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a)45-5 ( x - 2 ) = 20
5 ( x - 2 )=45-20
5 (x - 2 )=25
x - 2=25/5
x - 2=5
x = 5+2
x=7
b)3 (x-4)+24=96
3 (x-4)=96-24
3 (x-4)=72
x-4=72/3
x-4=24
x=24+4
x=28
6, \(x=\dfrac{9}{2}+\dfrac{1}{2}=\dfrac{10}{2}=5\)
7, \(x=-\dfrac{3}{4}-\dfrac{5}{4}=-\dfrac{8}{4}=-2\)
8, \(x=\dfrac{5}{7}-\dfrac{2}{7}=\dfrac{3}{7}\)
9, \(x=\dfrac{1}{3}+\dfrac{4}{3}=\dfrac{5}{3}\)
10, \(x=-\dfrac{7}{2}-\dfrac{5}{2}=-\dfrac{12}{2}=-6\)
\(6,x-\dfrac{1}{2}=\dfrac{9}{2}\Leftrightarrow x=5\)
\(7,x+\dfrac{5}{4}=\dfrac{-3}{4}\Leftrightarrow x=-2\)
\(8,x+\dfrac{2}{7}=\dfrac{5}{7}\Leftrightarrow x=\dfrac{3}{7}\)
\(9,x-\dfrac{4}{3}=\dfrac{1}{3}\Leftrightarrow x=\dfrac{5}{3}\)
\(10,x+\dfrac{5}{2}=\dfrac{-7}{2}\Leftrightarrow x=-6\)
a) \(\left(9^4.8+9^4.5\right):\left(9^2.\left(10-1\right)\right)\)
=\(9^4.13:9^3=13.9=117\)
b) 100-(75-25)=100-50=50
a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)
y = \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)
y = \(\dfrac{4}{3}\)
b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)
y - 0,5 + 0,5 = \(\dfrac{3}{4}\)
y = \(\dfrac{3}{4}\)
c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2
0,8 - 0,4y = 0,2
0,4y = 0,8 - 0,2
0,4y = 0,6
y = 1,5
d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{14}{9}\)
y = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)
y = \(\dfrac{29}{36}\)
e, y : \(\dfrac{5}{4}\) = \(\dfrac{9}{5}\) + \(\dfrac{1}{2}\)
y : \(\dfrac{5}{4}\) = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{8}\)
f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\)) = \(\dfrac{4}{5}\)
2y = \(\dfrac{4}{5}\)
y = \(\dfrac{2}{5}\)
\(1,\\ a,\Leftrightarrow4^{5-x}=4^2\Leftrightarrow5-x=2\Leftrightarrow x=3\\ b,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x+1=3\Leftrightarrow x=2\\ 2,\\ a,3^{100}=\left(3^2\right)^{50}=9^{50}\\ b,2^{98}=\left(2^2\right)^{49}=4^{49}< 9^{49}\\ c,5^{30}=5^{29}\cdot5< 6\cdot5^{29}\\ d,3^{30}=\left(3^3\right)^{10}=27^{10}>8^{10}\\ 4,\\ a,\Leftrightarrow5\left(x-10\right)=10\\ \Leftrightarrow x-10=2\Leftrightarrow x=12\\ b,\Leftrightarrow3\left(70-x\right)+5=92\\ \Leftrightarrow3\left(70-x\right)=87\\ \Leftrightarrow70-x=29\\ \Leftrightarrow x=41\\ c,\Leftrightarrow16+x-5=315-230=85\\ \Leftrightarrow x=74\\ d,\Leftrightarrow2^x-5+74=707:\left(16-9\right)=707:7=101\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)
\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)
\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)
\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)
\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)
\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)
\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)
\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)
\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)
\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)
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1) \(2^{x+2}-2x=96\)
\(2^x.2^2-2^x=96\)
\(2^x\left(2^2-1\right)=96\)
\(2^x=32=2^5\)
Vậy x = 5
2) \(2^x\le32\)
Để \(2^x\le32\) thì 2x phải nhỏ hơn hoặc = 32
Mà 25 = 32
Nên x = 5
3) \(2^{x+1}\le64\)
Để: \(2^{x+1}\le64\) thì 2x+1 phải nhỏ hơn hoặc = 64
Mà 2x+1 => 25+1 = 26 = 64
Vậy x = 5
4) \(9< 3^x< 81\)
Để: \(9< 3^x< 81\) thì: 3x phải lớn hơn 9 và nhỏ hơn 81
Mà 3x => 33 = 27
Vậy x = 3
5) \(10^x+48=y^2\)
Nếu x = 0
=> y2 = 48 + 1 = 49
=> y = \(\pm7\)
Nếu x khác 0
=> \(10^x=.........0\)
=> \(y^2=........0+48\)
\(=...................8\)
Mà số chính phương không có chữ số tận cùng là 8
Vậy \(x=0;y=\pm7\)
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