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15 tháng 9 2017
a) 3*2x =48 \(\Leftrightarrow\)2x=16 \(\Leftrightarrow\)x=4 b) (2x +1 )3 =53 \(\Leftrightarrow\)2x +1 =5 \(\Leftrightarrow\)2x =4 \(\Leftrightarrow\)x=2 c) 36+x=36 \(\Leftrightarrow\)x=0 d) 1+x-15=27-1 \(\Leftrightarrow\)x=40
F
17 tháng 9 2017
d)19900+(x-15)=3199:3196-12000
1+(x-15)=26
x-15=26-1=25
x=25+15=40
c) (90:15)2+x=26-22.7
36+x=-16320
x=-16320-36=-16356
b) (2x+1)3=125
(2x+1)3=53
2x+1=5
2x=5-1=4=22
x=2
a) ko hiểu đề bài
17 tháng 9 2017
a) \(3.2^x-3=45\)
\(3\left(2^x-1\right)=45\)
\(2^x-1=15\)
\(2^x=15+1\)
\(2^x=16\)
\(2^x=2^4\)
=>x=4
b) \(\left(2^x+1\right)^3=125\)
\(\left(2^x+1\right)^3=5^3\)
=> \(2^x+1=5\)
Làm tương tự câu a, đc x = 2
c) \(\left(90:15\right)^2+x=2^6-2^{2.7}\)
\(6^2+x=64-16384\)
\(x=-16320-36\)
x=-16356
d) \(1990^0+\left(x-15\right)=3^{199}:3^{196}-1^{2000}\)
1 + x - 15 = 33-1
x= 27-1-1+15
x=40
14 tháng 7 2016
b./ \(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)(b)
Mà \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)
(b) \(\Leftrightarrow x+2010=0\Leftrightarrow x=-2010\)
14 tháng 7 2016
a./
\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0.\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)(a)
Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)
(a) \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
ND
1
3 tháng 10 2015
3x+1+2.3x-1=99
=>3x-1(32+2)=99
=>3x-1.11=99
=>3x-1=99:11=9=32
=.x-1=2
=>x=3
Vậy x=3
b,3.2x+2+3.2x=120
=>3.2x(22+1)=120
=>3.2x.5=120
=>3.2x=120:5
=>2x=24:3=8=23
=>x=3
Vậy x=3
=
PN
3 tháng 7 2017
Bài 2:
a) \(\left(x-3\right)^3+27=0\)
\(\Leftrightarrow\left(x-3\right)^3=0-27\)
\(\Leftrightarrow\left(x-3\right)^3=-27\)
\(\Leftrightarrow\left(x-3\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x-3=-3\)
\(\Leftrightarrow x=\left(-3\right)+3\)
\(\Leftrightarrow x=0\)
b) \(-125-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x+1\right)^3=-125-0\)
\(\Leftrightarrow\left(x+1\right)^3=-125\)
\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x+1=-5\)
\(\Leftrightarrow x=\left(-5\right)-1\)
\(\Leftrightarrow x=-6\)
c) \(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{16}=0\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=0+\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Leftrightarrow2x-\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{4}+\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{2}:2\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
d) \(2^x+2^{x+1}=24\)
\(\Leftrightarrow2^x+2^x.2=24\)
\(\Leftrightarrow2^x\left(1+2\right)=24\)
\(\Leftrightarrow2^x.3=24\)
\(\Leftrightarrow2^x=24:3\)
\(\Leftrightarrow2^x=8\)
\(\Leftrightarrow2^x=2^3\)
\(\Rightarrow x=3\)
e) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=1\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=1+\dfrac{1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=-\dfrac{3}{2}\\x+\dfrac{1}{5}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
g) \(\left|x-3\right|+2x=10\)
\(\Leftrightarrow\left|x-3\right|=10-2x\)
\(\Leftrightarrow\left|x-3\right|=2.5-2x\)
\(\Leftrightarrow\left|x-3\right|=2\left(5-x\right)\)
(không chắc có nên làm tiếp câu g không, thấy đề cứ là lạ, có j sai sai...)
PN
3 tháng 7 2017
Bài 1:
a) \(2^7+2^9⋮10\)
Ta có: \(2^7+2^9=2^{4.1}.2^3+2^{4.2}.2\)
\(\Leftrightarrow\overline{A6}.2^3+\overline{B6}.2\)
\(\Leftrightarrow\overline{A6}.8+\overline{B6}.2\)
\(\Leftrightarrow\overline{C8}+\overline{D2}\)
\(\Leftrightarrow\overline{E0}\)
Mà \(\overline{E0}⋮10\) \(\Rightarrow2^7+2^9⋮10\)
b) \(8^{24}.25^{10}⋮2^{36}.5^{20}\)
Ta có: \(8^{24}.25^{10}=\left(2^3\right)^{24}.\left(5^2\right)^{10}\)
\(\Leftrightarrow2^{72}.5^{20}\)
Do \(2^{72}⋮2^{36}\) và \(5^{20}⋮5^{20}\) \(\Rightarrow8^{24}.25^{10}⋮2^{36}.5^{20}\)
c) \(3^{10}+3^{12}⋮30\)
Ta có: \(3^{10}+3^{12}=3^{4.2}.3^2+3^{4.3}\)
\(\Leftrightarrow\overline{A1}.3^2+\overline{B1}\)
\(\Leftrightarrow\overline{A1}.9+\overline{B1}\)
\(\Leftrightarrow\overline{C9}+\overline{B1}\)
\(\Leftrightarrow\overline{D0}⋮10\)
(Chứng minh chia hết cho 10 rồi chứng minh chia hết cho 3, mình chưa tìm được cách làm, chờ chút)
A
20 tháng 8 2020
a, \(\frac{x}{5}=\frac{2}{3}\Leftrightarrow x=\frac{10}{3}\)
b, \(\frac{x}{-24}=\frac{20}{42}\Leftrightarrow x=-\frac{80}{7}\)
c, \(\frac{x+3}{15}=\frac{1}{3}\Leftrightarrow3x+9=15\Leftrightarrow x=2\)
d, \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\Leftrightarrow x\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}\Leftrightarrow-\frac{5}{6}x=\frac{5}{12}\Leftrightarrow x=-\frac{1}{2}\)
AH
11 tháng 8 2016
a) 2 - ( \(5\frac{3}{8}\)x X - \(\frac{5}{24}\)) = \(\frac{5}{12}\)
\(5\frac{3}{8}\)x X - \(\frac{5}{24}\)= \(\frac{19}{12}\)
\(5\frac{3}{8}\)x X = \(\frac{43}{24}\)
X = \(\frac{1}{3}\)
b) \(1\frac{2}{9}\): ( \(3\frac{1}{3}\)x X + \(\frac{1}{6}\)) = \(\frac{22}{23}\)
\(3\frac{1}{3}\)x X + \(\frac{1}{6}\) = \(\frac{23}{18}\)
\(3\frac{1}{3}\)x X = \(\frac{10}{9}\)
X =\(\frac{1}{3}\)
C) \(\frac{4}{5}\)x X - \(\frac{1}{2}\)x X + \(\frac{3}{4}\)x X = \(\frac{7}{40}\)
( \(\frac{4}{5}-\frac{1}{2}+\frac{3}{4}\)) x X = \(\frac{7}{40}\)
\(\frac{21}{20}\) x X = \(\frac{7}{40}\)
X =\(\frac{1}{6}\)