Ta có: \(32^{27}=\left(2^5\right)^{27}=2^{135}\)
\(16^{39}=\left(2^4\right)^{39}=2^{156}\)
mà \(2^{135}< 2^{156}\)
nên \(32^{27}< 16^{39}\)
mà \(16^{39}< 18^{39}\)
nên \(32^{27}< 18^{39}\)
\(\Leftrightarrow-32^{27}>-18^{39}\)
\(\Leftrightarrow\left(-32\right)^{27}>\left(-18\right)^{39}\)
So Sánh : \(\left(\dfrac{1}{16}\right)^{200}\)và\(\left(\dfrac{1}{2}\right)^{1000}\)
\(\dfrac{\left(-27\right)^{10}.16^{25}}{6^{30}.\left(-32\right)^{15}}\)
1.so sánh:
a)\(5^6\)và\(\left(-2\right)^{14}\)
b)\(9^5\)và \(27^3\)
c) \(\left(\frac{1}{8}\right)^6\) và \(\left(\frac{1}{32}\right)^4\)
So sánh
a)\(\left(\dfrac{1}{12}\right)^{27}\) và\(\left(\dfrac{1}{12}\right)^{18}\)
b)\(10^{20}\) và \(\left(-9\right)^{10}\)
c)\(64^3\) và \(16^2\)
d)\(\left(-16\right)^3\) và \(\left(-64\right)^2\)
Tìm tích của xyz biết x,y và z \(\in\) R+ để \(x\left(y+z\right)=32\) ; \(y\left(x+z\right)=27\) ; \(z\left(x+y\right)=35\)
A=\(\dfrac{2^{30}\cdot5^7+3^{13}\cdot5^{27}}{2^{27}\cdot5^7+2^{10}\cdot5^{27}}\)
M=\(\left(x-4\right)^{\left(x-5\right)^{\left(x-6\right)^{\left(x+6\right)^{\left(x+5\right)}}}}\) tại x=7
1. So sánh các tích sau hợp lý nhất:
P = \(\left(-\dfrac{21}{34}\right)\)\(\left(-\dfrac{13}{27}\right)\)\(\dfrac{7}{9}\)
Q = \(\left(-0,5\right)\)\(\dfrac{3}{19}\)\(\left(-\dfrac{24}{41}\right)\)\(\left(\dfrac{8}{31}\right)\)
M = \(\dfrac{-6}{17}.\dfrac{-5}{17}.\) ... \(.\dfrac{5}{17}.\dfrac{6}{17}\)
2. Tính hợp lý:
A = \(\left(-\dfrac{3}{11}\right).\dfrac{7}{10}.\dfrac{11}{-13}.\left(-20\right)\)
B = \(\dfrac{-4}{9}:\dfrac{2}{13}+\left(\dfrac{-5}{18}\right).\dfrac{2}{13}\)
C = \(\left(2\dfrac{1}{15}.\dfrac{3}{19}.\dfrac{2}{31}\right):\left(-\dfrac{5}{19}\right)\)
Help me!
1) Tính: \(M=\left[\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}:\left(\frac{1}{18}-\frac{4}{9}\right)\right]:\left(\frac{494949}{525252}-1\right)\)
Tìm số tự nhiên x
a) \(\dfrac{1}{27}.9^x=3^x\)
b) \(2^{x-1}+5.2^{x-2}=\dfrac{7}{32}\)
c) \(\left(\dfrac{4}{5}\right)^{2x+7}=\dfrac{625}{256}\)
d) \(\left(4x-3\right)^4=\left(4x-3\right)^2\)
