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26 tháng 9 2018
\(a)\)\(\left(\frac{3}{5}\right)^{2x+1}=\frac{81}{625}\)
\(\Leftrightarrow\)\(\left(\frac{3}{5}\right)^{2x+1}=\left(\frac{3}{5}\right)^4\)
\(\Leftrightarrow\)\(2x+1=4\)
\(\Leftrightarrow\)\(x=\frac{3}{2}\)
Vậy \(x=\frac{3}{2}\)
\(b)\)\(\left(\frac{2}{3}\right)^x.\left(\frac{2}{3}\right)^3=\frac{32}{243}\)
\(\Leftrightarrow\)\(\left(\frac{2}{3}\right)^{x+3}=\left(\frac{2}{3}\right)^5\)
\(\Leftrightarrow\)\(x+3=5\)
\(\Leftrightarrow\)\(x=2\)
Vậy \(x=2\)
\(c)\)\(\left(2x-1\right)^2=\left(2x-1\right)^3\)
\(\Leftrightarrow\)\(\left(2x-1\right)^3-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\)\(\left(2x-1\right)^2\left(2x-1-1\right)=0\)
\(\Leftrightarrow\)\(\left(2x-1\right)^2\left(2x-2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(2x-1\right)^2=0\\2x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}}\)
Vậy \(x=\frac{1}{2}\) hoặc \(x=1\)
Chúc bạn học tốt ~
VM
13 tháng 10 2019
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!
13 tháng 10 2019
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
NG
8 tháng 10 2020
\(\text{1, }\frac{27^4.9^3}{81^2}=\frac{\left(3^3\right)^4.\left(3^2\right)^3}{\left(3^4\right)^2}=\frac{3^{12}.3^6}{3^8}=3^{10}\)
\(\text{2, }\left(\frac{1}{5}\right)^{2002}.\left(-5\right)^{2000}=\frac{1}{5^{2002}}.5^{2000}=\frac{5^{2000}}{5^{2002}}=\frac{1}{5^2}=\frac{1}{5^2}\)
\(\text{3, }\frac{4^{11}.4^5}{2^{31}}=\frac{2^{22}.2^{10}}{2^{31}}=\frac{2^{32}}{2^{31}}=2\)
\(\text{4, }3^2.\frac{1}{243}.81^2.\frac{1}{3^2}=\frac{3^2.81^2}{3^5.3^2}=\frac{3^2.3^8}{3^7}=\frac{3^{10}}{3^7}=3^3=27\)
\(\text{5, }4^2.\frac{25^2}{2^3.5^2}+32.125=\frac{2^4.5^4}{2^3.5^2}+2^5.5^3=2.5^2+2^5.5^2=5^2.\left(2+2^5.5\right)=25.\left(2+32.5\right)=25.162=4050\)
18 tháng 8 2019
\(a;3^2\cdot\frac{1}{243}\cdot81^2\cdot\frac{1}{3^3}\)
\(=3^2\cdot\frac{1}{3^5}\cdot3^4\cdot\frac{1}{3^3}\)
\(=\left(3^2\cdot3^4\right)\cdot\left(\frac{1}{3^5}\cdot\frac{1}{3^3}\right)\)
\(=3^6\cdot\frac{1}{3^8}\)
\(=\frac{3^6}{3^8}\)
\(=\frac{1}{3^2}=\frac{1}{9}\)
18 tháng 8 2019
\(3^2.\frac{1}{243}.81^2.\frac{1}{3^3}\)
= \(9.\frac{1}{243}.6561.\frac{1}{27}\)
= \(9\)
b ) \(\left(4,2\right)^5:\left(2^3.\frac{1}{16}\right)\)
= \(\left(\frac{21}{5}\right)^5:\left(8.\frac{1}{16}\right)\)
= \(130691232:\frac{1}{2}\)
= \(130691232\times2\)
= 261382464
Chúc bạn học tốt !!!
=
25 tháng 8 2016
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)..........\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}............\frac{99}{100}\)
\(=\frac{1.2.3.........99}{2.3...........100}\)
\(=\frac{1}{100}\)
25 tháng 8 2016
Ta có: \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{100}\right)\)
=\(\frac{1}{2}.\frac{2}{3}....\frac{99}{100}\)
=\(\frac{1.2....99}{2.3...100}\)
=\(\frac{1}{100}\)
NN
24 tháng 4 2020
1.a) Sửa lại đề: \(\frac{11}{17}\)ở mẫu chuyển thành \(\frac{11}{7}\)
\(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)\(=\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}=\frac{3}{11}\)
( vì \(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\ne0\))
2.a) \(\frac{3}{5}+\frac{3}{2}.x=\frac{-5}{7}\)\(\Leftrightarrow\frac{3}{2}.x=\frac{-5}{7}-\frac{3}{5}\)
\(\Leftrightarrow\frac{3}{2}.x=\frac{-46}{35}\)\(\Leftrightarrow x=\frac{-46}{35}:\frac{3}{2}\)\(\Leftrightarrow x=\frac{-92}{105}\)
Vậy \(x=\frac{-92}{105}\)
b) \(\left(4x-\frac{1}{3}\right).\left(\frac{3}{2}x+\frac{5}{6}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{3}=0\\\frac{3}{2}x+\frac{5}{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=\frac{1}{3}\\\frac{3}{2}x=\frac{-5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-5}{9}\end{cases}}\)
Vậy \(x=\frac{-5}{9}\)hoặc \(x=\frac{1}{12}\)
VM
8 tháng 2 2020
b) \(\left(x+\frac{1}{2}\right)^3:3=-\frac{1}{81}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^3=\left(-\frac{1}{81}\right).3\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow x+\frac{1}{2}=-\frac{1}{3}\)
\(\Rightarrow x=\left(-\frac{1}{3}\right)-\frac{1}{2}\)
\(\Rightarrow x=-\frac{5}{6}\)
Vậy \(x=-\frac{5}{6}.\)
c) \(\frac{x-2}{2}=\frac{8}{x-2}\left(x\ne2\right).\)
\(\Rightarrow\left(x-2\right).\left(x-2\right)=8.2\)
\(\Rightarrow\left(x-2\right)^2=16\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm4\right)^2\)
\(\Rightarrow x-2=\pm4.\)
\(\Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+2\\x=\left(-4\right)+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
Vậy \(x\in\left\{6;-2\right\}.\)
Chúc bạn học tốt!
\(3^2\times\dfrac{1}{243}\times81^2\times\dfrac{1}{3^3}\)
\(=3^2\times\dfrac{1}{3^5}\times\left(3^4\right)^2\times\dfrac{1}{3^3}\)
\(=\dfrac{3^2}{3^5}\times3^{4.2}\times\dfrac{1}{3^3}\)
\(=\dfrac{3^2}{3^5}\times3^8\times\dfrac{1}{3^3}\)
\(=\dfrac{1}{3^3}\times\dfrac{3^8}{3^3}\)
\(=\dfrac{3^5}{3^3}=3^2\)