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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.
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
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
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=(8 2/7-4 2/7)-3 4/9`
`=8+2/7-4-2/7-3-4/9`
`=4-3-4/9`
`=1-4/9=5/9`
`B=(10 2/9-6 2/9)+2 3/5`
`=10+2/9-6-2/9+2+3/5`
`=4+2+3/5`
`=6+3/5=33/5`
Bài 2:
`a)5 1/2*3 1/4`
`=11/2*13/4`
`=143/8`
`b)6 1/3:4 2/9`
`=19/3:38/9`
`=19/3*9/38=3/2`
`c)4 3/7*2`
`=31/7*2`
`=62/7`
Bài 1:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
a, Ta có: \(3^{21}>3^{20}\left(1\right)\)
\(2^{31}>2^{30}\)(2)
Mà \(\left\{{}\begin{matrix}3^{20}=3^{2.10}=\left(3^2\right)^{10}=9^{10}\\2^{30}=2^{3.10}=\left(2^3\right)^{10}=8^{10}\end{matrix}\right.\)
Do \(9>8\Rightarrow9^{10}>8^{10}\Rightarrow3^{20}>2^{30}\left(3\right)\)
Từ (1);(2) và (3) ta suy ra \(3^{21}>2^{31}\)
a)\(3^{21}=\left(3^2\right)^{10}.3=9^{10.3}\)
\(2^{31}=\left(2^3\right)^{10}.2=8^{10}.2\)
Vì \(9^{10}.3>8^{10}.2\Rightarrow3^{21}>2^{31}\)
b)\(A=\dfrac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}\)
\(A=\dfrac{1+5+5^2+...+5^8}{1+5+5^2+...+5^8}+\dfrac{5^9}{1+5+5^2+...+5^8}\)
\(A=1+\dfrac{5^9}{1+5+5^2+..+5^9}\)
A=\(1+1:\dfrac{1+5+5^2+...+5^9}{5^9}\)
\(A=1+1:\left(\dfrac{1}{5^9}+\dfrac{1}{5^8}+\dfrac{1}{5^7}+...+\dfrac{1}{5}\right)\)
Tương tự \(B=1+1:\left(\dfrac{1}{3^9}+\dfrac{1}{3^8}+\dfrac{1}{3^7}+...+\dfrac{1}{3}\right)\)
Vì \(\dfrac{1}{5^9}+\dfrac{1}{5^8}+\dfrac{1}{5^7}+....+\dfrac{1}{5}< \dfrac{1}{3^9}+\dfrac{1}{3^8}+...+\dfrac{1}{3}\)
\(\Rightarrow A>B\)
\(C=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2^2\right)\\ C=\dfrac{6}{7}+\dfrac{5}{8}.\dfrac{1}{5}-\dfrac{3}{16}.\left(-4\right)\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{16}.\left(-4\right)\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{16}.\dfrac{-4}{1}\\ C=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{-3}{4}\\ C=\dfrac{48}{56}+\dfrac{7}{56}-\dfrac{-42}{56}\\ C=\dfrac{97}{56}\)
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 ...