Tính giá trị :
\(J=4^{1-2\log_2\sqrt[4]{7}}-36^{\log_62}+81^{0,25-0,5\log_97}\)
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a) \(\sqrt{0,01}-\sqrt{0,25}=\sqrt{\left(0,1\right)^2}-\sqrt{\left(0,5\right)^2}=0,1-0,5=-0,4\)b) \(0,5\sqrt{100}-\sqrt{\frac{1}{4}}=0,5\sqrt{10^2}-\sqrt{\left(\frac{1}{2}\right)^2}=0,5.10-\frac{1}{2}=5-\frac{1}{2}=\frac{9}{2}\)
a) \(\sqrt{0,01}+\sqrt{0,25}=0,1\cdot0,5=0,05\)
b) \(0,5\sqrt{100}-\frac{\sqrt{1}}{4}=0,5\cdot10-\frac{1}{4}=5-\frac{1}{4}=\frac{19}{4}\)
a)\(\sqrt{0,01}-\sqrt{0,25}\)
=\(\sqrt{\left(0,1\right)^2}-\sqrt{\left(0,5\right)^2}\)
= 0,1 - 0,5 = - 0,4
b)\(0,5.\sqrt{100}-\sqrt{\dfrac{1}{4}}\)
=0,5.\(\sqrt{10^2}-\sqrt{\left(\dfrac{1}{2}\right)^2}\)
=0,5.10−\(\dfrac{1}{2}\)
= 5 - 0,5
= 4,5.
a) 0,1 - 0,5 = -0,4
b)0,5 . 10 - 0,5 = 5 - 0,5 = 4,5
thầy thông cảm máy em không có dấu căn.
\(\)a)
\(\sqrt{0,01}-\sqrt{0,25}=\sqrt{\frac{1}{100}}-\sqrt{\frac{1}{4}}=\frac{1}{10}-\frac{1}{2}=\frac{1}{10}-\frac{5}{10}=\frac{-4}{10}=\frac{-2}{5}\)
b)
\(0,5.\sqrt{100}-\sqrt{\frac{1}{4}}=0,5.10-\frac{1}{2}=5-\frac{1}{2}=\frac{10}{2}-\frac{1}{2}=\frac{9}{2}\)
ai k mình mình k lại
a) \(\sqrt{0,01}=0,1;\sqrt{0,25}=0,5\)= 0,1-0,5 = -0,4
b) = 0,5 x 10 - \(\frac{1}{2}\)= 4,5 . ( Đơn giản nhỉ :) )
\(a,\left(\dfrac{1}{9}\right)^{x+1}>\dfrac{1}{81}\\ \Leftrightarrow\left(\dfrac{1}{9}\right)^{x+1}>\left(\dfrac{1}{9}\right)^2\\ \Leftrightarrow x+1< 2\\ \Leftrightarrow x< 1\)
\(b,\left(\sqrt[4]{3}\right)^x\le27\cdot3^x\\ \Leftrightarrow3^{\dfrac{x}{4}}\le3^{x+3}\\ \Leftrightarrow\dfrac{x}{4}\le3=x\\ \Leftrightarrow-\dfrac{3}{4}x\le3\\ \Leftrightarrow x\ge-4\)
c, ĐK: \(\left\{{}\begin{matrix}x+1>0\\2-4x>0\end{matrix}\right.\Leftrightarrow-1< x< \dfrac{1}{2}\)
\(log_2\left(x+1\right)\le log_2\left(2-4x\right)\\ \Leftrightarrow x+1\le2-4x\\ \Leftrightarrow5x\le1\\ \Leftrightarrow x\le\dfrac{1}{5}\)
Kết hợp với ĐKXĐ, ta được: \(-1< x\le\dfrac{1}{5}\)
2:
=1-1+1-1=0
3:
a: =>34*(100+1)/2:a=17
=>a=101
b: =>5/3(x-1/2)=5/4
=>x-1/2=5/4:5/3=3/4
=>x=5/4
1a, \(\dfrac{2005}{2001}\) = 1+\(\dfrac{4}{2001}\); \(\dfrac{2009}{2005}\)=1+\(\dfrac{4}{2005}\)vì\(\dfrac{4}{2001}\)>\(\dfrac{4}{2005}\)nên\(\dfrac{2005}{2001}\)>\(\dfrac{2009}{2005}\)
1b,\(\dfrac{1313}{1515}\)=\(\dfrac{1313:101}{1515:101}\)= \(\dfrac{13}{15}\); \(\dfrac{131313}{151515}\)=\(\dfrac{131313:10101}{151515:10101}\)=\(\dfrac{13}{15}\)
Vậy \(\dfrac{13}{15}\)=\(\dfrac{1313}{1515}\)=\(\dfrac{131313}{151515}\)
0,5 x 4 x 2 x0,25
= (0,5 x 2) x (4 x 0,25)
= 1 x 1
=1
\(\frac{1}{2}:0,5-\frac{1}{4}:0,25+\frac{1}{8}:0,125-\frac{1}{10}:0,1\)
\(=\frac{1}{2}\times2-\frac{1}{4}\times4+\frac{1}{8}\times8-\frac{1}{10}\times10\)
\(=1-1+1-1\)
\(=0\)
`#3107.101107`
a)
`2/5 \sqrt{25} - 1/2 \sqrt{4}`
`= 2/5 * \sqrt{5^2} - 1/2 * \sqrt{2^2}`
`= 2/5*5 - 1/2*2`
`= 2 - 1`
`= 1`
b)
`0,5*\sqrt{0,09} + 5*\sqrt{0,81}`
`= 0,5*\sqrt{(0,3)^2} + 5*\sqrt{(0,9)^2}`
`= 0,5*0,3 + 5*0,9`
`= 0,15 + 4,5`
`= 4,65`
c)
`2/5\sqrt{25/36} - 5/2\sqrt{4/25}`
`= 2/5*\sqrt{(5^2)/(6^2)} - 5/2*\sqrt{(2^2)/(5^2)}`
`= 2/5*5/6 - 5/2*2/5`
`= 1/3 - 1`
`= -2/3`
d)
`-2 \sqrt{(-36)/(-16)} + 5 \sqrt{(-81)/(-25)}`
`= -2*\sqrt{36/16} + 5*\sqrt{81/25}`
`= -2*\sqrt{(6^2)/(4^2)} + 5*\sqrt{(9^2)/(5^2)}`
`= -2*6/4 + 5*9/5`
`= -3 + 9`
`= 6`
\(J=4^{1-2\log_2\sqrt[4]{7}}-36^{\log_62}+81^{0,25-0,5\log_97}=\left(2^2\right)^{1-2\log_2\sqrt[4]{7}}+\left(6^2\right)^{\log_62}+\left(3^4\right)^{0,25-\frac{1}{2}\log_{3^2}7}\)
\(=\frac{2^2}{2^{4\log_2\sqrt[4]{7}}}+6^{\log_64}+\frac{3}{3^{\log_37}}=\frac{4}{7}-4+\frac{3}{7}=-3\)