\(\left(\frac{1}{3}\right)^{50}X\left(-9\right)^{25}-\frac{2}{3}X4\)
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23 tháng 8 2016
a) \(\left(\frac{2}{3}\right)^x=\left(\frac{4}{9}\right)^{50}\)
\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2^2}{3^2}\right)^{50}\)
\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2}{3}\right)^{100}\)
\(\Rightarrow x=100\)
Vậy x = 100
b) \(\left(\frac{2}{3}-x\right)^2=\frac{1}{36}\)
\(\Rightarrow\left(\frac{2}{3}-x\right)^2=\left(\frac{1}{6}\right)^2\)
\(\Rightarrow\frac{2}{3}-x=\frac{1}{6}\)
\(\Rightarrow x=\frac{2}{3}-\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
2)
Ta có:
\(74^{m+1}+74^m=74^m.74^1+74^m=74^m.\left(74+1\right)=74^m.75⋮25\)
( vì \(75⋮25\) )
\(\Rightarrowđpcm\)
MP
1
KB
14 tháng 12 2017
\(\left(\frac{1}{3}\right)^{50}.\left(-9\right)^{25}-\frac{2}{3}:4\)
\(=\frac{1}{3^{50}}.\left(-9\right)^{25}-\frac{2}{3}:4\)
\(=\left(-1\right)-\frac{1}{6}=-\frac{7}{6}\)
LT
1
S
9 tháng 3
\(a;\dfrac{2}{3}x-50\%x-\left(-\dfrac{4}{5}\right):1\dfrac{3}{5}=-0,12+1\dfrac{3}{25}\\ \dfrac{1}{6}x+\dfrac{1}{2}=1\Rightarrow\dfrac{1}{6}x=1-\dfrac{1}{2}=\dfrac{1}{2}\\ \Rightarrow x=\dfrac{1}{2}:\dfrac{1}{6}=3\\ b;\left(-1\dfrac{1}{6}+\dfrac{2}{3}-\dfrac{3}{4}\right):x+\left(-1\dfrac{11}{12}\right)\cdot1\dfrac{21}{23}=-6\dfrac{1}{3}\\ -\dfrac{5}{4}:x-\dfrac{11}{3}=-\dfrac{19}{3}\\ -\dfrac{5}{4}:x=-\dfrac{19}{3}+\dfrac{11}{3}=-\dfrac{8}{3}\\ x=-\dfrac{5}{4}:\left(-\dfrac{8}{3}\right)=\dfrac{15}{32}\\ c;50\%x-\dfrac{1}{3}x-\left(-\dfrac{2}{3}\right)^2\cdot\left(-1\dfrac{1}{8}\right)=-119\dfrac{3}{4}+120\dfrac{5}{6}\\ \dfrac{1}{6}x+\dfrac{1}{2}=\dfrac{13}{12}\Rightarrow\dfrac{1}{6}x=\dfrac{13}{12}-\dfrac{1}{2}=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{1}{6}=\dfrac{7}{2}\)
5 tháng 11 2019
(1/3)50. (-9)25 - 2/3 : 4
= (1/3)50 . [(-3)2]25 - 2/3 . 1/4
= (1/3)50.(-3)50- 1/6
= (1/3 . -3 )50 - 1/6
= (-1)50- 1/6
= 1 - 1/6
= 5/6
10 tháng 11 2019
theo đề ta có
=\(\left(\frac{1}{3^{ }}\right)^{50}.\left(-9\right)^{25}-\frac{2}{3}.\frac{1}{4}\)
=\(\left(\frac{1}{3}\right)^{50}.\left(\frac{1}{3}\right)^{25}.\left(-27\right)^{25}-\frac{1}{6}\)
=\(\left(\frac{1}{3}\right)^{50+25}.\)
= -11/3
chắc 100%