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Giải:
a) Biến đổi tử:
Đặt:
\(C=1+5+5^2+5^3+...+5^9\)
\(\Leftrightarrow5C=5+5^2+5^3+5^4...+5^{10}\)
\(\Leftrightarrow5C-C=5^{10}-1\)
\(\Leftrightarrow4C=5^{10}-1\)
\(\Leftrightarrow C=\dfrac{5^{10}-1}{4}\)
Tương tự ta có mẫu là:
\(\dfrac{5^9-1}{4}\)
Đặt vào A, được:
\(A=\dfrac{1+5+5^2+5^3+...+5^9}{1+5+5^2+5^3+...+5^8}\)
\(\Leftrightarrow A=\dfrac{\dfrac{5^{10}-1}{4}}{\dfrac{5^9-1}{4}}\)
\(\Leftrightarrow A=\dfrac{5^{10}-1}{5^9-1}\)
Vậy ...
b) Tương tự câu a, ta được:
\(B=\dfrac{\dfrac{3^{10}-1}{2}}{\dfrac{3^9-1}{2}}\)
\(\Leftrightarrow B=\dfrac{3^{10}-1}{3^9-1}\)
Vậy ...
\(C=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2^2\right)\\ C=\dfrac{6}{7}+\dfrac{5}{8}.\dfrac{1}{5}-\dfrac{3}{16}.\left(-4\right)\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{16}.\left(-4\right)\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{16}.\dfrac{-4}{1}\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{-3}{4}\\ C=\dfrac{48}{56}+\dfrac{7}{56}-\dfrac{-42}{56}\\ C=\dfrac{97}{56}\)
\(A=15.\left(\dfrac{3}{5}-\dfrac{2}{3}\right)+1\\ A=15.\left(\dfrac{9}{15}-\dfrac{10}{15}\right)+1\\ A=15.\dfrac{-1}{15}+1\\ A=-1+1\\ A=0\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}\\ C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{9}.\dfrac{9}{11}+\dfrac{12}{7}\\ C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\\ C=\dfrac{-5}{7}.1+\dfrac{12}{7}\\ C=\dfrac{-5}{7}+\dfrac{12}{7}\\ C=1\)
Sửa đề : \(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\)
\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{7}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2}{7}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}\right)}\right):\dfrac{2021}{2020}\\ =\left(\dfrac{2}{7}-\dfrac{2}{7}\right):\dfrac{2021}{2022}=0\)
a: =-1/3+1/3=0
b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)
c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)
d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)
a: =-1/3+1/3=0
b: =411(−27−47−17)=411⋅(−1)=−411=411(−27−47−17)=411⋅(−1)=−411
c: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167
d: =13+74−74+45=13+45=5+1215=1715
a) \(\dfrac{8}{9}x=\dfrac{2}{7}-\dfrac{2}{3}=-\dfrac{8}{21}\)
\(x=-\dfrac{8}{21}:\dfrac{8}{9}=-\dfrac{3}{7}\)
b) \(\dfrac{2}{5}x=\dfrac{2}{5}-\dfrac{2}{5}=0\)
\(x=0:\dfrac{2}{5}=0\)
c)\(\dfrac{7}{8}x=\dfrac{2}{9}-\dfrac{1}{3}=-\dfrac{1}{9}\)
\(x=-\dfrac{1}{9}:\dfrac{7}{8}=-\dfrac{8}{63}\)
a) 2/7 - 8/9 . x = 2/3
⇒ 8/9 . x = 2/7 - 2/3
⇒ 8/9 .x = -8/21
⇒ x = -8/21 : 8/9
⇒ x = -3/7.
Vậy...
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
bài 2:
\(A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=9.\left(1-\dfrac{1}{100}\right)=9.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=\dfrac{891}{100}\)
bài 3:
\(=>\dfrac{x}{3}=\dfrac{5}{8}+\dfrac{1}{8}=\dfrac{8}{8}=1=\dfrac{3}{3}\)
\(=>x=3\)
c: Ta có: \(\dfrac{5}{3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{101\cdot103}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{101\cdot103}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{102}{103}\)
\(=\dfrac{255}{103}\)
Theo bài ra, ta có:
+) A = \(\dfrac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}\)
= \(\dfrac{1+5+5^2+...+5^8}{1+5+5^2+...+5^8}\)+ \(\dfrac{5^9}{1+5+5^2+...+5^8}\)
= 1 + \(\dfrac{1}{\dfrac{1+5+5^2+...+5^8}{5^9}}\)
+) B = \(\dfrac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
= \(\dfrac{1+3+3^2+...+3^8}{1+3+3^2+...+3^8}\)+ \(\dfrac{3^9}{1+3+3^2+...+3^8}\)
= 1 + \(\dfrac{1}{\dfrac{1+3+3^2+...+3^8}{3^9}}\)
Nhận xét:
+) \(\dfrac{1+5+5^2+...+5^8}{5^9}\) = \(\dfrac{1}{5^9}\) + \(\dfrac{1}{5^8}\) + ... + \(\dfrac{1}{5^{ }}\)
+) \(\dfrac{1+3+3^2+...+3^8}{3^9}\) = \(\dfrac{1}{3^9}\) + \(\dfrac{1}{3^8}\) + ... + \(\dfrac{1}{3}\)
Có: \(\dfrac{1}{5^9}\) < \(\dfrac{1}{3^9}\) ; \(\dfrac{1}{5^8}\) < \(\dfrac{1}{3^8}\) ; ... ; \(\dfrac{1}{5^{ }}\) < \(\dfrac{1}{3}\)
⇒ \(\dfrac{1+5+5^2+...+5^8}{5^9}\) < \(\dfrac{1+3+3^2+...+3^8}{3^9}\)
⇒ \(\dfrac{1}{\dfrac{1+5+5^2+...+5^8}{5^9}}\) > \(\dfrac{1}{\dfrac{1+3+3^2+...+3^8}{3^9}}\)
⇒ A > B
Vậy A > B.