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a: \(\left(5^{19}:5^{17}+3\right):7=\left(5^2+3\right):7=4\)
b: \(7^9:7^7-3^2+2^3\cdot5^2\)
\(=7^2-3^2+8\cdot25\)
=49-9+200
=240
c: \(1200:2+6^2\cdot2^1+18\)
\(=600+72+18\)
=690
d) Ta có: \(-8+\left(-14\right)+34+\left(-12\right)\)
\(=-8-14+34-12\)
\(=-20+20\)
\(=-0\)
e) Ta có: \(\left(-7\right)+\left(-20\right)+57+\left(-30\right)\)
\(=\left(-7+57\right)+\left(-20-30\right)\)
\(=50-50=0\)
f) Ta có: \(300-\left(-180\right)-120-42\)
\(=300+180-120-42\)
\(=258+60=318\)
d, Ta có : \(\left(-8\right)+\left(-14\right)+34+\left(-12\right)\)
\(=-\left(8+12\right)+34-14=-20+20=0\)
e, Ta có : \(\left(-7\right)+\left(-20\right)+57+\left(-30\right)\)
\(=-\left(30+20\right)+57-7=-50+50=0\)
f, Ta có : \(300-\left(-180\right)-120-42=318\)
\(4^3\cdot4^{x-1}=64\)
\(\Leftrightarrow4^{x-1}=1\)
\(\Leftrightarrow x-1=0\)
hay x=1
a)3\(^{11}\):3\(^9\)−147:7\(^2\)
=3\(^{11}\)\(^{-9}\)−147:49
=3\(^2\)−147:49
=9−3
=6
b)295−(31−22×5)w)295-(31-22×5)
=295−(31−2\(^2\).5)\(^2\)
=295−11\(^2\)
=295−121
=174
\(180:\left\{\left(-30\right)+\left[4^2.5-\left(14+3^{11}:3^9\right)\right]\right\}\)
\(=180:\left\{\left(-30\right)+\left[16.5-\left(14+3^{11}.\dfrac{1}{3^9}\right)\right]\right\}\)
\(=180:\left\{\left(-30\right)+\left[80-\left(14+3^2\right)\right]\right\}\)
\(=180:\left\{\left(-30\right)+\left[80-\left(14+9\right)\right]\right\}\)
\(=180:\left[\left(-30\right)+80-23\right]\)
\(=180:27\\ =\dfrac{20}{3}\)