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\(2^{35}< 2^{36}=8^{12}\)
\(3^{24}=\left(3^2\right)^{12}=9^{12}\)
\(8^{12}< 9^{12}\)
=>\(2^{36}< 3^{24}\)
=>\(2^{35}< 3^{24}\)
tạm thời cái đề bài là tìm x vậy
tiếp tục thui
=) (6x-10)-(6x-3)\(⋮\)2x-1
=)6x-10-6x+3\(⋮\)2x-1
=) (6x-6x)-(10-3)\(⋮\)2x-1
=)0-7\(⋮\)2x-1
=)-7\(⋮\)2x-1=)2x-1\(\in\)Ư(-7)={-7;-1;1;7}
=)2x\(\in\){-6;0;2;8}
=)x\(\in\){-3;0;1;4}
A = 1-1/3 + 1/3 - 1/5 +...+ 1/2017 - 1/2019
A= 1-1/2019
A= 2018/2019
| x - \(\frac{1}{2}\)| = 1
TH1:x - \(\frac{1}{2}\) = 1 TH2:x - \(\frac{1}{2}\) = -1
x = 1 + \(\frac{1}{2}\) x = -1 + \(\frac{1}{2}\)
x = \(\frac{3}{2}\) x = \(\frac{-1}{2}\)
Vậy x thuộc \(\frac{3}{2}\)và \(\frac{-1}{2}\)
+) Ta có: \(15-x+17=-\left(-6\right)+\left|-12\right|\)
\(\Leftrightarrow-x+32=6+12\)
\(\Leftrightarrow-x=18-32\)
\(\Leftrightarrow x=14\)
Vậy......
+) Ta có: \(x+20=-\left(-23\right)\)
\(\Leftrightarrow x+20=23\)
\(\Leftrightarrow x=3\)
Vậy........
+) Ta có: \(-\left|-5\right|-\left(-x\right)+4=3-\left(-25\right)\)
\(\Leftrightarrow-5+x+4=3+25\)
\(\Leftrightarrow x=29\)
Vậy.....
Chúc e hok tốt và giữ lời hứa nhá ^_^
a)\(\frac{18}{-5}.\frac{-5}{6}=\frac{18}{6}=3\)
c)\(\frac{115}{30}\div\frac{-5}{6}=\frac{115}{30}.\frac{6}{-5}=\frac{-23}{5}\)
d)\(-3x+\frac{1}{2}=3\)
\(-3x=3-\frac{1}{2}\)
\(-3x=\frac{5}{2}\)
\(x=\frac{5}{2}\div3\)
\(x=\frac{5}{6}\)
a)= 18/-5 . -5/6 , bỏ -5=3
c)= 115/30 : 5/6 = 23/6 :5/6 =23/6 . 6/5 bỏ 6 = 23/5
d) 3.x +1/2 = 3:
3.x = 3-1/2 =5/2
x =5/2 : 3 = 5/6
nhanh con mẹ mày
\(\frac{1}{3}\times\frac{5}{7}-\frac{7}{27}\times\frac{36}{14}\)\(=\frac{1\times5}{3\times7}-\frac{7\times36}{27\times14}\)\(=\frac{5}{21}-\frac{252}{378}=\frac{5}{21}-\frac{2}{3}=\frac{5}{21}-\frac{14}{21}\)\(=\frac{5-14}{21}=\frac{-9}{21}=\frac{-3}{7}\)
\(\frac{1}{3}.\frac{5}{7}-\frac{7}{27}.\frac{36}{14}=\frac{5}{21}-\frac{2}{3}=-\frac{9}{21}=-\frac{3}{7}\)
a, \(3-\left|2x+1\right|=\left(-5\right)\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow2x+1=\pm8\)
TH1 : \(2x+1=8\Leftrightarrow x=\frac{7}{2}\)
TH2 : \(2x+1=-8\Leftrightarrow x=-\frac{9}{2}\)
b, \(12+\left|3-x\right|=9\Leftrightarrow\left|3-x\right|=3\Leftrightarrow3-x=\pm3\)
TH1 : \(3-x=3\Leftrightarrow x=0\)
TH2 : \(3-x=-3\Leftrightarrow x=6\)
a,3- | 2x + 1 | = ( - 5 )
| 2x + 1 | = 8
TH1: 2x+1=8
2x=7
x=7/2
TH2:2x+1=-8
2x=-9
x=-9/2
\(\frac{8}{35}:\frac{-3}{\frac{9}{20}}=\frac{8}{35}:\frac{-20}{3}=\frac{8}{35}.\frac{3}{-20}=\frac{-6}{175}\)