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a) Nhân 2 vế của đẳng thức đầu cho 23---> \(1^3.2^3+2^3.2^3+...+10^3.2^3=3025.2^3\)
\(\Rightarrow2^3+4^3+...+20^3=3025.8=24200\)
b Chia 2 vế của đẳng thức đầu cho 23---> \(\frac{1^3}{2^3}+\frac{2^3}{2^3}+...+\frac{10^3}{2^3}=\frac{3025}{2^3}\)
\(\Rightarrow0,5^3+1^3+2^3+...+5^3=\frac{3025}{8}=378,125\)
((3\(^2\)))\(^2\) - ((-5\(^2\)))\(^2\) + ((-2\(^3\)))\(^2\)
= 81 - 625 + 64
= -544+ 64
= -480
2\(^4\) + 8[(-2)\(^2\) :\(\dfrac{1}{2}\)]\(^0\) - 2\(^{-2}\). 4 + (-2)\(^2\)
= 16+ 8.1 - \(\dfrac{1}{4}\). 4 + 4
= 16+ 8- 1+4
= 27
2\(^4\) + 3(\(\dfrac{1}{2}\))\(^0\) + 2\(^{-2}\).8 + [(-2)\(^3\). \(\dfrac{1}{2^4}\)].2 - \(\dfrac{1}{2}\)
= 16 + 3.1 +\(\dfrac{1}{4}\).8 + [(-8).\(\dfrac{1}{16}\)].2 -\(\dfrac{1}{2}\)
= 16 + 3+ 2 + \(\dfrac{-1}{2}\).2- \(\dfrac{1}{2}\)
= 21 + (-1)- \(\dfrac{1}{2}\)
= 20-\(\dfrac{1}{2}\) = \(\dfrac{40}{2}\) - \(\dfrac{1}{2}\)= \(\dfrac{39}{2}\)
\(\dfrac{15^{10}.5^{10}}{75^{10}}\) + \(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}\)
= \(\dfrac{\left(15.5\right)^{10}}{75^{10}}\) + \(\dfrac{\left(0,4.2\right)^5}{\left(0.4\right)^6}\)
= \(\dfrac{75^{10}}{75^{10}}\) + \(\dfrac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}\)
= 1 + \(\dfrac{2^5}{0,4}\) = 1+ 80 = 81
\(\dfrac{2^{13}.9^4}{6^3.8^3}\)
= \(\dfrac{2^{13}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}\) = \(\dfrac{2^{13}.3^8}{2^3.3^3.2^9}\)
= \(\dfrac{2^4.3^5}{2^3}\) = 2.3\(^5\) = 486
a: \(A=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{-5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\)
\(=\dfrac{-3}{5}+\dfrac{3}{5}=0\)
b: \(=3^4-\left(-8\right)^2-\left(-25\right)^2\)
\(=81-64-625=-608\)
c: \(=2^3+3\cdot1\cdot\dfrac{1}{4}\cdot4+\left[4:\dfrac{1}{2}\right]:8\)
\(=8+3+4\cdot2:8=11+1=12\)
a: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
\(\Leftrightarrow2^x=2^{10}\cdot5:\dfrac{5}{2}=2^{10}\cdot5\cdot\dfrac{2}{5}=2^{11}\)
=>x=11
b: \(\Leftrightarrow3^x\cdot\dfrac{1}{3}+3^x\cdot9=3^{13}\cdot28\)
\(\Leftrightarrow3^x=3^{13}\cdot28:\dfrac{28}{3}=3^{14}\)
hay x=14
a,(=)\(3^{x+1}.\left(3+4\right)=7.3^6\)
(=)\(3^{x+1}=3^6\)
=>x+1=6(=)x=5
b
8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)
=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)
=7.(-7)
=-49
Bài 1:
a) \(\left(\frac{1}{2}\right)^2\) và \(\left(\frac{1}{2}\right)^5\)
Ta có: \(\left(\frac{1}{2}\right)^2=\frac{1}{4}.\)
\(\left(\frac{1}{2}\right)^5=\frac{1}{32}.\)
Vì \(\frac{1}{4}< \frac{1}{32}.\)
=> \(\left(\frac{1}{2}\right)^2< \left(\frac{1}{2}\right)^5.\)
b) \(\left(2,4\right)^3\) và \(\left(2,4\right)^2\)
Ta có: \(\left(2,4\right)^3=13,824.\)
\(\left(2,4\right)^2=5,76.\)
Vì \(13,284>5,76.\)
=> \(\left(2,4\right)^3>\left(2,4\right)^2.\)
c) \(\left(-1\frac{1}{2}\right)^2\) và \(\left(-1\frac{1}{2}\right)^3\)
Ta có: \(\left(-1\frac{1}{2}\right)^2=\left(-\frac{3}{2}\right)^2=\frac{9}{4}.\)
\(\left(-1\frac{1}{2}\right)^3=\left(-\frac{3}{2}\right)^3=-\frac{27}{8}.\)
Vì số dương luôn lớn hơn số âm nên \(\frac{9}{4}>-\frac{27}{8}.\)
=> \(\left(-1\frac{1}{2}\right)^2>\left(-1\frac{1}{2}\right)^3.\)
Chúc bạn học tốt!
\(\frac{1}{2}+\frac{2}{3}-\frac{3}{4}+\frac{4}{5}-\frac{5}{6}+\frac{6}{7}+\frac{5}{6}-\frac{4}{5}+\frac{3}{4}-\frac{2}{3}+\frac{1}{2}\)
\(=\left(\frac{1}{2}+\frac{1}{2}+\frac{6}{7}\right)+\left(\frac{2}{3}-\frac{2}{3}\right)+\left(\frac{-3}{4}+\frac{3}{4}\right)+\left(\frac{4}{5}-\frac{4}{5}\right)+\left(\frac{-5}{6}+\frac{5}{6}\right)\)
\(=\frac{13}{7}+0+0+0+0\)
\(=\frac{13}{7}\)
\(\frac{1}{2}+\frac{2}{3}-\frac{3}{4}+\frac{4}{5}-\frac{5}{6}+\frac{6}{7}+\frac{5}{6}-\frac{4}{5}+\frac{3}{4}-\frac{2}{3}+\frac{1}{2}.\)
\(=\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{2}{3}-\frac{2}{3}\right)-\left(\frac{3}{4}-\frac{3}{4}\right)+\left(\frac{4}{5}-\frac{4}{5}\right)-\left(\frac{5}{6}-\frac{5}{6}\right)+\frac{6}{7}\)
\(=1+0-0+0+\frac{6}{7}\)
\(=1+\frac{6}{7}=1\frac{6}{7}\)
a: \(2^2+4^2+...+24^2\)
\(=2^2\left(1^2+2^2+...+12^2\right)\)
\(=4\cdot650=2600\)
b: \(2^3+4^3+...+20^3\)
\(=2^3\left(1^3+2^3+...+10^3\right)\)
\(=8\cdot3025=24200\)
24200