\(\frac{1}{4}:\left(\frac{5}{3}\right)^{x-1}=0,09\)
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\(a)-3\frac{1}{2}+\frac{1}{3}.\left(x-1\right)=-1\frac{1}{3}:2\frac{1}{3}\)
\(-\frac{7}{2}+\frac{1}{3}.\left(x-1\right)=-\frac{4}{3}:\frac{7}{3}\)
\(-\frac{7}{2}+\frac{1}{3}.\left(x-1\right)=-\frac{4}{7}\)
\(\frac{1}{3}.\left(x-1\right)=-\frac{4}{7}-\frac{-7}{2}\)
\(\frac{1}{3}.\left(x-1\right)=\frac{41}{14}\)
\(\Rightarrow x-1=\frac{41}{14}:\frac{1}{3}\)
\(\Rightarrow x-1=\frac{123}{14}\)
\(\Rightarrow x=\frac{123}{14}+1\)
\(\Rightarrow x=\frac{137}{14}\)
a)
\(\frac{{{4^3}{{.9}^7}}}{{{{27}^5}{{.8}^2}}} = \frac{{{{\left( {{2^2}} \right)}^3}.{{\left( {{3^2}} \right)}^7}}}{{{{\left( {{3^3}} \right)}^5}.{{\left( {{2^3}} \right)}^2}}} =\frac{2^{2.3}.3^{2.7}}{3^{3.5}.2^{2.3}}= \frac{{{2^6}{{.3}^{14}}}}{{{3^{15}}{{.2}^6}}} = \frac{1}{3}\)
b)
\(\frac{{{{\left( { - 2} \right)}^3}.{{\left( { - 2} \right)}^7}}}{{{{3.4}^6}}} =\frac{(-2)^{3+7}}{3.(2^2)^6}= \frac{{{{\left( { - 2} \right)}^{10}}}}{{3.{{\left( {{2^{2.6}}} \right)}}}} = \frac{{{2^{10}}}}{{{{3.2}^{12}}}} = \frac{1}{{{{3.2}^2}}} = \frac{1}{{12}}\)
c)
\(\begin{array}{l}\frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,09} \right)}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} = \frac{{{{\left( {0,2} \right)}^5}.{{\left[ {{{\left( {0,3} \right)}^2}} \right]}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} = \frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,3} \right)}^6}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}}\\ = \frac{{{{\left( {0,3} \right)}^2}}}{{{{\left( {0,2} \right)}^2}}} = \frac{{0,9}}{{0,4}} = \frac{9}{4}\end{array}\)
d)
Cách 1: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{8 + 16 + 32}}{{49}} = \frac{{56}}{{49}} = \frac{8}{7}\)
Cách 2: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{2^3.(1+2+2^2)}}{{7^2}} = \frac{{2^3.7}}{{7^2}} = \frac{8}{7}\)
\(M=\frac{3:\frac{2}{5}-0,09\left(0,15-2\frac{1}{2}\right)}{0,32+0,03-\left(5,3-3,88\right)+0,67}\)
\(\Leftrightarrow M=\frac{\frac{15}{2}+\frac{9}{235}}{-0,4}\)
\(\Leftrightarrow M=-\frac{3543}{188}\)
\(N=\frac{\left(2,1-1,965\right):\left(1,2.0,045\right)}{0,00325:0,013}-\frac{1:0,25}{1,6.0,625}\)
\(\Leftrightarrow N=\frac{0,135:0,054}{0,25}-4\)
\(\Leftrightarrow N=\frac{2,5}{0,25}-4\)
\(\Leftrightarrow N=10-4=6\)
Ta có:\(\Leftrightarrow\frac{3}{4}M=\frac{3}{4}.-\frac{3543}{188}=-\frac{10629}{752}\)
\(\Leftrightarrow\frac{1}{3}N=\frac{1}{3}.6=2\)
\(\Rightarrow M+N=-\frac{10629}{752}+2=-\frac{9125}{752}\)
Do đó ta được:\(\frac{12}{100}\) của tổng là:\(\frac{12}{100}.\frac{-9125}{752}=-\frac{1095}{752}\)
\(\frac{1}{4}:\left(\frac{5}{3}\right)^{x-1}=0,09\)
\(\frac{1}{4}:\left(\frac{5}{3}\right)^{x-1}=\frac{9}{100}\)
\(\left(\frac{5}{3}\right)^{x-1}=\frac{1}{4}:\frac{9}{100}\)
\(\left(\frac{5}{3}\right)^{x-1}=\frac{1}{4}\cdot\frac{100}{9}\)
\(\left(\frac{5}{3}\right)^{x-1}=\frac{100}{36}\)
\(\left(\frac{5}{3}\right)^{x-1}=\frac{25}{9}\)
\(\left(\frac{5}{3}\right)^{x-1}=\left(\frac{5}{3}\right)^2\)
\(\Rightarrow x-1=2\Rightarrow x=3\)
Vậy x cần tìm bằng 3
=> (5/3)x-1=25/9
=> (5/3)x-1=(5/30)2
=> x-1=2
=> x=3