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1020 = 102.10 = ( 102)10
Vì ( 102)10 > 910 nên 1020 > 910
648 = 644.4 = ( 644)4
1612 = 163.4 = ( 163)4
Vì ( 644)4 > ( 163)4 nên 648 > 1612
Ta có : A = |x - 500| + |x - 300| \(\ge\left|x-500+x-300\right|=-200\)
Vậy Amin = -200 khi 300 \(\le\) x \(\le\) 500
a, Ta có :
\(\left(\frac{1}{2}\right)^{50}=\left(\left(\frac{1}{2}\right)^5\right)^{10}=\left(\frac{1}{32}\right)^{10}\)
bạn so sánh nha :)
b,
T/c : \(99^{20}=\left(\left(99\right)^2\right)^{10}=9801^{10}\)
tiếp đây thì bạn tự làm nha có gì k hiểu ibx mk
a) 102 và 90^ 10
Ta có : 9010 = (905)2
Vì 905 > 10 => 90^10 > 10^2
b) (-5)^30 và (-3)^50
Ta có : (-5)^30= (-5^3)^10= -125^10
(-3)^50= (-3^5)^ 10= -243^10
Vì -125>-243 => (-3)^50 < (-5)^30
c) (-1)^10/16 và 1^50/2
Ta có: (-1)^10/16 = 1/16 = 1/2^4= 2(-4)
1^50/2 = 1/2= 2(-1)
Vì 2(-1) < 2(-4) => 1^50/2 < (-1)^10/16
\(1.a)\) Ta có: \(\left\{{}\begin{matrix}64^8=\left(8^2\right)^8=8^{16}\\16^{12}=8^{12}.2^{12}=8^{12}.\left(2^3\right)^4=8^{12}.8^4=8^{16}\end{matrix}\right.\)
Có: \(8^{16}=8^{16}\Rightarrow64^8=16^{12}\)
Vậy...
\(b)\) Ta có: \(\left\{{}\begin{matrix}\left(-5\right)^{30}=\left[\left(-5\right)^3\right]^{10}=\left(-125\right)^{10}\\\left(-3\right)^{50}=\left[\left(-3\right)^5\right]^{10}=\left(-243\right)^{10}\end{matrix}\right.\)
Có: \(\left(-125\right)^{10}< \left(-243\right)^{10}\Rightarrow\left(-5\right)^{30}< \left(-3\right)^{50}\)
Vậy...
\(c)\) Ta có: \(\left\{{}\begin{matrix}2^{27}=\left(2^3\right)^9=8^9\\3^{18}=\left(3^2\right)^9=9^9\end{matrix}\right.\)
Có: \(8^9< 9^9\Rightarrow2^{27}< 3^{18}\)
Vậy...
\(d)\) Ta có: \(\left\{{}\begin{matrix}\left(\dfrac{1}{25}\right)^{10}=\left[\left(\dfrac{1}{5}\right)^2\right]^{10}=\left(\dfrac{1}{5}\right)^{20}\\\left(\dfrac{1}{125}\right)^8=\left[\left(\dfrac{1}{5}\right)^3\right]^8=\left(\dfrac{1}{5}\right)^{24}\end{matrix}\right.\)
Có: \(\left(\dfrac{1}{5}\right)^{20}< \left(\dfrac{1}{5}\right)^{24}\Rightarrow\left(\dfrac{1}{24}\right)^{10}< \left(\dfrac{1}{125}\right)^8\)
Vậy...
\(e)\)Có: \(32^9=\left(2^5\right)^9=2^{45}< 2^{52}=\left(2^4\right)^{13}=16^{13}< 18^{13}\)
\(\Rightarrow32^9< 18^{13}\)
Vậy...
a) Ta có: 1020= (102)10=10010>9010
\(\Rightarrow\)1020>9010
b) Ta có: (-5)30 = (-53)10 =(-125)10
và (-3)50 = (-35)10 = (-243)10
Mà (-125)10 < (-243)10 => (-5)10 < (-3)50
c)- 0,320=(0,32)10=0,0910.
Do 0,09<0,1 =>0,0910<0,110.
=>0,110>0,320.
d) Ta có : \(\left(\dfrac{1}{16}\right)^{10}=\left(\dfrac{1}{2^4}\right)^{10}=\dfrac{1}{2^{40}}\)
\(\left(\dfrac{1}{2}\right)^{50}=\dfrac{1}{2^{50}}\)
Vì \(2^{40}< 2^{50}\Rightarrow\dfrac{1}{2^{40}}>\dfrac{1}{2^{50}}\Rightarrow\left(\dfrac{1}{16}\right)^{10}>\left(\dfrac{1}{2}\right)^{50}\)
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
cảm ơn nhìu
1)Ta có \(\left(\frac{1}{16}\right)^{10}\)=\(\left[\left(\frac{1}{2}\right)^4\right]^{10}\)=\(\left(\frac{1}{2}\right)^{40}\)
Vì \(2^{40}\)<\(2^{50}\)=>\(\left(\frac{1}{2}\right)^{40}\)>\(\left(\frac{1}{2}\right)^{50}\)
1) \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1^4}{2^4}\right)^{10}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{2}\right)^{40}\)
Vì \(\left(\frac{1}{2}\right)^{40}< \left(\frac{1}{2}\right)^{50}\) nên \(\left(\frac{1}{16}\right)^{10}< \left(\frac{1}{2}\right)^{50}\)
2) \(64^8=\left(4^3\right)^8=4^{24}\)
\(16^{12}=\left(4^2\right)^{12}=4^{24}\)
Vì \(4^{24}=4^{24}\) nên \(64^8=16^{12}\)
\(\left(\frac{1}{16}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)
Ta có: \(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}\)
Do \(\frac{1}{6}>\frac{1}{32}\Rightarrow\left(\frac{1}{6}\right)^{10}>\left(\frac{1}{32}\right)^{10}\)
Vậy \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
a) \(10^{20}\) và \(9^{10}\)
Vì 10 > 9 ; 20 > 10
nên \(10^{20}>9^{10}\)
Vậy \(10^{20}>9^{10}\)
b) \(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)
Ta có: \(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)
\(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)
Vì 243 > 125 nên \(125^{10}< 243^{10}\)
Vậy \(\left(-5\right)^{30}< \left(-3\right)^{50}\)
c) \(64^8\) và \(16^{12}\)
Ta có: \(64^8=\left(4^3\right)^8=4^{24}\)
\(16^{12}=\left(4^2\right)^{12}=4^{24}\)
Vậy \(64^8=16^{12}\left(=4^{24}\right)\)
d) \(\left(\frac{1}{6}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)
Ta có: \(\left(\frac{1}{6}\right)^{10}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{2}\right)^{40}\)
Vì 40 < 50 nên \(\left(\frac{1}{2}\right)^{40}< \left(\frac{1}{2}\right)^{50}\)
Vậy \(\left(\frac{1}{16}\right)^{10}< \left(\frac{1}{2}\right)^{50}\)