:
\(7^{\text{4}}A=7^{32}+7^{28}+...+7^8+7^{\text{4}}+7^0\)
\(20\text{4}1A-A=\left(7^{32}+7^{28}+7^{2\text{4}}+...+7^8+7^{\text{4}}\right)-\left(7^{28}+7^{2\text{4}}+...+7^{\text{4}}+7^0\right)\)
\(2\text{4}00A=7^{32}-7^0=7^{32}-1\)
\(\Rightarrow A=\left(7^{32}-1\right):2\text{4}00\)
\(7^2B=\left(7^{32}+7^{30}+7^{28}+...+7^{\text{4}}+7^2\right)\)
49B-B= ....tự..điền......như A nhé.....
48B=732-1 =>B=[7232-1]:48
=>\(C=\frac{A}{B}=\frac{\left(7^{32}-1\right):2\text{4}00}{\left(7^{32}-1\right):\text{4}8}\)
Tui nghĩ vậy đc r á
p/s:ko chắc
.
Tách phần lử trên ra sao cho có thể rút gọn với phần ơn dưới
\(a,\left(10\frac{2}{9}.2\frac{3}{5}\right)-6\frac{2}{9}=\frac{1196}{45}-\frac{56}{9}=\frac{1196}{45}-\frac{280}{45}=\frac{916}{45}\)
\(b,\frac{6}{7}+\frac{1}{7}.\frac{2}{7}+\frac{1}{7}.\frac{5}{7}=\frac{1}{7}\left(6+\frac{2}{7}+\frac{5}{7}\right)=\frac{1}{7}.7=1\)
\(c,3.136.8+4.14.6-14.150=3264+336-2100=1500\)
\(d,\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
\(e,\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)
a) \(\frac{4.7}{9.32}\)=\(\frac{28}{288}\)=\(\frac{7}{72}\)
b)\(\frac{3.21}{14.15}\)=\(\frac{63}{210}\)=\(\frac{3}{10}\)
c)\(\frac{2.5.13}{26.35}\)=\(\frac{130}{910}\)=\(\frac{1}{7}\)
d)\(\frac{9.6-9.3}{18}\)=\(\frac{27}{18}\)=\(\frac{3}{2}\)
e)\(\frac{17.5-17}{3-20}\)=\(\frac{68}{-17}\)=\(-4\)
f)\(\frac{49+7.49}{49}\)=\(\frac{392}{49}\)=\(8\)
H=\(\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot3\cdot6+2\cdot6\cdot12+3\cdot9\cdot18+5\cdot15\cdot30}=\frac{1.2.3+2^3.\left(1.2.3\right)+3^3.\left(1.2.3\right)+5^3.\left(1.2.3\right)}{1.3.6+2^3.\left(1.3.6\right)+3^3.\left(1.3.6\right)+5^3.\left(1.3.6\right)}=\frac{1.2.3.\left(1+2^3+3^3+5^3\right)}{1.3.6.\left(1+2^3+3^3+5^3\right)}=\frac{2}{6}=\frac{1}{3}\)
A=\(\frac{1.2.3.4...8.9}{2.3.4.5...9.10}\)
A=\(\frac{1}{10}\)
mình làm đc 1 câu thôi. Bạn thông cảm nhé
Bài làm
\(A=\frac{2^2.10+2^3.6}{2^2.15-2^4}\)
\(A=\frac{2^2.10+2.2^2.6}{2^2.15-2^2.2^2.1}\)
\(A=\frac{2^2.\left(10+6\right).2}{2^2.\left(15-1\right).2^2}\)
\(A=\frac{2^2.16.2}{2^2.14.2^2}\)
\(A=\frac{16}{14.2}\)
\(A=\frac{8}{7.2}\)
\(A=\frac{8}{14}\)
\(A=\frac{4}{7}\)
Vậy \(A=\frac{4}{7}\)
\(B=\frac{2^9.15^{17}.75^3}{18^8.5^{24}.9^2}\)
\(B=\frac{2^9.\left(3.5\right)^{17}.\left(3.5^2\right)^3}{\left(2.3^2\right)^8.5^{24}.\left(3^2\right)^2}\)
\(B=\frac{2^9.3^{17}.5^{17}.3^3.5^6}{2.3^{19}.5^{24}.3^4}\)
\(B=\frac{2^8.1.1.1.5}{1.3^2.1.3}\)
\(B=\frac{2^8.5}{3^3}\)
\(B=\frac{1280}{27}\)