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a, Ta có: \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
=> \(\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\)=> \(\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
b, Ta có: \(\left(\dfrac{1}{3}\right)^{75}=\left[\left(\dfrac{1}{3}\right)^3\right]^{25}=\left(\dfrac{1}{27}\right)^{25}\)
\(\left(\dfrac{1}{5}\right)^{50}=\left[\left(\dfrac{1}{5}\right)^2\right]^{25}\)\(=\left(\dfrac{1}{25}\right)^{25}\)
Do \(\left(\dfrac{1}{27}\right)^{25}< \left(\dfrac{1}{25}\right)^{25}=>\left(\dfrac{1}{3}\right)^{75}< \left(\dfrac{1}{5}\right)^{50}\)
Kiểm tra lại bài nhé, học tốt!!
Bài 2:
a: \(\Leftrightarrow x=\dfrac{29}{60}\cdot\dfrac{-7}{5}=\dfrac{-203}{300}\)
b: \(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29-45}{60}=\dfrac{-16}{60}=\dfrac{-8}{30}\)
\(\Leftrightarrow x=\dfrac{-8}{30}:\dfrac{2}{5}=\dfrac{-8\cdot5}{30\cdot2}=\dfrac{-40}{60}=-\dfrac{2}{3}\)
Bài 1:
a, 2225 = (23)75 = 875
3150 = (32)75 = 975
Vì 875 < 975 nên 2225 < 3150
b, 212 = (24)3 = 163 ; 418 = (42)9 = 169
Bài 2:
a, 3300 = (33)100 = 27100
5200 = (52)100 = 25100
Vì 27100 > 25100 nên 3300 > 5200
b, Do \(\hept{\begin{cases}\left(x-3\right)^2\ge0\\\left|y^2-25\right|\ge0\end{cases}\forall x,y\Rightarrow\left(x-3\right)^2+\left|y^2-25\right|\ge0}\) (1)
Mà \(\left(x-3\right)^2+\left|y^2-25\right|=0\) (2)
Từ (1) và (2) => \(\hept{\begin{cases}\left(x-3\right)^2=0\\\left|y^2-25\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=5\end{cases}}\)
Bài 3:
2x = -3y = 4z
=> \(\frac{2x}{12}=\frac{-3y}{12}=\frac{4z}{12}\)
=> \(\frac{x}{6}=\frac{-y}{4}=\frac{z}{3}\)
=> \(\frac{x}{6}=\frac{-2y}{8}=\frac{3z}{9}=\frac{x-2y-3z}{6+8-9}=\frac{30}{5}=6\)
=> x = 36, y = -24, z = 18
`a)2^{300}=(2^3)^100=8^100`
`3^200=(3^2)^100=9^100`
Vì `9^100>8^100`
`=>2^300<3^200`
`b)3xx24^10`
`=3.(3.8)^10`
`=3^{11}.8^10`
`=3^{11}.2^30`
`2^300=2^{30}.2^{270}`
`=2^{30}.8^{90}`
Vì `3^11<8^90`
`=>3^{11}.2^30<8^{90}.2^30=2^300`
`=>3xx24^{10}<2^300+3^20+4^30`
câu 1
a(0,125)3x83=(0,125x8)3=13=1
b,2-(\(\frac{-3}{2}\))0+\(\frac{16}{4}:\frac{1}{2}\)=2-1+4:\(\frac{1}{2}\)=1+8=9
c\(^{3^5\cdot\frac{9}{3^7}\cdot2^0}\)=\(3^5\cdot\frac{3}{1}\cdot1=3^5\cdot3\cdot1=3^6\)
d,\(\frac{3}{2}-\frac{5}{6}:\left(\frac{1}{2}\right)^2=\frac{3}{2}-\frac{5}{6}:\frac{1}{4}=\frac{3}{2}-\frac{10}{3}=\frac{9}{6}-\frac{20}{6}=\frac{-11}{6}\)
câu 2
a\(\frac{x}{2}=\frac{4}{5}=\Rightarrow x\cdot5=2\cdot4\Rightarrow x=\frac{2.4}{5}=1,6\)
\(^{3^x}\)=27
\(^{3^x}\)=\(3^3\)
=>x =3
a) \(2^{300}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=\left(3^2\right)^{100}=9^{100}\)
Vì \(8^{100}< 9^{100}\left(8< 9\right)\)
Nên \(2^{300}< 3^{200}\)
b) \(3^x=27\)
\(3^x=3^3\)
Mà \(3^3=27\)
\(\Rightarrow\)\(x=3\)
Vậy x = 3