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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---> \(\frac{1^3}{2^3}+\frac{2^3}{2^3}+...+\frac{10^3}{2^3}=\frac{3025}{2^3}\)
\(\Rightarrow0,5^3+1^3+2^3+...+5^3=\frac{3025}{8}=378,125\)
Giải:
a) \(A=1+3+3^2+3^3+...+3^{99}+3^{100}\)
\(\Leftrightarrow3A=3+3^2+3^3+3^4+...+3^{100}+3^{101}\)
\(\Leftrightarrow3A-A=2A=3^{101}-1\)
\(\Leftrightarrow A=\dfrac{3^{101}-1}{2}\)
b) \(B=1-3+3^2+3^3+...+3^{99}+3^{100}\)
\(\Leftrightarrow3B=3-3^2+3^3+3^4+...+3^{100}+3^{101}\)
\(\Leftrightarrow3B-B=2B=3^{101}-1-6-18=3^{101}--25\)
\(\Leftrightarrow B=\dfrac{3^{101}-25}{2}\)
Chúc bạn học tốt!
\(A=1+3+3^2+3^3+...+3^{99}+3^{100}\)
\(3A=3+3^2+3^3+3^4+...+3^{100}+3^{101}\)
\(3A-A=\left(3+3^2+3^3+3^4+...+3^{100}+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{99}\right)\)
\(2A=3^{101}-1\Leftrightarrow A=\dfrac{3^{101}-1}{2}\)
B đề sai
BT1: \(\left(3^2\right)^2-\left(-2^3\right)^2-\left(-5^2\right)^2=81-64-625=-608\)
BT2: a, \(\dfrac{1}{9}.27^x=3^x\)
\(3^{3x-2}=3^x\)
\(\Rightarrow3x-2=x\Rightarrow x=\dfrac{1}{2}\)
b, \(3^{-2}.3^4.3^x=3^7\)
\(3^{2+x}=3^7\Rightarrow2+x=7\)
\(\Rightarrow x=5\)
c, \(2^{-1}.2^x+4.2^x=9.2^5\)
\(2^x\left(2^{-1}+4\right)=288\)
\(\Rightarrow2^x=288:4,5=64=2^6\)
\(\Rightarrow x=6\)
d, \(\left(2x-3\right)^2=16=4^2\)
\(\Rightarrow2x-3=4\Rightarrow x=\dfrac{7}{2}\)
e, \(\left(3x-2\right)^5=-243=-3^5\)
\(\Rightarrow3x-2=-3\Rightarrow x=\dfrac{-1}{3}.\)
BT1: \(a,3^2.\dfrac{1}{243}.81^2.\dfrac{1}{33}=3^2.3^{-5}.3^8.3^{-1}\dfrac{1}{11}\)
\(=3^4.\dfrac{1}{11}=\dfrac{81}{11}\)
b, \(\left(4.5^3\right):\left(2^3.\dfrac{1}{10}\right)=100.5.\dfrac{1}{8}.10=625\)
thui tyuwj lamf ddi
1.
a) \(3^3.9^{-1}\)
\(=27.\frac{1}{9}\)
\(=3.\)
b) \(25.5^{-1}.5^0\)
\(=25.\frac{1}{5}.1\)
\(=5.1\)
\(=5.\)
c) \(3^2.\frac{1}{243}.81^2.3^{-3}\)
\(=9.\frac{1}{243}.6561.\frac{1}{27}\)
\(=\frac{1}{27}.6561.\frac{1}{27}\)
\(=243.\frac{1}{27}\)
\(=9.\)
Chúc bạn học tốt!
a, \(3^3.9^{-1}\)
\(=27.\frac{1}{9}\)
\(=\frac{27}{9}=3\)
b, \(25.5^{-1}.5^0\)
\(=25.\frac{1}{5}.1\)
\(=\frac{25}{5}.1\)
\(=5.1\)
\(=5\)
c, \(3^2.\frac{1}{143}.81^2.3^{-3}\)
\(=9.\frac{1}{143}.6561.\frac{1}{3^3}\)
\(=9.\frac{1}{143}.6561.\frac{1}{27}\)
\(=9.\frac{1}{143}\left(6561.\frac{1}{27}\right)\)
\(=9.\frac{1}{143}.243\)
\(=\frac{9}{143}.243\)
\(=\frac{2187}{143}\)
Câu d tương tự các câu trên
e) \(\frac{1}{7}.\frac{-3}{8}+\frac{-13}{8}.\frac{1}{7}\)
\(=\frac{1}{7}.\left[\left(-\frac{3}{8}\right)+\left(-\frac{13}{8}\right)\right]\)
\(=\frac{1}{7}.\left(-2\right)\)
\(=-\frac{2}{7}.\)
Chúc bạn học tốt!
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!
Cau a la 1
Cau b la 1215
Cau c la 768
Cau d la \(\frac{4185}{13}\)
\(\frac{4^2.4^3}{2^{10}}=\frac{4^{2+3}}{\left(2^2\right)^5}=\frac{4^5}{4^5}=1\)
\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2.3\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2\right)^5.3^5}{\left(0,2\right)^6}=\frac{3^5}{0,2}=1215\)
\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^2.2^5.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^2.2^5.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}\)
\(\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+3.\left(2.3\right)^2+3^3}{-13}=\frac{2^3.3^3+3.2^2.3^2+3^3}{-13}=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=\left(-3\right)^3=-27\)
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: \(=2^3+3\cdot1\cdot\dfrac{1}{4}\cdot4+\left[4:\dfrac{1}{2}\right]:8\)
\(=8+3+4\cdot2:8=11+1=12\)
Tính:
2) \(\left(\frac{2}{3}\right)^3-\left(\frac{3}{4}\right)^2.\left(-1\right)^5\)
\(=\frac{8}{27}-\frac{9}{16}.\left(-1\right)\)
\(=\frac{8}{27}-\left(-\frac{9}{16}\right)\)
\(=\frac{371}{432}.\)
Xin lỗi, anh chỉ làm câu này thôi em.
Chúc em học tốt!
= \(\frac{1}{27}\)
3-3 = \(\left(\frac{1}{3}\right)^3\)