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a) \({( - 2)^4} \cdot {( - 2)^5} = {\left( { - 2} \right)^{4 + 5}} = {\left( { - 2} \right)^9}\)
\({( - 2)^{12}}:{( - 2)^3} = {\left( { - 2} \right)^{12 - 3}} = {\left( { - 2} \right)^9}\)
Vậy \({( - 2)^4} \cdot {( - 2)^5}\) = \({( - 2)^{12}}:{( - 2)^3}\);
b) \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6} = {\left( {\frac{1}{2}} \right)^{2 + 6}} = {\left( {\frac{1}{2}} \right)^8}\)
\({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2} = {\left( {\frac{1}{2}} \right)^{4.2}} = {\left( {\frac{1}{2}} \right)^8}\)
Vậy \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6}\) = \({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2}\)
c) \({(0,3)^8}:{(0,3)^2} = {\left( {0,3} \right)^{8 - 2}} = {\left( {0,3} \right)^6}\)
\({\left[ {{{(0,3)}^2}} \right]^3} = {\left( {0,3} \right)^{2.3}} = {\left( {0,3} \right)^6}\)
Vậy \({(0,3)^8}:{(0,3)^2}\)= \({\left[ {{{(0,3)}^2}} \right]^3}\).
d) \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3} = {\left( { - \frac{3}{2}} \right)^{5 - 3}} = {\left( { - \frac{3}{2}} \right)^2} = {\left( {\frac{3}{2}} \right)^2}\)
Vậy \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3}\) = \({\left( {\frac{3}{2}} \right)^2}\).
a ) \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)và \(x+z=18\)
Áp dụng t/c dãy tỏ số bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+z}{2+4}=\frac{18}{6}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=3\\\frac{y}{3}=3\\\frac{z}{4}=3\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=6\\y=9\\z=12\end{cases}}\)
b ) \(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}\) và \(y-x=39\)
Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}=\frac{y-x}{-6-5}=\frac{39}{-11}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=\frac{39}{-11}\\\frac{y}{-6}=\frac{39}{-11}\\\frac{z}{7}=\frac{39}{-11}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=\frac{195}{11}\\y=-\frac{234}{11}\\z=\frac{273}{11}\end{cases}}\)
a.=> \(\frac{\left(\frac{1}{3}\right).x}{\frac{2}{3}}=\frac{\frac{7}{4}}{\frac{2}{5}}\)
=> \(\frac{1}{3}.x=\frac{7}{4}.\frac{2}{3}:\frac{2}{5}\)
=>\(\frac{1}{3}.x=\frac{35}{12}\)
=> x\(=\frac{35}{12}:\frac{1}{3}\)
Vậy x=\(\frac{35}{4}\).
b. => \(\frac{4,5}{0,3}=\frac{2,25}{0,1.x}\)
=>\(0,1.x=\frac{2,25.0,3}{4,5}\)
=>\(0,1.x=0,15\)
=>\(x=0,15:0,1\)
Vậy x=1,5
c. =>\(\frac{8}{\frac{1}{4}.x}=\frac{2}{0,02}\)
=>\(\frac{1}{4}.x=\frac{8.0,02}{2}\)
=>\(\frac{1}{4}.x=0,08\)
=>\(x=0,08:\frac{1}{4}\)
Vậy x=0,32.
d. =>\(\frac{3}{\frac{9}{4}}=\frac{\frac{3}{4}}{6.x}\)
=>\(3.6x=\frac{9}{4}.\frac{3}{4}\)
=>\(18x=\frac{27}{16}\)
=>\(x=\frac{27}{16}:18\)
Vậy x=\(\frac{3}{32}\)
3) \(\frac{x-3}{x-2}=\frac{4}{7}\)
\(\Rightarrow\left(x-3\right).7=\left(x-2\right).4\)
\(\Rightarrow7x-21=4x-8\)
\(\Rightarrow7x-4x=\left(-8\right)+21\)
\(\Rightarrow3x=13\)
\(\Rightarrow x=13:3\)
\(\Rightarrow x=\frac{13}{3}\)
Vậy \(x=\frac{13}{3}.\)
4) \(\left|x\right|+3,5=0\)
\(\Rightarrow\left|x\right|=0-3,5\)
\(\Rightarrow\left|x\right|=-3,5\)
Vì \(\left|x\right|\ge0\) \(\forall x.\)
\(\Rightarrow\left|x\right|>-3,5\)
\(\Rightarrow\left|x\right|\ne-3,5\)
Vậy \(x\in\varnothing.\)
5) \(\left|x+\frac{1}{3}\right|-4=1\)
\(\Rightarrow\left|x+\frac{1}{3}\right|=1+4\)
\(\Rightarrow\left|x+\frac{1}{3}\right|=5.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{3}=5\\x+\frac{1}{3}=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5-\frac{1}{3}\\x=\left(-5\right)-\frac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{14}{3}\\x=-\frac{16}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{14}{3};-\frac{16}{3}\right\}.\)
Chúc bạn học tốt!
\(1\frac{2}{3}:\frac{x}{4}=6:0,3\)
\(\frac{5}{3}:\frac{x}{4}=20\)
\(\Rightarrow\frac{x}{4}=\frac{5}{3}:20\)
\(\Rightarrow\frac{x}{4}=\frac{1}{12}\Rightarrow x=\frac{4}{12}=\frac{1}{3}\)
\(\left|\frac{5}{-4}\right|-\left|\frac{1}{-3}\right|+-\frac{5}{6}-4\frac{1}{2}\)
\(=\left|-\frac{5}{4}\right|-\left|\frac{-1}{3}\right|+\frac{-5}{6}-\frac{9}{2}\)
\(=\frac{5}{4}-\frac{1}{3}+\frac{-5}{6}-\frac{9}{2}=-\frac{53}{12}\)
\(\frac{5}{8}-\left|-\frac{1}{12}\right|-3\frac{1}{4}+\left|-\frac{5}{6}\right|\)
\(=\frac{5}{8}-\frac{1}{12}-\frac{13}{4}+\frac{5}{6}=-\frac{15}{8}\)
\(\frac{3}{-7}+\left|-\frac{5}{12}\right|+3\frac{1}{4}+\left|-\frac{5}{6}\right|\)
\(=\frac{-3}{7}+\frac{5}{12}+\frac{13}{4}+\frac{5}{6}=\frac{57}{14}\)
\(1\frac{3}{5}-\left|\frac{1}{-4}\right|+\frac{2}{-3}-\left|-\frac{1}{2}\right|\)
\(=\frac{8}{5}-\left|\frac{-1}{4}\right|+\frac{-2}{3}-\frac{1}{2}\)
\(=\frac{8}{5}-\frac{1}{4}+\frac{-2}{3}-\frac{1}{2}\)
\(=\frac{27}{20}+\frac{-7}{6}=\frac{27}{20}-\frac{7}{6}=\frac{11}{60}\)