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\(a.\left(\frac{-2}{3}+\frac{4}{15}\right)^3=\left(\frac{-10+4}{15}\right)^3=\frac{-6^3}{15}=\frac{-8}{125}\)
\(b.\left(\frac{3}{21}-\frac{2}{7}\right)^2=\left(\frac{3}{21}-\frac{6}{21}\right)^2=\frac{-1}{7}^2=\frac{-1}{49}\)
\(d.3-\left(-3.15\right)^0+\left(0.5\right)^2:2=3-1+0.25=2+0.25=2.25\)
\(e.81-3^2:\left(0.375\right)^2=81-9:0.140625=81-64=17\)
- a) =\(\left(\frac{-10}{15}+\frac{4}{15}\right)^3\)=\(\left(\frac{-6}{15}\right)^3\) = \(\left(\frac{-2}{5}\right)^3\) =\(\frac{-8}{125}\)
b) \(=\left(\frac{3}{27}-\frac{6}{21}\right)^2=\left(\frac{-3}{21}\right)^2=\left(\frac{-1}{7}\right)^2=\frac{1}{49}\)
d) \(=3-1+1:2=2+\frac{1}{2}=\frac{5}{2}\)
e) \(=81-9:\left(\frac{3}{8}\right)^2=72:\frac{9}{64}=72.\frac{64}{9}=512\)
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\(H=\frac{8}{1^2\cdot3^2}+\frac{16}{3^2\cdot5^2}+...+\frac{48}{11^2\cdot13^2}\)
\(H=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{5^2}+...+\frac{1}{11^2}-\frac{1}{13^2}\)
\(H=1-\frac{1}{13^2}\)
\(H=\frac{168}{169}\)
Phương thiếu bước nhé
\(H=\frac{8}{1^2.3^2}+\frac{16}{3^2.5^2}+\frac{24}{5^2.7^2}+...+\frac{48}{11^2.13^2}\)
\(H=\frac{3^2-1^2}{1^2.3^2}+\frac{5^2-3^2}{3^2.5^2}+\frac{7^2-5^2}{5^2.7^2}+...+\frac{13^2-11^2}{11^2.13^2}\)
\(H=\frac{3^2}{1^2.3^2}-\frac{1^2}{1^2.3^2}+\frac{5^2}{3^2.5^2}-\frac{3^2}{3^2.5^2}+\frac{7^2}{5^2.7^2}-\frac{5^2}{5^2.7^2}+...+\frac{13^2}{11^2.13^2}-\frac{11^2}{11^2.13^2}\)
\(H=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{5^2}+\frac{1}{5^2}-\frac{1}{7^2}+...+\frac{1}{11^2}-\frac{1}{13^2}\)
\(H=1-\frac{1}{13^2}=1-\frac{1}{169}=\frac{168}{169}\)
Chúc bạn học tốt ~
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Đặt A=\(\frac{11}{5^2\cdot6^2}+\frac{13}{6^2\cdot7^2}+\frac{15}{7^2\cdot8^2}+...+\frac{39}{19^2\cdot20^2}\)
A=\(\frac{11}{25\cdot36}+\frac{13}{36\cdot49}+\frac{15}{49\cdot64}+...+\frac{39}{361\cdot400}\)
A=\(\frac{1}{25}-\frac{1}{36}+\frac{1}{36}-\frac{1}{49}+\frac{1}{49}-\frac{1}{64}+...+\frac{1}{361}-\frac{1}{400}\)
A=\(\frac{1}{25}-\frac{1}{400}\)
A=\(\frac{3}{80}\)
\(\Rightarrow\)A không phải là số nguyên
\(\frac{11}{5^2.6^2}+\frac{13}{6^2.7^2}+\frac{15}{7^2.8^2}+...+\frac{39}{19^2.20^2}\)
\(=\frac{11}{25.36}+\frac{13}{36.49}+\frac{15}{49.64}+...+\frac{39}{361.400}\)
\(=\frac{1}{25}-\frac{1}{36}+\frac{1}{36}-\frac{1}{49}+\frac{1}{49}-\frac{1}{64}+...+\frac{1}{361}-\frac{1}{400}\)
\(=\frac{1}{25}-\frac{1}{400}\)
\(=\frac{3}{80}\)
Mà \(\frac{3}{80}\notin Z\)
\(\Rightarrow\frac{11}{5^2+6^2}+\frac{13}{6^2.7^2}+\frac{15}{7^2.8^2}+...+\frac{39}{19^2.20^2}\notin Z\)
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\(x=\frac{10^2.5^3.15^6}{3^6.5^{10}}=\frac{\left(2.5\right)^2.\left(3.5\right)^6}{3^6.5^7}=\frac{2^2.5^2.3^6.5^6}{3^6.5^7}=2^2.5=20\)
Vậy x = 20
\(=\frac{7^2.3^3-3^2.5^2.7+7.3^4}{3.3^2.5^2-2^3.5^3}\)\(=\frac{7.3^2\left(7.3-5^2+3^2\right)}{5^2\left(3^3-2^3.5\right)}\)\(=\frac{7.3^2.5}{5^2.\left(-13\right)}=\frac{7.3^2}{5.\left(-13\right)}=\frac{63}{-65}\)