Tìm x biết 6(x-2)-(x-3)=31
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3/2+5/4+9/8/+17/16+33/32-6+x-1/x+1=31/32-2/2015
=(1+1/2)+(1+1/4)+(1+1/8)+(1+1/16)+(1+1/32-6+x-1/x+1=31/32-2/2015
=(1/2+1/4+1/8+1/16+1/32)+(1+1+1+1+1)-6+x-1/x+1=31/32-2/2015
=31/32+5-6+x-1/x+1=31/32-2/2015
=5-6+x-1/x+1=31/32-2/2015-31/32
=-1+x-1/x+1=-2/2015
=x-1/x+1=-2/2015- -1
=x-1/x+1=2013/2015
=>x=2014
a) \(\left(2x+3\right)^3=\left(2x+3\right)^8\)
TH1 \(2x+3=1\)
\(2x=1-3=-2\)
\(x=-1\)
TH2 \(2x+3=0\)
\(2x=-3\Rightarrow x=-\frac{3}{2}\)
b) ? sai đề
c) \(\left|5-3\right|=\left|11+2x\right|\Rightarrow\left|2\right|=\left|11+2x\right|\)
\(\hept{\begin{cases}11+2x=-2\\11+2x=2\end{cases}\Rightarrow}\hept{\begin{cases}2x=13\\2x=9\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{13}{2}\\x=\frac{9}{2}\end{cases}}\)
d) \(\left(x-5\right)^4=\left(x-5\right)^6\Rightarrow\hept{\begin{cases}x-5=0\\x-5=1\end{cases}}\Rightarrow\hept{\begin{cases}x=5\\x=6\end{cases}}\)
Có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}=\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}...\frac{30}{2.31}.\frac{31}{2.32}=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}...\frac{1}{2}.\frac{1}{2}.\frac{1}{32}\)
\(=\frac{1}{2^{31}.2^5}=\frac{1}{2^{36}}=2^x\)\(\Rightarrow1=2^x.2^{36}=2^{36+x}\)\(\Rightarrow2^{36+x}=2^0\Rightarrow36+x=0\Rightarrow x=-36\)
x - 4 : (6 +7 ) x 2 =31
x - 4 : (13 x 2) = 31
x - 4 : 26 = 31
x : 26= 31 x 4
x : 26 = 124
x=124 : 26
x = 4
t i c k nha
thiếu đè bạn ơi
x - 4 thiếu đây nè (6 +7 ) x 2 =31
a, - 2 .( x + 6 ) + 6 . ( x - 10 ) = 8
- 2x - 12 + 6x - 60 = 8
4x - 72 = 8
4x = 8 + 72
4x = 80
x = 20
b, - 4 . ( 2x + 9 ) - ( - 8x + 3 ) - ( x + 13 ) = 0
- 8x - 36 + 8x - 3 - x - 13 = 0
- x - 52 = 0
x = - 52
c, 7x . ( 2 + x ) - 7x . ( x + 3 ) = 14
7x . ( 2 + x - x - 3 ) = 14
7x . ( - 1 ) = 14
7x = 14 : ( - 1 )
7x = - 14
x = - 2
d, 2 . ( 5 + 3x ) + x = 31
10 + 6x + x = 31
10 + 7x = 31
7x = 31 - 10
7x = 21
x = 3
a)-2(x+6)+6(x-10)=8
-2x+-12+6x+-60=8
4x+-72=8
4x=80
x=80:4
x=20
b)-4(2x+9)-(-8x+3)-(x+13)=0
-8x+-36+8x-3+x-13=0
(-8x+8x)+-36+-3+x-13=0
0+-52+x=0
x=0-(-52)
x=-52
c)7x(2+x)-7x(x+3)=14
14x+7x2
=>\(1\cdot\dfrac{2}{4}\cdot\dfrac{3}{6}\cdot...\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=2^x\)
=>\(2^x=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\left(\dfrac{1}{2}\right)^{30}\cdot\left(\dfrac{1}{2}\right)^6=\dfrac{1}{2^{36}}\)
=>x=-36
6(x - 2) - (x - 3) = 31
6x - 12 - x + 3 = 31
6x - x = 31 - 3 + 12
5x = 40
x = 8