Toán Lớp 7: A)12^20/9^10.4^20
B)-3|x-1/6|=-1/2
C)(x-3^2)^3=(3^2)^3
D)x:0,5=32:x
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Toán Lớp 7: A)12^20/9^10.4^20
B)-3|x-1/6|=-1/2
C)(x-3^2)^3=(3^2)^3
D)x:0,5=32:x
Comments ( 1 )
a)\dfrac{{{{12}^{20}}}}{{{9^{10}}{{.4}^{20}}}} = \dfrac{{{{\left( {{2^2}.3} \right)}^{20}}}}{{{{\left( {{3^2}} \right)}^{10}}.{{\left( {{2^2}} \right)}^{20}}}}\\
= \dfrac{{{2^{40}}{{.3}^{20}}}}{{{3^{20}}{{.2}^{40}}}}\\
= 1\\
b) – 3.\left| {x – \dfrac{1}{6}} \right| = – \dfrac{1}{2}\\
\Leftrightarrow \left| {x – \dfrac{1}{6}} \right| = \dfrac{1}{6}\\
\Leftrightarrow \left[ \begin{array}{l}
x – \dfrac{1}{6} = \dfrac{1}{6}\\
x – \dfrac{1}{6} = – \dfrac{1}{6}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{3}\\
x = 0
\end{array} \right.\\
Vậy\,x = 0;x = \dfrac{1}{3}\\
c){\left( {x – {3^2}} \right)^3} = {\left( {{3^2}} \right)^3}\\
\Leftrightarrow {\left( {x – 9} \right)^3} = {9^3}\\
\Leftrightarrow x – 9 = 9\\
\Leftrightarrow x = 18\\
Vậy\,x = 18\\
d)x:0,5 = 32:x\\
\Leftrightarrow \dfrac{x}{{0,5}} = \dfrac{{32}}{x}\\
\Leftrightarrow {x^2} = 32.0,5\\
\Leftrightarrow {x^2} = 16\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = – 4
\end{array} \right.\\
Vậy\,x = 4;x = – 4
\end{array}$