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b bạn gõ lại đề giúp mình nhé
a.\(2^{225}=\left(2^3\right)^{75}=8^{75}\left(1\right)\)
\(3^{150}=\left(3^2\right)^{75}=9^{75}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow2^{225}< 3^{150}\)
b.\(2^{91}=2^{7.13}=\left(2^{13}\right)^7=8192^7\left(1\right)\)
\(5^{35}=5^{7.5}=\left(5^5\right)^7=3125^7\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow5^{35}< 2^{91}\)
c.\(9999^{10}=\left(99.101\right)^{10}=99^{10}.101^{10}>99^{10}.99^{10}=99^{20}\)
\(\Rightarrow9999^{10}>99^{20}\)
Bài 2:
\(b.\)\(75^{20}=\left(3.5^2\right)^{20}=\left(3^{20}.5^{10}\right).5^{30}=\left[̣\left(3^2\right)^{10}.5^{10}\right].5^{30}=45^{10}.5^{30}\)
\(\Rightarrowđpcm\)
Bài 3:
a,\(25^4.2^8=\left(5^2\right)^4.2^8=5^8.2^8=10^8\)
b. \(\frac{27^2}{25^3}=\frac{\left(3^3\right)^2}{\left(5^2\right)^3}=\frac{3^6}{5^6}=\left(\frac{3}{5}\right)^3\)\(:v\)
\(27^2.25^3=3^6.5^6=15^6\)( đề phòng you viết sai đề )
a) Ta có : \(31^5< 32^5=\left(2^5\right)^5=2^{25}< 2^{28}=\left(2^4\right)^7=16^7< 17^7\)
\(\Rightarrow31^5< 17^7\)
b) Ta có : \(8^{12}=\left(2^3\right)^{12}=2^{36}>2^{32}=\left(2^4\right)^8=16^8>12^8\)
\(\Rightarrow8^{12}>12^8\)
c) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2A=1-\frac{1}{99}\)
\(A=\frac{1-\frac{1}{99}}{2}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
a) \(31^5< 34^5=2^5.17^5=32.17^5\)
\(17^7=17^2.17^5=289.17^5\)
\(\Rightarrow31^5< 17^7\)
b) \(12^8< 16^8=\left(2^4\right)^8=2^{32}\)
\(8^{12}=\left(2^3\right)^{12}=2^{36}\)
\(\Rightarrow8^{12}>12^8\)
c) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3A-A=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{1}{3^{98}}-\frac{1}{3^{98}}\right)-\frac{1}{3^{99}}\)
\(\Rightarrow2A=1-\frac{1}{3^{99}}< 1\Rightarrow A< \frac{1}{2}\)
Câu a)
\(A=\sqrt{20+1}+\sqrt{40+2}+\sqrt{60+3}\)
\(=\sqrt{1\left(20+1\right)}+\sqrt{2\left(20+1\right)}+\sqrt{3\left(20+1\right)}\)
\(=\sqrt{20+1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
\(B=\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{20}+\sqrt{40}+\sqrt{60}\)
\(=1\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{1}\cdot\sqrt{20}+\sqrt{2}\cdot\sqrt{20}+\sqrt{3}\cdot\sqrt{20}\right)\)
\(=\sqrt{1}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)+\sqrt{20}\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
\(=\left(\sqrt{20}+\sqrt{1}\right)\left(\sqrt{1}+\sqrt{2}+\sqrt{3}\right)\)
Ta thấy: \(\hept{\begin{cases}\left(\sqrt{20+1}\right)^2=20+1\\\left(\sqrt{20}+\sqrt{1}\right)^2=20+1+2\sqrt{20}\end{cases}}\)
\(\Rightarrow\left(\sqrt{20+1}\right)^2< \left(\sqrt{20}+\sqrt{1}\right)^2\Rightarrow\sqrt{20+1}< \sqrt{20}+\sqrt{1}\)
Vậy A < B.
Nhiều thế :( Làm 1,2 câu thôi nhé
a) \(\frac{1}{3}+\frac{1}{4}=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}\) (bị mất nét nhưng vẫn nhìn ra là số 12 nhỉ?)
b) \(\frac{-2}{5}+\frac{7}{21}=\frac{-42}{105}+\frac{35}{105}=\frac{-7}{105}=\frac{-1}{15}\)
\(A=1+3+3^2+3^3+...+3^{2016}\)
\(A=1+3\left(1+3^2+...+3^{2015}\right)\)
\(A=1+3\left(A-3^{2016}\right)\)
\(A=1+3A-3^{2017}\)
\(2A=3^{2017}-1\Rightarrow A=\frac{3^{2017}-1}{2}\)
\(A< B\)
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