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b) Ta có : \(17^{20}=\left(17^2\right)^{10}=289^{10}>71^{10}\)
\(\Rightarrow\) \(17^{20}>71^{10}\)
\(71^5<71^{10}\)
\(\Rightarrow\) \(17^{20}>71^{10}>71^5\)
Vậy \(17^{20}>71^5\)
a, 1 - 7x = 3x - 4
=> -7x - 3x = - 4 - 1
=> - 10x = - 5
=> x = 1/2
vậy_
b, đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=1-\frac{1}{3^{99}}\)
\(A=\frac{1-\frac{1}{3^{99}}}{2}\)
mk chỉ bt lm mấy phần hui à!
d)\(\frac{5}{17}+\frac{-4}{7}-\frac{20}{31}+\frac{12}{17}-\frac{11}{31}\)\(=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}-\frac{11}{31}\right)+\frac{-4}{7}\)
\(=\frac{17}{17}+\frac{-31}{31}+\frac{-4}{7}\)\(=1+\left(-1\right)+\frac{-4}{7}\)\(=0+\frac{-4}{7}\)\(=-\frac{4}{7}\)
e)\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{20}{7}-\frac{13}{3}+\frac{13}{23}}\)
\(\frac{8^{11}.3^{17}}{27^{10}.9^{15}}=\frac{8^{11}.3^{17}}{3^{30}.3^{30}}=\frac{8^{11}}{3^{13}.3^{30}}=\frac{8^{11}}{3^{43}}\)
\(\frac{\left(5^4-5^3\right)^3}{125^4}=\frac{[\left(5-1\right).5^3]^3}{5^{12}}=\frac{\left(4.5^3\right)^3}{5^{12}}=\frac{64.5^9}{5^{12}}=\frac{64}{5^3}=\left(\frac{4}{5}\right)^3\)
\(\frac{4^{20}-2^{20}+6^{20}}{6^{20}-3^{20}+9^{20}}=\frac{2^{40}-2^{20}+6^{20}}{6^{20}-3^{20}+3^{40}}=\frac{2^{20}.\left(2^{20}-1+3^{30}\right)}{3^{20}.\left(2^{20}-2+3^{20}\right)}=\frac{2^{20}}{3^{20}}=\left(\frac{2}{3}\right)^{20}\)
c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Leftrightarrow x=0+2010\)
\(\Rightarrow x=2010\)
Vậy \(x=2010.\)
Mình chỉ làm câu c) thôi nhé.
Chúc bạn học tốt!
1/ so sánh
a) 812 và 128
Ta có: \(8^{12}=\left(8^3\right)^4=512^4\\ 12^8=\left(12^2\right)^4=144^4\)
vì 5124>1444 nên 812>128
b) (0,4)60và (-0,8)30
Gọi A= (0,4)60 và B= (-0,8)30
\(\Rightarrow\frac{A}{B}=\frac{\left(0,4\right)^{60}}{\left(-0,8\right)^{30}}=\frac{\left(0,1.2^2\right)^{60}}{\left(0,1.2^3\right)^{30}}=\frac{0,1^{60}.2^{120}}{0,1^{30}.2^{90}}=0,1^{30}.2^{30}=0,2^{30}>1\\ \Rightarrow A< B\)
e)\(A=\frac{20^{15}+1}{20^{16}+1}vàB=\frac{20^{16}+1}{20^{17}+1}\\ 20.A=20.\frac{20^{15}+1}{20^{16}+1}=\frac{20^{16}+20}{20^{16}+1}=\frac{20^{16}+1+19}{20^{16}+1}=\frac{20^{16}+1}{20^{16}+1}+\frac{19}{20^{16}+1}=1+\frac{19}{20^{16}+1}\left(1\right)\)
\(20.B=20.\frac{20^{16}+1}{20^{17}+1}=\frac{20^{17}+20}{20^{17}+1}=\frac{20^{17}+1+19}{20^{17}+1}=\frac{20^{17}+1}{20^{17}+1}+\frac{19}{20^{17}+1}=1+\frac{19}{20^{17}+1}\left(2\right)\)
Từ (1) và (2) ⇒ A>B
\(A=\frac{17^{20}+2}{17^{20}-1}=\frac{17^{20}-1+3}{17^{20}-1}=1+\frac{3}{17^{20}-1}\)
\(B=\frac{17^{20}-2}{17^{20}-5}=\frac{17^{20}-5+3}{17^{20}-5}=1+\frac{3}{17^{20}-5}\)
Vì \(17^{20}-1>17^{20}-5\)
\(=>\frac{3}{17^{20}-1}<\frac{3}{17^{20}-5}\)
\(=>1+\frac{3}{17^{20}-1}<1+\frac{3}{17^{20}-5}\)
=>A<B